Eigenvalues and Eigenvectors of Symmetric Matrices - Linear Algebra

The first step into solving for eigenvalues, is adding in a along the main diagonal.

Now the next step to take the determinant.

Now lets FOIL, and solve for .

Now lets use the quadratic equation to solve for .

So our eigen values are

In this problem, we will get three eigen values and eigen vectors since it's a symmetric matrix.

To find the eigenvalues, we need to minus lambda along the main diagonal and then take the determinant, then solve for lambda.

This can be factored to

Thus our eigenvalues are at

Now we need to substitute into or matrix in order to find the eigenvectors.

For .

Now we need to get the matrix into reduced echelon form.

This can be reduced to

This is in equation form is , which can be rewritten as . In vector form it looks like, .

We need to take the dot product and set it equal to zero, and pick a value for , and .

Let , and .

Now we pick another value for , and so that the result is zero. The easiest ones to pick are , and .

So the orthogonal vectors for are , and .

Now we need to get the last eigenvector for .

After row reducing, the matrix looks like

So our equations are then

, and , which can be rewritten as , .

Then eigenvectors take this form, . This will be orthogonal to our other vectors, no matter what value of , we pick. For convenience, let's pick , then our eigenvector is.

Any real value of makes a symmetric matrix with real entries. It holds that any eigenvalues of must be real regardless of the value of .

An eigenvalue of is a zero of the characteristic equation formed from the determinant of , so find this determinant as follows:

Subtracting elementwise:

Set the determinant to 0 and solve for :

The determinant can be found by taking the upper-left-to-lower-right product and subtracting the upper-right-to-lower-left product:

,

so the eigenvalues of this matrix are 0 and .

The first step into solving for eigenvalues, is adding in a along the main diagonal.

Now the next step to take the determinant.

Now lets FOIL, and solve for .

Now lets use the quadratic equation to solve for .

So our eigen values are

In this problem, we will get three eigen values and eigen vectors since it's a symmetric matrix.

To find the eigenvalues, we need to minus lambda along the main diagonal and then take the determinant, then solve for lambda.

This can be factored to

Thus our eigenvalues are at

Now we need to substitute into or matrix in order to find the eigenvectors.

For .

Now we need to get the matrix into reduced echelon form.

This can be reduced to

This is in equation form is , which can be rewritten as . In vector form it looks like, .

We need to take the dot product and set it equal to zero, and pick a value for , and .

Let , and .

Now we pick another value for , and so that the result is zero. The easiest ones to pick are , and .

So the orthogonal vectors for are , and .

Now we need to get the last eigenvector for .

After row reducing, the matrix looks like

So our equations are then

, and , which can be rewritten as , .

Then eigenvectors take this form, . This will be orthogonal to our other vectors, no matter what value of , we pick. For convenience, let's pick , then our eigenvector is.