Finding Eigenvalues and Eigenvectors Example 1

In this example we will find the eigenvalues and a corresponding eigenvector for each eigenvalue for the matrix

$$A=\left(\begin{array}{cc}7& -5\\ 10& -8\end{array}\right)$$

First we introducte the matrix and we store it in the variable A

A = matrix([[ 7, -5], \
            [10, -8]])

Eigenvalues

To find the eigenvalues we will calulate

$$A-\lambda I.$$

However, in SageMath (as well as in Python), lambda is a reserved word. So we will use the letter $l$ instead. We store the matrix into the variable AmlI:

l = var('l', latex_name=r'\lambda')
AmlI = A - l*identity_matrix(2)

Here, we have used the function identity_matrix with argument 2 to create the corresponding identity matrix.

To find the eigenvalues we need to calculate the determinant of $A-\lambda I$, which gives us the characteristic equation. We achieve that with the function determinant from the matrix class. We will create the characteristic equation $p(\lambda )$.

p(l) = AmlI.determinant()

We can check the equation by using the function show.

show(p(l).expand())

The output is

$${\lambda }^2+\lambda -6$$

Finding the zeros of this equation will give us the eigenvalues.

In our case:

• $\lambda =2$ is a zero since $p(2)=0$.
• $\lambda =-3$ is a zero since $p(-3)=0$.

The eigenvalues are $\lambda =2$, $\lambda =-3$.

Eigenvectors

To find the eigenvectors, we need to solve the system

$$(A-\lambda I)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$$

Let’s do it!

Eigenvector with associated eigenvalue $\lambda =2$.

We substitue $\lambda =2$ into $A-\lambda I$ and we will store the result into the variable L1:

L1 = AmlI(l=2)

Now we multiply the matrix by the column vector $\left(\begin{array}{c}x\\ y\end{array}\right)$. To do that, we need to create the variable y

NOTE: Although we call it “column vector”, it should be a matrix in SageMath.

y = var('y')
L1xy = L1*matrix([[x],[y]])

If we use the function show, we can visualize the matrix L1xy:

$$\left(\begin{array}{c}5x-5y\\ 10x-10y\end{array}\right)$$

Now we can see that one equation is a multiple of the other. Therefore, we could choose, for example, the first equation

$$5x-5y=0$$

And we obtain the line $y=x$. Now, subsituting $y=x$ into the vector $\left(\begin{array}{c}x\\ y\end{array}\right)$ we have

$$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}x\\ x\end{array}\right)=x\left(\begin{array}{c}1\\ 1\end{array}\right)$$

And we can choose ${\mathbf{v}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$ as an eigenvector.

Eigenvector with associated eigenvalue $\lambda =-3$.

We repeat the same procedure above.

We substitue $\lambda =-3$ into $A-\lambda I$ and we will store the result into the variable L2:

L2 = AmlI(l=-3)

Now we multiply the matrix by the column vector $\left(\begin{array}{c}x\\ y\end{array}\right)$. To do that, we need to create the variable y

y = var('y')
L2xy = L2*matrix([[x],[y]])

If we use the function show, we can visualize the matrix L2xy:

$$\left(\begin{array}{c}10x-5y\\ 10x-5y\end{array}\right)$$

Now we can see that one equation is a multiple of the other. Therefore, we could choose, for example, the first equation

$$10x-5y=0$$

And we obtain the line $y=2x$. Now, subsituting $y=2x$ into the vector $\left(\begin{array}{c}x\\ y\end{array}\right)$ we have

$$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}x\\ 2x\end{array}\right)=x\left(\begin{array}{c}1\\ 2\end{array}\right)$$

And we can choose ${\mathbf{v}}_2=\left(\begin{array}{c}1\\ 2\end{array}\right)$ as an eigenvector.