# Lecture 5: Eigenvectors and Eigenvalues

### Instructions

• This section covers the concepts listed below. You can click the button to go directly to that topic.
• After each video, there are notes for the material in the video.
• At the end of the page, there are exercises covering the material in the section.
• When you have finished the material below, you can go to the next section or return to the main page.

### Concepts

Finding Eigenvalues and Eigenvectors

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#### Video Notes

We will use the theorems from the previous Lecture to find a method to determine the eigenvalues of a matrix. Recall that $$\lambda$$ is an eigenvalue of the matrix $$M$$ if there is a non–zero vector $$\boldsymbol{v}$$ such that:

\begin{align*}
M \boldsymbol{v} = \lambda\boldsymbol{v}.
\end{align*}
Using properties of matrix–vector multiplication, this can be re–written as:
\begin{align*}
(M - \lambda I)\boldsymbol{v} = \boldsymbol{0}.
\end{align*}
Recall that the null space of a matrix $$A$$ is the set of solutions to $$A\boldsymbol{v} = \boldsymbol{0}$$. Note that the null space of $$A$$ always contains the zero vector. So $$\lambda$$ is an eigenvalue if and only if the null space of $$M-\lambda I$$ contains more than the zero vector and that happens if and only if $$\det(M-\lambda I) = 0$$.

Definition. The null space of a $$A$$ is the set of solutions to $$A\boldsymbol{v} = \boldsymbol{0}$$.

This is called a "space" because it is a vector space in its own right. We won't discuss this too much, but a practical application is that it has a basis. This means that there is a finite set of vectors $$\boldsymbol{u}_1, \ldots, \boldsymbol{u}_m$$ such that every vector in the null space can be written as a linear combination of those vectors. To make this a little more clear, here is an example:

Example. Find a basis for the null space of $$\begin{pmatrix}1&2\\-2&-4\end{pmatrix}$$.

Solution. We row reduce:
\begin{align*}
\begin{pmatrix}1&2\\-2&-4\end{pmatrix}
\sim\begin{pmatrix}1&2\\0&0\end{pmatrix}.
\end{align*}
So $$x_2$$ is the free variable and $$x_1 = -2x_2$$. So the set of solutions are vectors of the form:
\begin{align*}
\begin{pmatrix}-2t\\t\end{pmatrix} = t\begin{pmatrix}-2\\1\end{pmatrix}.
\end{align*}
That is, the vector $$\begin{pmatrix}-2\\1\end{pmatrix}$$ is a basis for the null space since any vector in the null space can be written as a linear combination (in this case with only one basis vector, "linear combination"means "multiple") of this vector.

#### Example

Find a basis for the null space of $$\begin{pmatrix}1&0&1\\0&2&1\\1&2&2\end{pmatrix}.$$

We row–reduce (the steps in the row–reduction are in the video):
\begin{align*}
\begin{pmatrix}1&0&1\\0&2&1\\1&2&2\end{pmatrix}
\sim\begin{pmatrix}1&0&1\\0&1&1/2\\0&0&0\end{pmatrix}.
\end{align*}
Since $$x_3$$ is the third variable, we sole for $$x_1$$ and $$x_2$$ in terms of $$x_3$$ and we let $$x_3 = t$$. So the set of solutions is the set of vectors of the form:
\begin{align*}
\begin{pmatrix}
-t\\-t/2\\t
\end{pmatrix}
=t\begin{pmatrix}-1\\1/2\\1\end{pmatrix}.
\end{align*}
Above, we can factor out a $$1/2$$ to get:
\begin{align*}
t\begin{pmatrix}-1\\1/2\\1\end{pmatrix}
=\frac{t}{2}\begin{pmatrix}-2\\1\\2\end{pmatrix}.
\end{align*}
That is, the vector $$\begin{pmatrix}-2\\1\\2\end{pmatrix}$$ is a basis for this null space since all vectors in the null space of this matrix are multiples of it.

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Linear Algebra for Math 308: L5V2

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#### Video Notes

We saw above that $$\lambda$$ is an eigenvalue of $$M$$ if and only if $$\det(M-\lambda I) = 0$$. So to find eigenvalues, we compute $$\det(M-\lambda I)$$ and see for which values this is zero. Then to find the corresponding eigenvectors, we want to find a basis for the null space of $$M-\lambda I$$; any non–zero vector in the null space of $$M-\lambda I$$ is an eigenvector of $$M$$ associated to $$\lambda$$. If $$\lambda$$ is an eigenvalue, then we call the null space of $$M-\lambda I$$ the eigenspace associated to $$\lambda$$ and denote it by $$E_\lambda$$.

#### Examples

1. Find all eigenvalues of the matrices.
1. $$M=\begin{pmatrix}3&-1\\4&-2\end{pmatrix}$$
2. $$M=\begin{pmatrix}3&-2\\4&-1\end{pmatrix}$$

1. We need to compute $$\det{M - \lambda I}$$ and find where this is $$0$$:
\begin{align*}
\det(M - \lambda I)
&= \det\begin{pmatrix}3 - \lambda & -1\\ 4 & -2-\lambda\end{pmatrix}
\\&= -(3-\lambda)(2+\lambda) +4
\\&= \lambda^2 - \lambda -2
\\&= (\lambda +1)(\lambda - 2).
\end{align*}
Since the eigenvalues are the zeros of $$\det(M- \lambda I)$$, this means the eigenvalues of this matrix are $$-1$$ and $$2$$.
2. Proceeding as above, we compute the determinant:
\begin{align*}
\det(M - \lambda I)
&= \det\begin{pmatrix}3-\lambda & -2 \\ 4 & -1-\lambda\end{pmatrix}
\[4mm]\&= (\lambda - 3)(\lambda + 1) + 8
\\&= \lambda^2 - 2\lambda +5
\\&= (\lambda -1)^2 +4.
\end{align*}
Now we see that the eigenvalues are $$1 \pm 2i$$. So, we can have complex eigenvalues even if the matrix has only real entries. The good news is that if the matrix has real entries, then eigenvalues will always come in complex conjugate pairs.

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Linear Algebra for Math 308: L5V4

2. Find all eigenvalues of $$M=\begin{pmatrix}1 & 0 & 0\\ 2 & 1 & -2\\ 3 & 2 & 1\end{pmatrix}$$

This is just like the previous examples, except that the determinant is slightly more complicated:
\begin{align*}
\det(M-\lambda I)
&= \det\begin{pmatrix}1-\lambda&0&0\\ 2&1-\lambda&-2\\ 3&2&1-\lambda\end{pmatrix}
\\[4mm]&= (1-\lambda)\det\begin{pmatrix}1-\lambda & 02\\ 2 & 1-\lambda\end{pmatrix}
\\[4mm]&= (1-\lambda)\left((\lambda-1)^2 +4\right)
\\&= -(\lambda-1)\left(\lambda^2 -2\lambda + 5\right).
\end{align*}
So $$\lambda = 1$$ is a solution as well as $$\lambda = 1 \pm 2i$$. So, in this case, there are three eigenvalues; one is real and the other two are complex conjugate pairs.

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Linear Algebra for Math 308: L5V5

If $$M$$ is a matrix, and $$\lambda$$ is an eigenvalue, then $$E_\lambda$$ is defined to be the null space of $$M-\lambda I$$ (note that this does include the zero vector). This is a vector space in it own right – called the Eigenspace – and its dimension is the number of free variables there are after row-reducing $$M-\lambda I$$. A basis for $$E_\lambda$$ is the set of vectors that are multiplied by the free variables.

#### Examples

1. Consider the matrix $$M=\begin{pmatrix}3&-1\\4&-2\end{pmatrix}.$$
1. Find all the eigenvectors.
2. Find a basis for $$E_{-1}$$ and $$E_2.$$

We saw above that the eigenvalues are $$-1$$ and $$2$$. The eigenvectors are solutions to $$(M -\lambda I)\boldsymbol{v} = \boldsymbol{0}$$. So we want to find the null space of $$M - \lambda I$$. For the eigenvalue $$-1$$ we do this via row--reduction:
\begin{align*}
\begin{pmatrix}
3 + 1 & -1\\ 4 & -1
\end{pmatrix}
=\begin{pmatrix}
4 & -1\\ 4 & -1
\end{pmatrix}
\sim
\begin{pmatrix}
4 & -1\\ 0 & 0
\end{pmatrix}.
\end{align*}
This is the same as the equation $$4x_1 - x_2 = 0$$ or $$x_2 = 4x_1$$. So then any vector of the form:
\begin{align*}
\begin{pmatrix}t\\4t\end{pmatrix}
=t\begin{pmatrix}1\\4\end{pmatrix}.
\end{align*}
is in the null space of $$M - \lambda I$$; the eigenvectors are the non--zero vectors. That is, any vector of this form for $$t\neq 0$$.

To find the eigenvalues associated with $$\lambda = 2$$ we find the null space of $$M-2I$$:
\begin{align*}
\begin{pmatrix}
3-2 & -1\\ 4 & -2 -2
\end{pmatrix}
=\begin{pmatrix}
1 & -1 \\ 4 & -4
\end{pmatrix}
\sim
\begin{pmatrix}
1 & -1\\ 0 & 0
\end{pmatrix}.
\end{align*}
And so the solution is $$x_1 - x_2 =0$$ or $$x_1 = x_2$$. So then the eigenvectors are non–zero vectors of the form:
\begin{align*}
\begin{pmatrix}t\\t\end{pmatrix}
=t\begin{pmatrix}1\\1\end{pmatrix}.
\end{align*}

The null space of $$M+I$$ was seen to be $$t\begin{pmatrix}1\\4\end{pmatrix}$$. There is one free variable, so the dimension of $$E_{-1}$$ is $$1$$, and $$\begin{pmatrix}1\\4\end{pmatrix}$$ is a basis.

Similarly, the null space for $$M - 2I$$ was seen to be $$t\begin{pmatrix}1\\1\end{pmatrix}$$ and so the dimension of $$E_{2}$$ is $$1$$ and a basis is $$\begin{pmatrix}1\\1\end{pmatrix}$$.

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Linear Algebra for Math 308: L5V6

2. Find a basis for the eigenspaces of $$M=\begin{pmatrix}1 & 0 & 0\\ 2 & 1 & -2\\ 3 & 2 & 1\end{pmatrix}.$$

We saw above that the eigenvalues are $$1$$ and $$1\pm 2i$$. We'll find $$E_{1}$$ first:
\begin{align*}
\begin{pmatrix}
1-1 & 0 & 0\\ 2 & 1-1 & -2\\ 3 & 2 & 1-1
\end{pmatrix}
&=\begin{pmatrix}
0 & 0 & 0\\ 1 & 0 & -2\\3 & 2 & 0
\end{pmatrix}\\[4mm]
&\sim
\begin{pmatrix}
0&0&0\\1&0&-2\\0&2&6
\end{pmatrix}\\[4mm]
&\sim
\begin{pmatrix}
0&0&0\\1&0&-2\\0&1&3
\end{pmatrix}.
\end{align*}
There is one free-variable (corresponding to $$x_3$$) and this is the same as the system:
\begin{align*}
x_1 - 2x_3 &= 0\\
x_2 + 3x_3 &= 0.
\end{align*}
So the solutions are $$x_1 = 2x_3$$ and $$x_2 = -3x_3$$ and $$x_3 = t$$ is free. So vectors of the form:
\begin{align*}
\begin{pmatrix}2t\\-3t\\t\end{pmatrix} = t\begin{pmatrix}2\\-3\\1\end{pmatrix}.
\end{align*}
There is one free variable so the dimension of $$E_{1}$$ is $$1$$ and a basis is $$\begin{pmatrix}2\\-3\\1\end{pmatrix}$$.

Now we find $$E_{1+2i}$$:
\begin{align*}
\begin{pmatrix}
1-(1+2i) & 0 & 0\\ 2 & 1-(1+2i) & -2\\ 3 & 2 & 1-(1+2i)
\end{pmatrix}
&=\begin{pmatrix}
-2i & 0 & 0\\ 1 & -2i & -2\\3 & 2 & -2i
\end{pmatrix}\\[4mm]
&\sim
\begin{pmatrix}
1 & 0 & 0\\ 1 & -2i & -2\\3 & 2 & -2i
\end{pmatrix}\\[4mm]
&\sim
\begin{pmatrix}
1 & 0 & 0\\ 0 & -2i & -2\\0 & 2 & -2i
\end{pmatrix}.
\end{align*}
Note that the bottom row is $$i$$ times the middle row. And so we can continue as:
\begin{align*}
\begin{pmatrix}
1 & 0 & 0\\ 0 & -2i & -2\\0 & 2 & -2i
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & 0\\ 0 & -2i & -2\\0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0\\ 0 & 1 & -i\\0 & 0 & 0
\end{pmatrix}.
\end{align*}
This says $$x_1 = 0$$ and $$x_2 = ix_3$$. So the null space of $$M - (1+2i)I$$ is:
\begin{align*}
\begin{pmatrix}0\\ti\\t\end{pmatrix} = t\begin{pmatrix}0\\i\\1\end{pmatrix}.
\end{align*}
Since there is one free variable, then the dimension of $$E_{1+2i}$$ is $$1$$ and $$\begin{pmatrix}0\\i\\1\end{pmatrix}$$ is a basis.

We don't have to do much work to find $$E_{1-2i}$$. Indeed, if $$M$$ has real entries and if $$\boldsymbol{v}$$ is an eigenvalue associated to $$\lambda$$ then $$\overline{\boldsymbol{v}}$$ is an eigenvector associated to $$\overline{\lambda}$$. To see this:
\begin{align*}
M\overline{\boldsymbol{v}}
= \overline{M\boldsymbol{v}}
= \overline{\lambda\boldsymbol{v}}
= \overline{\lambda}\overline{\boldsymbol{v}}.
\end{align*}
So then $$\begin{pmatrix}0\\-i\\1\end{pmatrix}$$ is a basis for $$E_{1-2i}$$. This fact will come into use in systems of differential equations.
Basis for eigenspace $$E_1$$:

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Linear Algebra for Math 308: L5V7

Basis for eigenspaces $$E_{1+2i}$$ and $$E_{1-2i}:$$

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Linear Algebra for Math 308: L5V8

3. Find the eigenvalues and associated eigenspaces for the matrix $$M = \begin{pmatrix}3&-1&-2\\0&2&0\\1&-1&0\end{pmatrix}.$$

First we compute the eigenvalues:
\begin{align*}
\det(M-\lambda I)
&= \det \begin{pmatrix}3-\lambda&-1&-2\\0&2-\lambda&0\\1&-1&0-\lambda\end{pmatrix}
\\[4mm]&= (3-\lambda)\det\begin{pmatrix}2-\lambda&0\\-1&-\lambda\end{pmatrix}
- (-1)\det\begin{pmatrix}0&0\\1&-\lambda\end{pmatrix}
-2 \det\begin{pmatrix}0& 2-\lambda\\1&-1\end{pmatrix}
\\[4mm]&= (3-\lambda)(2-\lambda)(-\lambda) +2(2-\lambda)
\\&=(2-\lambda)((\lambda-3)\lambda +2)
\\&= (2-\lambda)(\lambda^2 -3\lambda +2)
\\&=(2-\lambda)(\lambda-2)(\lambda-1).
\end{align*}
So the eigenvalues are $$1$$ and $$2$$. We first compute $$E_1$$:
\begin{align*}
\begin{pmatrix}
3-1&-1&-2\\0&2-1&0\\1&-1&0-1
\end{pmatrix}
&=
\begin{pmatrix}
2&-1&-2\\0&1&0\\1&-1&-1
\end{pmatrix}\\[4mm]
&\sim
\begin{pmatrix}
1&-1&-1\\2&-1&-2\\0&1&0
\end{pmatrix}\\[4mm]
&\sim
\begin{pmatrix}
1&-1&-1\\0&-3&-4\\0&1&0
\end{pmatrix}\\[4mm]
&\sim
\begin{pmatrix}
1&0&-1\\0&0&0\\0&1&0
\end{pmatrix}
\end{align*}
There is one free variable, corresponding to $$x_3$$ and this system says:
\begin{align*}
x_1 - x_3 &= 0\\
x_2 &= 0.
\end{align*}
So the null space is $$t\begin{pmatrix}1\\0\\1\end{pmatrix}$$. Since there is one free variable, the dimension is $$1$$ and a basis is $$\begin{pmatrix}1\\0\\1\end{pmatrix}$$.

Now we compute $$E_2$$:
\begin{align*}
\begin{pmatrix}
3-2&-1&-2\\0&2-2&0\\1&-1&0-2
\end{pmatrix}
&\sim
\begin{pmatrix}
1&-1&-2\\0&0&0\\1&-1&-2
\end{pmatrix}\\[4mm]
&\sim
\begin{pmatrix}
1&-1&-2\\0&0&0\\0&0&0
\end{pmatrix}.
\end{align*}
Here there are two free variables corresponding to $$x_2$$ and $$x_3$$ and this says that $$x_1 - x_2 - 2x_3 = 0$$ or $$x_1 = x_2 + 2x_3$$. So the null space is given by:
\begin{align*}
\begin{pmatrix}t + 2s\\t\\s\end{pmatrix}
= t\begin{pmatrix}1\\1\\0\end{pmatrix} + s\begin{pmatrix}2\\0\\1\end{pmatrix}.
\end{align*}
Since there are two free variables, the dimension is $$2$$ and
\begin{align*}
\left\{\begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}2\\0\\1\end{pmatrix}\right\}
\end{align*}
is a basis for $$E_2.$$

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Linear Algebra for Math 308: L5V9

4. Find all eigenvalues and associated eigenspaces for the matrix $$M = \begin{pmatrix}1&1\\0&1\end{pmatrix}.$$

First the eigenvalues:
\begin{align*}
\det \begin{pmatrix}1-\lambda & 1\\0&1-\lambda\end{pmatrix}
=(1-\lambda)^2.
\end{align*}
So $$1$$ is the only eigenvalue. We now compute $$E_1$$:
\begin{align*}
\begin{pmatrix}
1-1&1\\0&1-1
\end{pmatrix}
\sim
\begin{pmatrix}
0&1\\0&0
\end{pmatrix}.
\end{align*}
There is one free variable corresponding to $$x_1$$ and this equation says $$x_2 = 0$$ and $$x_1$$ is free. So the null space of $$M-I$$ is:
\begin{align*}
\begin{pmatrix}t\\0\end{pmatrix} = t\begin{pmatrix}1\\0\end{pmatrix}.
\end{align*}
Since there is only one free variable, the dimension of $$E_1$$ is $$1$$ and a basis is $$\begin{pmatrix}1\\0\end{pmatrix}$$.

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Linear Algebra for Math 308: L5V10

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#### Video Notes

Definition 2. If $$\lambda$$ is an eigenvalue of a matrix then its algebraic multiplicity is the multiplicity of $$\lambda$$ as a root of the characteristic polynomial. Its geometric multiplicity is the dimension of $$E_\lambda.$$

We've seen above that sometimes the algebraic and geometric multiplicity are equal and sometimes the geometric multiplicity is less than algebraic multiplicity. We didn't see any examples where the geometric multiplicity is greater than the algebraic multiplicity because this can never happen.

We won't prove this fact, but if $$\boldsymbol{v}_1\ldots\boldsymbol{v}_k$$ are eigenvectors associated to the eigenvalues $$\lambda_1,\ldots, \lambda_k$$ and none of the eigenvalues are equal then the vectors $$\boldsymbol{v}_1\ldots\boldsymbol{v}_k$$ are linearly independent. This implies that if $$\lambda_1, \ldots, \lambda_m$$ are the distinct eigenvalues of the matrix $$M$$ (that is, we don't repeat them according to multiplicity) and if $$\sum_{k=1}^{m}\dim E_{\lambda_k} = n$$ then there is a basis for $$\mathbf{R}^n$$ that consists of eigenvectors for the matrix $$M$$.
Generalized Eigenvectors

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#### Video Notes

We've seen several examples in which the sum of dimension of the eigenspaces equals $$n$$ and the last example we saw, the sum of the dimension of eigenspaces did not equal $$n$$. In general, if the geometric multiplicity of an eigenvalue is less than the algebraic multiplicity, there are special vectors called generalized eigenvectors that can be found instead. These vectors aren't as nice as eigenvectors, but they are nonetheless useful.

If $$\dim E_{\lambda} = m$$ and the algebraic multiplicity of $$\lambda$$ is $$m+l$$ then there are $$l$$ linearly independent generalized eigenvectors. They are defined by:
\begin{align*}
(A - \lambda I)\boldsymbol{w}_k = \boldsymbol{w}_{k-1},
\end{align*}
for $$k=1,\ldots, l$$ and where $$\boldsymbol{w}_0$$ is an ordinary eigenvector associate to the eigenvalue $$\lambda$$. The generalized eigenvalues are linearly independent.

For reasons that are beyond the scope of these lectures, there might be some choices of $$\boldsymbol{w}_0$$ such that the equation $$(A-\lambda I)\boldsymbol{w}_1 = \boldsymbol{w}_0$$ has no solution. In this case, choose a different eigenvector as your $$\boldsymbol{w}_0$$. In the case of $$2\times 2$$ systems, this will never happen though.

#### Examples

1. Find the eigenvectors and generalized eigenvectors for $$\begin{pmatrix}1&1\\0&1\end{pmatrix}.$$

We saw above that $$1$$ is an eigenvalue with algebraic multiplicity $$2$$ and that $$\begin{pmatrix}1\\0\end{pmatrix}$$ is the basis of $$E_1$$. We will now find the generalized eigenvectors. The difference between geometric and algebraic multiplicity of the eigenvalue is $$1$$ so there is one generalized eigenvetor. We can find it by solving:
\begin{align*}
\begin{pmatrix}1\\0\end{pmatrix}
= (M - \lambda I)\boldsymbol{w}_1
= \begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
= \begin{pmatrix}x_2\\0\end{pmatrix}.
\end{align*}
This gives $$x_2 = 1$$ and $$x_1$$ is free (you can find this also by row-reducing, but this example is simple enough that that isn't needed). So, any non–zero vector of the form $$\begin{pmatrix}t\\1\end{pmatrix}$$ is a generalized eigenvector. Taking $$t=0$$ is the simplest thing to do. Thus, a generalized eigenvector is
\begin{align*}
\boldsymbol{w}_1 = \begin{pmatrix}0\\1\end{pmatrix}.
\end{align*}

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Linear Algebra for Math 308: L5V14

2. Find all eigenvectors and generalized eigenvectors for $$M = \begin{pmatrix}2 & 0 & 0\\0&2&1\\0&0&2\end{pmatrix}.$$

You may compute for yourself that $$2$$ is the only eigenvalue; since it is the only one, its algebraic multiplicity is $$3$$. To find the eigenvectors we find a basis for the null space of $$M - 2I$$:
\begin{align*}
\begin{pmatrix}
2 - 2 &0 & 0\\ 0 & 2-2 &1\\ 0 & 0 & 2-2
\end{pmatrix}
\sim
\begin{pmatrix}
0 &0 & 0\\ 0 &0&1\\ 0 & 0 & 0
\end{pmatrix}.
\end{align*}
There are two free variables corresponding to $$x_1$$ and $$x_2$$ and this last matrix indicates that $$x_3 = 0$$. So any vector of the form:
\begin{align*}
\begin{pmatrix}t\\s\\0\end{pmatrix}
=t\begin{pmatrix}1\\0\\0\end{pmatrix} + s\begin{pmatrix}0\\1\\0\end{pmatrix}
\end{align*}
is in the null space. And so a basis for $$E_2$$:
\begin{align*}
\{\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix}\}.
\end{align*}
The geometric multiplicity is $$2$$ so there is one generalized eigenvector. We can find it by solving:
\begin{align*}
\begin{pmatrix}
0\\1\\0
\end{pmatrix}
&=(M-2I)\boldsymbol{w}_1\\[4mm]
&=\begin{pmatrix}
0 &0 & 0\\ 0 &0&1\\ 0 & 0 & 0
\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\\[4mm]
&= \begin{pmatrix}0\\x_3\\0\end{pmatrix}.
\end{align*}
So $$x_3 = 1$$ and $$x_1$$ and $$x_2$$ are free. So any non–zero vector of the form:
\begin{align*}
\begin{pmatrix}a\\b\\1\end{pmatrix}
=a\begin{pmatrix}1\\0\\0\end{pmatrix} + b\begin{pmatrix}0\\1\\0\end{pmatrix} +
\begin{pmatrix}0\\0\\1\end{pmatrix}
\end{align*}
is a generalized eigenvector. So we can take
\begin{align*}
\boldsymbol{w}_1 = \begin{pmatrix}0\\0\\1\end{pmatrix}
\end{align*}

An important fact is the following: for any $$n\times n$$ matrix, we can find a basis for $$\mathbf{R}^n$$ that consists entierly of eigenvectors and generalized eigenvectors of the matrix $$M$$.

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Linear Algebra for Math 308: L5V15

3. Find all eigenvectors and generalized eigenvectors for $$M = \begin{pmatrix}0 & 1 & -2\\1&2&1\\2&-1&4\end{pmatrix}.$$

You may check that $$2$$ is the only eigenvalue. To find eigenvectors:
\begin{align*}
\begin{pmatrix}-2 & 1 & -2\\1 & 0 & 1\\ 2 & -1 & 2\end{pmatrix}
&\sim \begin{pmatrix}-2 & 1 & -2\\1 & 0 & 1\\ 0 & 0 & 0\end{pmatrix}\\[4mm]
&\sim \begin{pmatrix}1 & 0 & 1\\-2 & 1 & -2\\ 0 & 0 & 0\end{pmatrix}\\[4mm]
&\sim \begin{pmatrix}1&0&1\\0&1&0\\0&0&0\end{pmatrix}.
\end{align*}
So, $$x_3$$ is a free variable and $$x_1 = -x_3$$ and $$x_2=0$$. Since there is only one eigenvector, the dimension of $$E_2$$ is $$1$$ and a basis is:
\begin{align*}
\left\{\begin{pmatrix}1\\0\\-1\end{pmatrix}\right\}.
\end{align*}
This means there are two linearly independent generalized eigenvectors. The first we find by solving:
\begin{align*}
\begin{pmatrix}1\\0\\-1\end{pmatrix}
= \begin{pmatrix}-2 & 1 & -2\\1 & 0 & 1\\ 2 & -1 & 2\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}.
\end{align*}
So we row–reduce:
\begin{align*}
\left(\begin{array}{c c c | c}
-2 & 1 & -2 & 1\\ 1 & 0 & 1 & 0\\ 2 & -1 & 2 & -1
\end{array}\right)
&\sim \left(\begin{array}{c c c | c}
1 & 0 & 1 & 0\\ -2 & 1 & -2 & 1\\ 2 & -1 & 2 & -1
\end{array}\right)\\[4mm]
&\sim  \left(\begin{array}{c c c | c}
1 & 0 & 1 & 0\\ -2 & 1 & -2 & 1\\ 0 & 0 & 0 & 0
\end{array}\right)\\[4mm]
&\sim \left(\begin{array}{c c c | c}
1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 0
\end{array}\right)
\end{align*}
So then any non--zero vector of the form $$\begin{pmatrix}a\\1\\-a\end{pmatrix}$$ is a generalized eigenvector. We take $$\boldsymbol{w}_1 = \begin{pmatrix}0\\1\\0\end{pmatrix}$$. Now to find $$\boldsymbol{w}_2$$ we solve:
\begin{align*}
\begin{pmatrix}0\\1\\0\end{pmatrix}
= \begin{pmatrix}-2 & 1 & -2\\1 & 0 & 1\\ 2 & -1 & 2\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}.
\end{align*}
So we row-reduce:
\begin{align*}
\left(\begin{array}{c c c | c}
-2 & 1 & -2 & 0\\ 1 & 0 & 1 & 1\\ 2 & -1 & 2 & 0
\end{array}\right)
&\sim \left(\begin{array}{c c c | c}
1 & 0 & 1 & 1\\ -2 & 1 & -2 & 0\\ 2 & -1 & 2 & 0
\end{array}\right)\\[4mm]
&\sim  \left(\begin{array}{c c c | c}
1 & 0 & 1 & 1\\ -2 & 1 & -2 & 0\\ 0 & 0 & 0 & 0
\end{array}\right)\\[4mm]
&\sim \left(\begin{array}{c c c | c}
1 & 0 & 1 & 1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 0 & 0
\end{array}\right)
\end{align*}
This gives $$x_3$$ is free, $$x_1 + x_3 = 1$$ and $$x_2 = 2$$. So any non–zero vector of the form $$\begin{pmatrix}1-a\\2\\a\end{pmatrix}$$ is a generalized eigenvector. We take $$a=0$$ and so $$\boldsymbol{v}_2 = \begin{pmatrix}1\\2\\0\end{pmatrix}$$. So the eigenvector and generalized eigenvector are:
\begin{align*}
\boldsymbol{v}=\begin{pmatrix}1\\0\\-1\end{pmatrix},
\boldsymbol{w}_1=\begin{pmatrix}0\\1\\0\end{pmatrix},
\boldsymbol{w}_2=\begin{pmatrix}1\\2\\0\end{pmatrix}
\end{align*}

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L5V16

Exercises

1. Find the eigenvalues and eigenvectors of the following matrix:
\begin{align*}
\begin{pmatrix}
5 & -1\\
3 & 1
\end{pmatrix}\end{align*}

First we find the eigenvalues:
\begin{align*}

&=\det\begin{pmatrix}5-\lambda & -1\\ 3 & 1-\lambda\end{pmatrix}\\[4mm]
&=(\lambda - 1)(\lambda - 5) + 3 \\
&= \lambda^2 - 6\lambda + 8 \\
&= (\lambda -2)(\lambda - 4).
\end{align*}
So the eigenvalues are $$\lambda_1 = 2$$ and $$\lambda_2 = 4$$. Now we find the eigenspaces (that is, the eigenvectors) corresponding to each eigenvalue. First, for $$\lambda_1$$ we find (non–zero) solutions to:
\begin{align*}
\begin{pmatrix}
3 & -1\\3 & -1
\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
=\begin{pmatrix}0\\0\end{pmatrix}.
\end{align*}
The solutions are $$3x_1 - x_2 = 0$$ or $$x_2 = 3x_1$$. That is, vectors of the form:
\begin{align*}
\begin{pmatrix}t\\3t\end{pmatrix}=t\begin{pmatrix}1\\3\end{pmatrix}.
\end{align*}
So we may take$$\boldsymbol{v}_1 = \begin{pmatrix}1\\3\end{pmatrix}$$. Next, for $$\lambda_2$$ we find (non–-zero) solutions to:
\begin{align*}
\begin{pmatrix}
1 & -1\\3 & -3
\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
=\begin{pmatrix}0\\0\end{pmatrix}.
\end{align*}
The solutions are $$x_1 - x_2 = 0$$ or $$x_1 = x_2$$. That is, vectors of the form:
\begin{align*}
\begin{pmatrix}t\\t\end{pmatrix} = t\begin{pmatrix}1\\1\end{pmatrix}
\end{align*}
So we may take $$\boldsymbol{v}_2 = \begin{pmatrix}1\\1\end{pmatrix}.$$

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L5E1

2. Find the eigenvalues and eigenvectors of the following matrix:
\begin{align*}
\begin{pmatrix}
-2 & 1\\
1 & -2
\end{pmatrix}
\end{align*}

First we find the eigenvalues:
\begin{align*}

=\det\begin{pmatrix}-2-\lambda & 1\\ 1 & -2-\lambda\end{pmatrix}
=(\lambda +2)^2 - 1.
\end{align*}
So the eigenvalues are $$\lambda_1 = -3$$ and $$\lambda_2 = -1$$. Now we find the eigenspaces (that is, the eigenvectors) corresponding to each eigenvalue. First, for $$\lambda_1$$ we find (non–zero) solutions to:
\begin{align*}
\begin{pmatrix}
1 & 1\\1 & 1
\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
=\begin{pmatrix}0\\0\end{pmatrix}.
\end{align*}
The solutions are $$x_1 + x_2 = 0$$ or $$x_1 = -x_2$$. That is, vectors of the form:
\begin{align*}
\begin{pmatrix}t\\-t\end{pmatrix}=t\begin{pmatrix}1\\-1\end{pmatrix}.
\end{align*}
So we may take $$\boldsymbol{v}_1 = \begin{pmatrix}1\\-1\end{pmatrix}$$. Next, for $$\lambda_2$$ we find (non–zero) solutions to:
\begin{align*}
\begin{pmatrix}
-1 & 1\\1 & -1
\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
=\begin{pmatrix}0\\0\end{pmatrix}.
\end{align*}
The solutions are $$x_1 - x_2 = 0$$ or $$x_1 = x_2$$. That is, vectors of the form:
\begin{align*}
\begin{pmatrix}t\\t\end{pmatrix} = t\begin{pmatrix}1\\1\end{pmatrix}
\end{align*}
So we may take $$\boldsymbol{v}_2 = \begin{pmatrix}1\\1\end{pmatrix}.$$

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L5E2

3. Find the eigenvalues and eigenvectors of the following matrix:
\begin{align*}
\begin{pmatrix}
3 & -2\\
4 & -1
\end{pmatrix}
\end{align*}

We find the zeros of the characteristic polynomial:
\begin{align*}
\det\begin{pmatrix}3-\lambda & -2\\4&-1-\lambda\end{pmatrix}
&=(\lambda + 1)(\lambda - 3) + 8\\
&=\lambda^2 -2\lambda +5\\
&=(\lambda -1)^2 +4.
\end{align*}
So the eigenvalues are $$\lambda_1 = 1 +2i$$ and $$\lambda_2 = 1-2i$$. To find the eigenvectors, we solve:
\begin{align*}
\begin{pmatrix}
3-(1+2i) & -2\\
4 & -1 - (1+2i)
\end{pmatrix}
\begin{pmatrix}x_1\\x_2\end{pmatrix}
&=\begin{pmatrix}
2 - 2i & -2\\
4 & -2 - 2i
\end{pmatrix}\\[4mm]
&=\begin{pmatrix}0\\0\end{pmatrix}.
\end{align*}
That is: $$(1-i)x_1 - x_2 = 0$$ or $$x_2 = (1-i)x_1$$ so the solutions are vectors of the form:
\begin{align*}
\begin{pmatrix}t\\t(1-i)\end{pmatrix}
=t\begin{pmatrix}1\\1-i\end{pmatrix}.
\end{align*}
So then a basis for the eigenspace associated to $$1+2i$$ is $$\left\{\begin{pmatrix}1\\1-i\end{pmatrix}\right\}$$. So we say $$\boldsymbol{v}_1 = \begin{pmatrix}1\\1-i\end{pmatrix}$$.

Since the matrix has real entries, we know that the eigenvector associated to $$\lambda_2$$ is the complex conjugate of the eigenvector associated to $$\lambda_1$$. So we take $$\boldsymbol{v}_2 = \begin{pmatrix}1\\1+i\end{pmatrix}$$.

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L5E3

4. Find the eigenvalues and eigenvectors of the following matrix:
\begin{align*}
\begin{pmatrix}
2 & -5\\
1 & -2
\end{pmatrix}\end{align*}

Note: There is no video solution to this problem.

First we find the eigenvalues:
\begin{align*}

&=\det \begin{pmatrix}2-\lambda & -5\\1 & -2-\lambda\end{pmatrix}\\[4mm]
&= (\lambda-2)(\lambda + 2) +5\\
&= \lambda^2 +1.
\end{align*}
So the eigenvalues are $$\lambda_1 = i$$ and $$\lambda_2 = -i$$. To find $$\boldsymbol{v}_1$$ we solve:
\begin{align*}
\begin{pmatrix}
2-i & -5\\ 1 & -2-i
\end{pmatrix}
\begin{pmatrix}x_1\\x_2\end{pmatrix}
=\begin{pmatrix}0\\0\end{pmatrix}.
\end{align*}
That is $$(2-i)x_1 = 5x_2$$. If we multiply both sides by $$2+i$$ we get $$5x_1 = 5(2+i)x_2$$ or $$x_1 = (2+i)x_2$$.So then we can take $$\boldsymbol{v}_1 = \begin{pmatrix}2+i\\1\end{pmatrix}$$ as the eigenvector associated with $$\lambda_1 = i$$.

Similar to the previous problem, $$\boldsymbol{v}_2 = \begin{pmatrix}2+i\\1\end{pmatrix}$$.

5. Find the eigenvalues and eigenvectors of the following matrix:
\begin{align*}
\begin{pmatrix}
0 & 1\\
-4 & 4
\end{pmatrix}
\end{align*}

First we find the eigenvalues:
\begin{align*}
0
&=\det\begin{pmatrix}-\lambda & 1\\ -4 & 4-\lambda\end{pmatrix}
&=\lambda(\lambda - 4) + 4
&= \lambda^2 -4\lambda + 4
&= (\lambda -2)^2.
\end{align*}
So $$\lambda = 2$$ is the only eigenvalue and has algebraic multiplicity $$2$$. Now we find the eigenvectors by solving:
\begin{align*}
\begin{pmatrix}
-2 & 1\\-4&2
\end{pmatrix}
\begin{pmatrix}x_1\\x_2\end{pmatrix}
=\begin{pmatrix}0\\0\end{pmatrix}.
\end{align*}
That is $$-2x_1 + x_2 = 0$$ or $$x_2 = 2x_1$$. So then $$\boldsymbol{v}_1 = \begin{pmatrix}1\\2\end{pmatrix}$$ is the only linearly independent eigenvector associated to $$\lambda =2$$.

Next we find a generalized eigenvector by solving:
\begin{align*}
\begin{pmatrix}
-2 & 1\\ -4 & 2
\end{pmatrix}
\begin{pmatrix}x_1\\x_2\end{pmatrix}
=\begin{pmatrix}1\\2\end{pmatrix}
\end{align*}
or $$-2x_1 + x_2 = 1$$ or $$x_2 = 1 + 2x_2$$. So any vector of the form:
\begin{align*}
\boldsymbol{v}_2 = \begin{pmatrix}t\\1+2t\end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix}
+ t\begin{pmatrix}1\\2\end{pmatrix}
\end{align*}
is a generalized eigenvector associated to $$\lambda = 2$$. We'll take the one with $$t=0$$ so that $$\boldsymbol{w} = \begin{pmatrix}0\\1\end{pmatrix}$$.

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L5E5

6. Find the eigenvalues and eigenvectors of the following matrix:
\begin{align*}
\begin{pmatrix}
1 & 0 & 0\\
-3 & -2 &3\\
0 & 0 & 1
\end{pmatrix}
\end{align*}

First, we find the eigenvalues:
\begin{align*}
0
&=\det
\begin{pmatrix}
1-\lambda & 0 & 0\\
-3 & -2-\lambda & 3\\
0 & 0 & 1-\lambda
\end{pmatrix}\\[4mm]
&=(1-\lambda)\det\begin{pmatrix}-2-\lambda & 3\\0&1-\lambda\end{pmatrix}\\[4mm]
&=(1-\lambda)(\lambda+2)(\lambda-1).
\end{align*}
So the eigenvalues are $$\lambda_1 = -2$$ and $$\lambda_{2,3}=1$$. That is, $$1$$ is an eigenvalue with algebraic multiplicity $$2$$. First, we find the eigenvectors associated to $$-2$$ by row–reducing:
\begin{align*}
\begin{pmatrix}
3 & 0 & 0\\
-3 & 0 & 3\\
0 & 0 & 3
\end{pmatrix}\sim
\begin{pmatrix}
3 & 0 & 0\\
0 & 0 & 3\\
0 & 0 & 0
\end{pmatrix}.
\end{align*}
So $$x_1$$ and $$x_3$$ are basic variables and $$x_2$$ is free. This says $$x_1 = 0$$, $$x_3=0$$ and $$x_2$$ is free. So any non–zero vector of the form:
\begin{align*}
t\begin{pmatrix}
0\\1\\0
\end{pmatrix}
\end{align*}
is an eigenvector. So we take $$\boldsymbol{v}_1 = \begin{pmatrix}0\\1\\0\end{pmatrix}$$.

Next we find the eigenvectors associated to $$1$$ by row–reducing:
\begin{align*}
\begin{pmatrix}
0 & 0 & 0\\
-3 & -3 & 3\\
0 & 0 & 0
\end{pmatrix}.
\end{align*}
This is already in echelon form. It says $$x_1$$ is the basic variable and $$x_2$$ and $$x_3$$ are free. So $$x_1 = -x_2 + x_3$$ and $$x_2 = s$$ and
$$x_3 = t$$. So any non–zero vector of the form:
\begin{align*}
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}
=\begin{pmatrix}-s + t\\s\\t\end{pmatrix}
=s\begin{pmatrix}-1\\1\\0\end{pmatrix} +
t\begin{pmatrix}1\\0\\1\end{pmatrix}.
\end{align*}
So then we can take $$\boldsymbol{v}_2 = \begin{pmatrix}-1\\1\\0\end{pmatrix}$$ and $$\boldsymbol{v}_3=\begin{pmatrix}1\\0\\1\end{pmatrix}$$.

Note: these vectors are slightly different than the ones found in the video; but both answers are correct. Indeed, $$\boldsymbol{v}_2 + \boldsymbol{v}_3 = \begin{pmatrix}0\\1\\1\end{pmatrix}$$ which is the one found in the video.

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L5E6

7. Find the eigenvalues and eigenvectors of the following matrix:
\begin{align*}
\begin{pmatrix}
3 & 0 & -6\\
-6 & -3 &6\\
0 & 0 & -3
\end{pmatrix}
\end{align*}

Note: There is no video solution to this problem.

First we find the eigenvalues:
\begin{align*}
0&=
\det
\begin{pmatrix}
3-\lambda & 0 & -6\\
-6 & -3-\lambda &6\\
0 & 0 & -3-\lambda
\end{pmatrix}\\[4mm]
&=(3-\lambda)\det\begin{pmatrix}-3-\lambda &6\\0&-3-\lambda\end{pmatrix}
-6\det\begin{pmatrix}-6 & -3-\lambda\\0 & 0\end{pmatrix}\\[4mm]
&=(3-\lambda)(\lambda+3)^2.
\end{align*}
So the eigenvalues are $$\lambda_{1,2} = -3$$ and $$\lambda_3 = 3$$. That is, $$-3$$ is an eigenvalue with algebraic multiplicity $$2$$. We first find the
eigenvectors associated to the eigenvalue $$-3$$:
\begin{align*}
\begin{pmatrix}
6&0&-6\\
-6&0&6\\
0&0&0
\end{pmatrix}
\sim
\begin{pmatrix}
1&0&-1\\
0&0&0\\
0&0&0
\end{pmatrix}.
\end{align*}
Then $$x_2=s$$ and $$x_3=t$$ are free. This says $$x_1 = x_3 = t$$ and $$x_2 = s$$. So any non–zero vector of the form:
\begin{align*}
\begin{pmatrix}
x_1\\x_2\\x_3
\end{pmatrix}
\begin{pmatrix}t\\s\\t\end{pmatrix}
=s\begin{pmatrix}0\\1\\0\end{pmatrix}
=t\begin{pmatrix}1\\0\\1\end{pmatrix}.
\end{align*}
So we take:
\begin{align*}
\boldsymbol{v}_1 = \begin{pmatrix}0\\1\\0\end{pmatrix},
\boldsymbol{v}_2 = \begin{pmatrix}1\\0\\1\end{pmatrix}.
\end{align*}

Now we find the eigenvector associated to $$3$$:
\begin{align*}
\begin{pmatrix}
0 & 0 & -6\\
-6 & -6 & 6\\
0 & 0 & -6
\end{pmatrix}
&\sim
\begin{pmatrix}
-6 & -6 & 6\\
0 & 0 & -6\\
0 & 0 & -6
\end{pmatrix}\\[4mm]
&\sim
\begin{pmatrix}
1 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}.
\end{align*}
So $$x_1$$ and $$x_3$$ are basic variables and $$x_2=t$$ is free. This matrix says that $$x_1 = -x_2 = -t$$ and $$x_3=0$$ so any vector non–zero vector of the form:
\begin{align*}
\begin{pmatrix}x_1\\x_2\\x_2\end{pmatrix}
=\begin{pmatrix}-t\\t\\0\end{pmatrix}
=t\begin{pmatrix}-1\\1\\0\end{pmatrix}.
\end{align*}
So we take:
\begin{align*}
\boldsymbol{v}_3 = \begin{pmatrix}-1\\1\\0\end{pmatrix}.
\end{align*}

8. Find the eigenvalues and eigenvectors of the following matrix:
\begin{align*}
\begin{pmatrix}
0 & -1 & 1\\
-3 & 0 & 2\\
-5 & -3 & 5
\end{pmatrix}
\end{align*}

Note: There is no video solution to this problem.

First we find the eigenvalues:
\begin{align*}
0
&=\det\begin{pmatrix}
-\lambda & -1 & 1\\
-3 & -\lambda & 2\\
-5 & -3 & 5-\lambda
\end{pmatrix}
\\[4mm]&=-\lambda\det\begin{pmatrix}-\lambda & 2\\-3 & 5-\lambda\end{pmatrix}
+ \det\begin{pmatrix}-3&2\\-5&5-\lambda\end{pmatrix}
+ \det\begin{pmatrix}-3 & -\lambda\\-5 & -3\end{pmatrix}
\\[4mm]&=-\lambda(\lambda(\lambda - 5) + 6)
+ 3(\lambda-5) + 10
+ 9 - 5\lambda
\\&= \lambda(\lambda^2 - 5\lambda + 6)
- (-2\lambda +4)
\\&=\lambda(\lambda -3)(\lambda - 2) + 2(\lambda - 2)
\\&= (\lambda - 2)(\lambda^2 - 3\lambda +2)
\\&= (\lambda-2)(\lambda - 1)(\lambda - 2).
\end{align*}
So $$\lambda_1 = 1$$ and $$\lambda_{2,3} = 2$$. We find the eigenvectors associated to $$1$$ first:
\begin{align*}
\begin{pmatrix}
-1&-1&1\\
-3&-1&2\\
-5&-3&4
\end{pmatrix}
&\sim\begin{pmatrix}
1&1&-1\\
0&2&-1\\
0&2&-1
\end{pmatrix}
\\[4mm]&\sim
\begin{pmatrix}
1&1&-1\\
0&2&-1\\
0&0&0
\end{pmatrix}
\\[4mm]&\sim
\begin{pmatrix}
1&0&-1/2\\
0&1&-1/2\\
0&0&0
\end{pmatrix}.
\end{align*}
So, $$x_1$$ and $$x_2$$ are basic and $$x_3 = t$$ is free. Then any non–zero vector of the form:
\begin{align*}
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}
=\begin{pmatrix}-t/2\\-t/2\\t\end{pmatrix}
=t\begin{pmatrix}-1/2\\-1/2\\1\end{pmatrix}.
\end{align*}
So we can take:
\begin{align*}
\boldsymbol{v}_1 = \begin{pmatrix}-1\\-1\\2\end{pmatrix}.
\end{align*}
Now for the eigenvalue $$2$$:
\begin{align*}
\begin{pmatrix}
-2 & -1 & 1\\
-3 & -2 & 2\\
-5 & -3 & 3
\end{pmatrix}
&\sim \begin{pmatrix}
1 & 1/2 & -1/2\\
-3 & -2 & 2\\
-5 & -3 & 3
\end{pmatrix}
\\[4mm]&\sim
\begin{pmatrix}
1 & 1/2 & -1/2\\
0 & -1/2 & 1/2\\
0 & -1/2 & 1/2
\end{pmatrix}
\\[4mm]&\sim\begin{pmatrix}
1 & 1/2 & -1/2\\
0 & 1 & -1\\
0 & 0 & 0
\end{pmatrix}
\\[4mm]&\sim
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & -1\\
0 & 0 & 0
\end{pmatrix}.
\end{align*}
So $$x_3=t$$ is free. So $$x_1=0$$ and $$x_2 = x_3 = t$$. So any non–zero vector of the form:
\begin{align*}
\begin{pmatrix}x_1\\x_2\\x_2\end{pmatrix}
=\begin{pmatrix}0\\t\\t\end{pmatrix}
=t\begin{pmatrix}0\\1\\1\end{pmatrix}.
\end{align*}
That is, we can take
\begin{align*}
\boldsymbol{v}_2 = \begin{pmatrix}0\\1\\1\end{pmatrix}.
\end{align*}
There is only one linearly independent eigenvector associated to $$2$$. So now we find an eigenvector by solving $$(M-2I)\boldsymbol{w} = \boldsymbol{v}_2$$. We do
this by row–reducing:
\begin{align*}
\left(\begin{array}{ccc|c}
-2 & -1 & 1 & 0\\
-3 & -2 & 2 & 1\\
-5 & -3 & 3 & 1
\end{array}\right)
&\sim \left(\begin{array}{ccc|c}
1 & 1/2 & -1/2 & 0\\
0 & -1/2 & 1/2 & 1\\
0 & -1/2 & 1/2 &1
\end{array}\right)
\\[4mm]&\sim \left(\begin{array}{ccc|c}
1 & 0 & 0 & 1\\
0 & -1/2 & 1/2 & 1\\
0 & 0 & 0 &0
\end{array}\right)
\\[4mm]&\sim\left(\begin{array}{ccc|c}
1 & 0 & 0 & 1\\
0 & 1 & -1 & -2\\
0 & 0 & 0 &0
\end{array}\right).
\end{align*}
So $$x_1$$ and $$x_2$$ are basic and $$x_3$$ are free. So $$x_1 = 1$$ and $$x_2 = -2 + x_3$$ and $$x_3 = t$$. So any non–zero vector of the form:
\begin{align*}
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}
&=\begin{pmatrix}1\\-2 + t\\t\end{pmatrix}
\\[4mm]&=\begin{pmatrix}1\\-2\\0\end{pmatrix}
+t\begin{pmatrix}0\\1\\1\end{pmatrix}.
\end{align*}
So we can take the generalized eigenvector to be:
\begin{align*}
\boldsymbol{w}=
\begin{pmatrix}1\\-2\\0\end{pmatrix}.
\end{align*}