Eigenvalues and eigenvectors of a matrix

On this post we explain what the eigenvalues and the eigenvectors of a matrix are. You will also find examples of how to compute the eigenvalues and the eigenvectors of a matrix, and finally, you have problems with solutions solved step by step to practice.

What are the eigenvalues and the eigenvectors of a matrix?

Although the concept of eigenvalue and eigenvector is a bit difficult to understand, their definition is as follows:

The eigenvectors are the non-zero vectors of a linear map that, when the linear transformation is applied to them, result in a scalar multiple of them (they do not change direction). This scalar is the eigenvalue.

\text{Av} = \lambda\text{v}

Where \text{A} is the matrix of the linear mapping, \text{v} is the eigenvector and \lambda the eigenvalue.

An eigenvalue is also known as a characteristic value, and an eigenvalue as a characteristic vector.

How to find the eigenvalues and the eigenvectors of a matrix

To find the eigenvalues and eigenvectors of a matrix, apply the following procedure:

  1. Calculate the characteristic polynomial by taking the following determinant:
  2.  \displaystyle \text{det}(A-\lambda I)

  3. Find the roots of the characteristic polynomial obtained in step 1. These roots are the eigenvalues of the matrix.
  4.  \displaystyle \text{det}(A-\lambda I)=0 \ \longrightarrow \ \lambda

  5. Calculate the eigenvector associated with each eigenvalue by solving the following system of equations for each eigenvalue:
  6. \displaystyle (A-\lambda I)\text{v}=0

Remember that I is the identity matrix.

This is the method to find the eigenvalues and eigenvectors of a matrix, but here we also give you some tricks: 😉

Tricks: we can take advantage of the properties of the eigenvalues and eigenvectors to calculate them more easily:

The trace of the matrix (sum of the elements on the main diagonal) is equal to the sum of all the eigenvalues.

\displaystyle tr(A)=\sum_{i=1}^n \lambda_i

The product of all the eigenvalues is equal to the determinant of the matrix.

\displaystyle det(A)=\prod_{i=1}^n \lambda_i

If there is any linear combination between rows or columns, at least one eigenvalue of the matrix is 0.

Having seen the theory of eigenvalues and eigenvectors, let’s see an example of how to compute them.

Example of calculating the eigenvalues and eigenvectors of a matrix

  • Find the eigenvalues and the eigenvectors of the following 2×2 matrix:

\displaystyle A= \begin{pmatrix}1&0\\[1.1ex] 5&2\end{pmatrix}

First, we have to find the characteristic polynomial of the matrix. And, for this, we must compute the following determinant:

\displaystyle \text{det}(A-\lambda I)= \begin{vmatrix}1- \lambda &0\\[1.1ex] 5&2-\lambda \end{vmatrix} = \lambda^2-3\lambda +2

Now we calculate the roots of the characteristic polynomial. So we set the polynomial obtained equal to 0 and solve the equation:

\displaystyle \lambda^2-3\lambda +2 = 0

\lambda= \cfrac{-(-3)\pm \sqrt{(-3)^2-4\cdot 1 \cdot 2}}{2\cdot 1} = \cfrac{+3\pm 1}{2}=\begin{cases} \lambda = 1 \\[2ex] \lambda = 2 \end{cases}

The solutions of the equation are the eigenvalues of the matrix.

Once we know the eigenvalues of the matrix, we proceed to calculate its eigenvectors. To do this, we must solve the following system of equations for each eigenvalue:

\displaystyle (A-\lambda I)\text{v}=0

First we calculate the eigenvector associated with the eigenvalue 1:

\displaystyle (A-1 I)\begin{pmatrix}x \\[1.1ex] y \end{pmatrix} =}\begin{pmatrix}0 \\[1.1ex] 0 \end{pmatrix}

\displaystyle \begin{pmatrix}0&0\\[1.1ex] 5&1\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \end{pmatrix} =}\begin{pmatrix}0 \\[1.1ex] 0 \end{pmatrix}

\displaystyle \left.\begin{array}{l} 0x+0y = 0 \\[2ex] 5x+y = 0\end{array}\right\}

From these equations we obtain the following subspace:

\displaystyle y=-5x

Note that subspaces of eigenvectors are also called eigenspaces.

Now we must find a base of this eigenspace, so we give, for example, the value 1 to the variable x and we obtain the following eigenvector:

\displaystyle x = 1 \ \longrightarrow \ y=-5\cdot 1 = -5

\displaystyle \text{v}=\begin{pmatrix}1 \\[1.1ex] -5\end{pmatrix}

Once we have found the eigenvector associated with the eigenvalue 1, we repeat the process to calculate the eigenvector of the eigenvalue 2:

\displaystyle (A-\lambda I)\text{v}=0

\displaystyle (A-2I)\begin{pmatrix}x \\[1.1ex] y \end{pmatrix} =}\begin{pmatrix}0 \\[1.1ex] 0 \end{pmatrix}

\displaystyle \begin{pmatrix}-1&0\\[1.1ex] 5&0\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \end{pmatrix} =}\begin{pmatrix}0 \\[1.1ex] 0 \end{pmatrix}

\displaystyle \left.\begin{array}{l} -x+0y = 0 \\[2ex] 5x+0y = 0\end{array}\right\} \longrightarrow \ x=0

In this case, only the first component of the vector must be 0, so we could give any value to variable y. But it is better to put a 1:

\displaystyle \text{v}=\begin{pmatrix}0 \\[1.1ex] 1 \end{pmatrix}

In conclusion, the eigenvalues and the eigenvectors of the matrix are:

\displaystyle \lambda = 1 \qquad \text{v}=\begin{pmatrix}1 \\[1.1ex] -5 \end{pmatrix}

\displaystyle \lambda = 2 \qquad \text{v}=\begin{pmatrix}0 \\[1.1ex] 1 \end{pmatrix}

Practice problems on eigenvalues and eigenvectors

Problem 1

Calculate the eigenvalues and eigenvectors of the following square matrix of order 2:

\displaystyle A= \begin{pmatrix}3&1\\[1.1ex] 2&4\end{pmatrix}

First we calculate the determinant of the matrix minus λ on its main diagonal:

\displaystyle \text{det}(A-\lambda I)= \begin{vmatrix}3- \lambda &1\\[1.1ex] 2&4-\lambda \end{vmatrix} = \lambda^2-7\lambda +10

Now we calculate the roots of the characteristic polynomial:

\displaystyle \lambda^2-7\lambda +10=0 \ \longrightarrow \ \begin{cases} \lambda = 2 \\[2ex] \lambda = 5 \end{cases}

We find the eigenvector associated with the eigenvalue 2:

\displaystyle (A- 2I)\text{v}=0

\displaystyle \begin{pmatrix}1&1\\[1.1ex] 2&2\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \end{pmatrix} =}\begin{pmatrix}0 \\[1.1ex] 0 \end{pmatrix}

\displaystyle \left.\begin{array}{l} x+y = 0 \\[2ex] 2x+2y = 0\end{array}\right\} \longrightarrow \ x=-y

\displaystyle \text{v}=\begin{pmatrix}1 \\[1.1ex] -1 \end{pmatrix}

And then we calculate the eigenvector associated with the eigenvalue 5:

\displaystyle (A-5I)\text{v}=0

\displaystyle\begin{pmatrix}-2&1\\[1.1ex] 2&-1\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \end{pmatrix} =}\begin{pmatrix}0 \\[1.1ex] 0 \end{pmatrix}

\displaystyle\left.\begin{array}{l} -2x+y = 0 \\[2ex] 2x-y = 0\end{array}\right\} \longrightarrow \ y=2x

\displaystyle \text{v}=\begin{pmatrix}1 \\[1.1ex] 2 \end{pmatrix}

Thus, the eigenvalues and eigenvectors of matrix A are:

 \displaystyle \lambda = 2 \qquad \text{v}=\begin{pmatrix}1 \\[1.1ex] -1 \end{pmatrix}

 \displaystyle \lambda = 5 \qquad \text{v}=\begin{pmatrix}1\\[1.1ex] 2 \end{pmatrix}

 

Problem 2

Determine the eigenvalues and eigenvectors of the following 2×2 square matrix:

\displaystyle A= \begin{pmatrix}2&1\\[1.1ex] 3&0\end{pmatrix}

First we subtract λ from the entries on the main diagonal of the matrix and compute the determinant of the resulting matrix in order to obtain the characteristic equation:

\displaystyle \text{det}(A-\lambda I)= \begin{vmatrix}2- \lambda &1\\[1.1ex] 3&-\lambda \end{vmatrix} = \lambda^2-2\lambda -3

Now we find the roots of the characteristic polynomial:

 \displaystyle \lambda^2-2\lambda -3=0 \ \longrightarrow \ \begin{cases} \lambda = -1 \\[2ex] \lambda = 3 \end{cases}

We calculate the eigenvector associated with the eigenvalue -1:

\displaystyle (A-(-1)I)\text{v}=0

\displaystyle (A+1I)\text{v}=0

\displaystyle \begin{pmatrix} 3&1\\[1.1ex] 3&1\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \end{pmatrix} =}\begin{pmatrix}0 \\[1.1ex] 0 \end{pmatrix}

\displaystyle \left.\begin{array}{l} 3x+1y = 0 \\[2ex] 3x+1y = 0\end{array}\right\} \longrightarrow \ y=-3x

\displaystyle \text{v}=\begin{pmatrix}1 \\[1.1ex] -3 \end{pmatrix}

And then we calculate the eigenvector associated with the eigenvalue 3:

\displaystyle (A-3I)\text{v}=0

\displaystyle \begin{pmatrix}-1&1\\[1.1ex] 3&-3\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \end{pmatrix} =}\begin{pmatrix}0 \\[1.1ex] 0 \end{pmatrix}

\displaystyle \left.\begin{array}{l} -1x+1y = 0 \\[2ex] 3x-3y = 0\end{array}\right\} \longrightarrow \ y=x

\displaystyle \text{v}=\begin{pmatrix}1 \\[1.1ex] 1 \end{pmatrix}

Therefore, the eigenvalues and eigenvectors of matrix A are:

\displaystyle \lambda = -1 \qquad \text{v}=\begin{pmatrix}1 \\[1.1ex] -3 \end{pmatrix}

\displaystyle \lambda = 3 \qquad \text{v}=\begin{pmatrix}1\\[1.1ex] 1 \end{pmatrix}

 

Problem 3

Compute the eigenvalues and eigenvectors of the following square matrix of order 3:

\displaystyle A= \begin{pmatrix}1&2&0\\[1.1ex] 2&1&0\\[1.1ex] 0&1&2\end{pmatrix}

First of all, we have to solve the determinant of matrix A minus the identity matrix multiplied by lambda in order to obtain the characteristic polynomial:

\displaystyle \text{det}(A-\lambda I)= \begin{vmatrix}1-\lambda&2&0\\[1.1ex] 2&1-\lambda&0\\[1.1ex] 0&1&2-\lambda\end{vmatrix}

In this case, the last column of the determinant has two zeros, so we will evaluate the 3×3 determinant by cofactors through that column:

\displaystyle \begin{aligned} \begin{vmatrix}1-\lambda&2&0\\[1.1ex] 2&1-\lambda&0\\[1.1ex] 0&1&2-\lambda\end{vmatrix}& = (2-\lambda)\cdot \begin{vmatrix}1-\lambda&2\\[1.1ex] 2&1-\lambda \end{vmatrix} \\[3ex] & = (2-\lambda)[\lambda^2 -2\lambda -3] \end{aligned}

Now we have to calculate the roots of the characteristic polynomial. It is better not to multiply the parentheses since then we would obtain a third degree polynomial, instead, if the two factors are solved separately it is easier to get the eigenvalues:

\displaystyle (2-\lambda)[\lambda^2 -2\lambda -3]=0 \ \longrightarrow \ \begin{cases} 2-\lambda=0 \ \longrightarrow \ \lambda = 2 \\[2ex] \lambda^2 -2\lambda -3=0 \ \longrightarrow \begin{cases}\lambda = -1 \\[2ex] \lambda = 3 \end{cases} \end{cases}

Now we compute the eigenvector associated with the eigenvalue 2:

\displaystyle (A-2I)\text{v}=0

\displaystyle \begin{pmatrix} -1&2&0\\[1.1ex] 2&-1&0\\[1.1ex] 0&1&0\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \\[1.1ex] z \end{pmatrix} =\begin{pmatrix}0 \\[1.1ex] 0\\[1.1ex] 0 \end{pmatrix}

\displaystyle \left.\begin{array}{l} -x+2y = 0 \\[2ex] 2x-y = 0\\[2ex] y=0 \end{array}\right\} \longrightarrow \ \begin{array}{l} y=0 \\[2ex] x=y=0 \end{array}

\displaystyle \text{v}=\begin{pmatrix}0 \\[1.1ex] 0 \\[1.1ex] 1\end{pmatrix}

We calculate the eigenvector associated with the eigenvalue -1:

\displaystyle (A+I)\text{v}=0

\displaystyle \begin{pmatrix} 2&2&0\\[1.1ex] 2&2&0\\[1.1ex] 0&1&3\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \\[1.1ex] z \end{pmatrix} =\begin{pmatrix}0 \\[1.1ex] 0\\[1.1ex] 0 \end{pmatrix}

\displaystyle \left.\begin{array}{l} 2x+2y = 0 \\[2ex] 2x+2y = 0\\[2ex] y+3z=0 \end{array}\right\} \longrightarrow \ \begin{array}{l} x=-y \\[2ex] y=-3z \end{array}

\displaystyle \text{v}=\begin{pmatrix}3 \\[1.1ex] -3 \\[1.1ex] 1\end{pmatrix}

And we calculate the eigenvector associated with the eigenvalue 3:

\displaystyle (A-3I)\text{v}=0

\displaystyle \begin{pmatrix} -2&2&0\\[1.1ex] 2&-2&0\\[1.1ex] 0&1&-1\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \\[1.1ex] z \end{pmatrix} =\begin{pmatrix}0 \\[1.1ex] 0\\[1.1ex] 0 \end{pmatrix}

\displaystyle \left.\begin{array}{l} -2x+2y = 0 \\[2ex] 2x-2y = 0\\[2ex] y-z=0 \end{array}\right\} \longrightarrow \ \begin{array}{l} x=y \\[2ex] y=z \end{array}

\displaystyle \text{v}=\begin{pmatrix}1 \\[1.1ex] 1 \\[1.1ex] 1\end{pmatrix}

So the eigenvalues and eigenvectors of matrix A are:

\displaystyle \lambda = 2 \qquad \text{v}=\begin{pmatrix}0 \\[1.1ex] 0 \\[1.1ex] 1\end{pmatrix}

\displaystyle \lambda = -1 \qquad \text{v}=\begin{pmatrix}3 \\[1.1ex] -3 \\[1.1ex] 1\end{pmatrix}

\displaystyle \lambda = 3 \qquad \text{v}=\begin{pmatrix}1\\[1.1ex] 1 \\[1.1ex] 1\end{pmatrix}

 

Problem 4

Compute the eigenvalues and the eigenvectors of the following 3×3 square matrix:

\displaystyle A= \begin{pmatrix}2&1&3\\[1.1ex]-1&1&1\\[1.1ex] 1&2&4\end{pmatrix}

First we solve the determinant of the matrix minus λ on its main diagonal to obtain the characteristic polynomial:

\displaystyle \text{det}(A-\lambda I)= \begin{vmatrix}2-\lambda&1&3\\[1.1ex]-1&1-\lambda&1\\[1.1ex] 1&2&4-\lambda\end{vmatrix}=-\lambda^3+7\lambda^2-10\lambda

We factor out the characteristic polynomial and solve for λ each equation:

\displaystyle \lambda(-\lambda^2+7\lambda-10)=0 \ \longrightarrow \ \begin{cases} \lambda=0\\[2ex] -\lambda^2+7\lambda-10=0 \ \longrightarrow \begin{cases}\lambda = 2 \\[2ex] \lambda = 5 \end{cases} \end{cases}

We calculate the eigenvector associated with the eigenvalue 0:

\displaystyle (A-0I)\text{v}=0

\displaystyle \begin{pmatrix} 2&1&3\\[1.1ex]-1&1&1\\[1.1ex] 1&2&4\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \\[1.1ex] z \end{pmatrix} =\begin{pmatrix}0 \\[1.1ex] 0\\[1.1ex] 0 \end{pmatrix}

\displaystyle \left.\begin{array}{l} 2x+y+3z= 0 \\[2ex] -x+y+z= 0\\[2ex] x+2y+4z=0 \end{array}\right\} \longrightarrow \ \begin{array}{l} x=-\cfrac{2z}{3} \\[4ex] y=-\cfrac{5z}{3} \end{array}

\displaystyle \text{v}=\begin{pmatrix}-2 \\[1.1ex] -5\\[1.1ex] 3\end{pmatrix}

We calculate the eigenvector associated with the eigenvalue 2:

\displaystyle (A-2I)\text{v}=0

\displaystyle \begin{pmatrix} 0&1&3\\[1.1ex]-1&-1&1\\[1.1ex] 1&2&2\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \\[1.1ex] z \end{pmatrix} =\begin{pmatrix}0 \\[1.1ex] 0\\[1.1ex] 0 \end{pmatrix}

 \displaystyle \left.\begin{array}{l} y+3z = 0 \\[2ex] -x-y+z= 0\\[2ex] x+2y+2z=0 \end{array}\right\} \longrightarrow \ \begin{array}{l} y=-3z \\[2ex] x=4z \end{array}

\displaystyle \text{v}=\begin{pmatrix}4\\[1.1ex] -3 \\[1.1ex] 1\end{pmatrix}

We calculate the eigenvector associated with the eigenvalue 5:

\displaystyle (A-5I)\text{v}=0

\displaystyle \begin{pmatrix} -3&1&3\\[1.1ex]-1&-4&1\\[1.1ex] 1&2&-1\end{pmatrix}\begin{pmatrix}x \\[1.1ex] y \\[1.1ex] z \end{pmatrix} =\begin{pmatrix}0 \\[1.1ex] 0\\[1.1ex] 0 \end{pmatrix}

\displaystyle \left.\begin{array}{l} -3x+y+3z = 0 \\[2ex] -x-4y+z = 0\\[2ex] x+2y-z=0 \end{array}\right\} \longrightarrow \ \begin{array}{l} x=z \\[2ex] y=0 \end{array}

\displaystyle \text{v}=\begin{pmatrix}1 \\[1.1ex] 0 \\[1.1ex] 1\end{pmatrix}

Therefore, the eigenvalues and eigenvectors of matrix A are:

\displaystyle \lambda = 0 \qquad \text{v}=\begin{pmatrix}-2 \\[1.1ex] -5 \\[1.1ex] 3\end{pmatrix}

\displaystyle \lambda = 2 \qquad \text{v}=\begin{pmatrix}4 \\[1.1ex] -3 \\[1.1ex] 1\end{pmatrix}

\displaystyle \lambda = 5 \qquad \text{v}=\begin{pmatrix}1\\[1.1ex] 0 \\[1.1ex] 1\end{pmatrix}

 

Problem 5

Find the eigenvalues and eigenvectors of the following 4×4 matrix:

\displaystyle A=\begin{pmatrix}1&0&-1&0\\[1.1ex] 2&-1&-3&0\\[1.1ex] -2&0&2&0\\[1.1ex] 0&0&0&3\end{pmatrix}

First, we solve the determinant of the matrix minus λ on its main diagonal to obtain the characteristic polynomial:

\displaystyle \text{det}(A-\lambda I)= \begin{vmatrix}1-\lambda&0&-1&0\\[1.1ex] 2&-1-\lambda&-3&0\\[1.1ex] -2&0&2-\lambda&0\\[1.1ex] 0&0&0&3-\lambda\end{vmatrix}

In this case, the last column of the determinant are all zeros except one element, so we will compute the determinant using the cofactor method through that column:

\displaystyle \begin{aligned} \begin{vmatrix}1-\lambda&0&-1&0\\[1.1ex] 2&-1-\lambda&-3&0\\[1.1ex] -2&0&2-\lambda&0\\[1.1ex] 0&0&0&3-\lambda\end{vmatrix}& = (3-\lambda)\cdot \begin{vmatrix}1-\lambda&0&-1\\[1.1ex] 2&-1-\lambda&-3\\[1.1ex] -2&0&2-\lambda\end{vmatrix} \\[3ex] & = (3-\lambda)[-\lambda^3 +2\lambda^2 +3\lambda] \end{aligned}

Now we must calculate the roots of the characteristic polynomial. It is better not to multiply the parentheses since then we would obtain a fourth degree polynomial, instead, if the two factors are solved separately it is easier to calculate the eigenvalues:

\displaystyle (3-\lambda)[-\lambda^3 +2\lambda^2 +3\lambda]=0 \ \longrightarrow \ \begin{cases} 3-\lambda=0 \ \longrightarrow \ \lambda = 3 \\[2ex] -\lambda^3 +2\lambda^2 +3\lambda =0 \ \longrightarrow \ \lambda(-\lambda^2 +2\lambda +3) =0 \end{cases}

\displaystyle \lambda(-\lambda^2 +2\lambda +3)=0 \ \longrightarrow \ \begin{cases} \lambda=0  \\[2ex] -\lambda^2 +2\lambda +3=0 \ \longrightarrow \ \begin{cases} \lambda=-1 \\[2ex] \lambda = 3 \end{cases}\end{cases}

We calculate the eigenvector associated with the eigenvalue 0:

\displaystyle (A-0I)\text{v}=0

\displaystyle \begin{pmatrix} 1&0&-1&0\\[1.1ex] 2&-1&-3&0\\[1.1ex] -2&0&2&0\\[1.1ex] 0&0&0&3\end{pmatrix}\begin{pmatrix}w \\[1.1ex] x \\[1.1ex] y\\[1.1ex] z \end{pmatrix} =\begin{pmatrix}0 \\[1.1ex] 0\\[1.1ex] 0 \\[1.1ex] 0\end{pmatrix}

\displaystyle \left.\begin{array}{l} w-y = 0 \\[2ex] 2w-x-3y = 0\\[2ex] -2w+2y=0 \\[2ex] 3z=0 \end{array}\right\} \longrightarrow \ \begin{array}{l} w=y \\[2ex] x=-w  \\[2ex]z=0 \end{array}

\displaystyle \text{v}=\begin{pmatrix}1 \\[1.1ex] -1 \\[1.1ex] 1  \\[1.1ex]0 \end{pmatrix}

We calculate the eigenvector associated with the eigenvalue -1:

\displaystyle (A+1I)\text{v}=0

\displaystyle \begin{pmatrix} 2&0&-1&0\\[1.1ex] 2&0&-3&0\\[1.1ex] -2&0&3&0\\[1.1ex] 0&0&0&4\end{pmatrix}\begin{pmatrix}w \\[1.1ex] x \\[1.1ex] y\\[1.1ex] z \end{pmatrix} =\begin{pmatrix}0 \\[1.1ex] 0\\[1.1ex] 0 \\[1.1ex] 0\end{pmatrix}

\displaystyle \left.\begin{array}{l} 2w-y = 0 \\[2ex] 2w-3y = 0\\[2ex] -2w+3y=0 \\[2ex] 4z=0 \end{array}\right\} \longrightarrow \ \begin{array}{l} y=w=0  \\[2ex]z=0 \end{array}

\displaystyle \text{v}=\begin{pmatrix}0 \\[1.1ex] 1 \\[1.1ex] 0  \\[1.1ex]0 \end{pmatrix}

We calculate the eigenvector associated with the eigenvalue 3:

\displaystyle (A-3I)\text{v}=0

\displaystyle \begin{pmatrix} -2&0&-1&0\\[1.1ex] 2&-4&-3&0\\[1.1ex] -2&0&-1&0\\[1.1ex] 0&0&0&0\end{pmatrix}\begin{pmatrix}w \\[1.1ex] x \\[1.1ex] y\\[1.1ex] z \end{pmatrix} =\begin{pmatrix}0 \\[1.1ex] 0\\[1.1ex] 0 \\[1.1ex] 0\end{pmatrix}

\displaystyle \left.\begin{array}{l} -2w-y = 0 \\[2ex] 2w-4x-3y = 0\\[2ex] -2w-y=0 \\[2ex] 0=0 \end{array}\right\} \longrightarrow \ \begin{array}{l} y=-2w \\[2ex] x=2w  \end{array}

\displaystyle \text{v}=\begin{pmatrix}1 \\[1.1ex] 2 \\[1.1ex] -2  \\[1.1ex]0 \end{pmatrix}

The algebraic multiplicity of eigenvalue 3 is 2 (it is repeated twice). So we must find another eigenvector that satisfies the same equations:

\displaystyle \text{v}=\begin{pmatrix}0 \\[1.1ex] 0 \\[1.1ex] 0  \\[1.1ex]1 \end{pmatrix}

Therefore, the eigenvalues and eigenvectors of matrix A are:

\displaystyle \lambda = 0 \qquad \text{v}=\begin{pmatrix}1 \\[1.1ex] -1 \\[1.1ex] 1  \\[1.1ex]0\end{pmatrix}

\displaystyle \lambda = -1 \qquad \text{v}=\begin{pmatrix}0 \\[1.1ex] 1 \\[1.1ex] 0  \\[1.1ex]0 \end{pmatrix}

\displaystyle \lambda = 3 \qquad \text{v}=\begin{pmatrix}1 \\[1.1ex] 2 \\[1.1ex] -2  \\[1.1ex]0\end{pmatrix}

\displaystyle \lambda = 3 \qquad \text{v}=\begin{pmatrix}0 \\[1.1ex] 0 \\[1.1ex] 0  \\[1.1ex]1\end{pmatrix}

 

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