Eigenvectors of repeated eigenvalues. Hello friends, today it’s all about the eigenvectors of repeated eigenvalues. Have a look!!

Eigenvectors of repeated eigenvalues_compressed

**Eigenvectors of repeated eigenvalues**

We all know that for any 3 × 3 matrix, the number of eigenvalues is 3.

Similarly, for any 4 × 4 matrix, the number of eigenvalues is 4 and so on.

Earlier on, I have also mentioned that it is possible to get the eigenvalues by solving the characteristic equation of the matrix.

Moreover, one can also find out the eigenvector corresponding to each eigenvalue.

You can also check out: 1. Eigenvalues of a matrix

So far, it’s all quite good.

Suppose a matrix has the same eigenvalues. That means ‘repeated eigenvalues’.

The question is: how to get the eigenvectors for them?

For example, any 3 × 3 matrix has 3 eigenvalues. Out of these three eigenvalues, two are the same.

Now can one get two different eigenvectors for the same eigenvalue? Yes, one can.

And the next question is: how?

So, here I’ll solve an example for you. I hope that will help!!

**Example on the eigenvectors of the repeated eigenvalues**

Disclaimer: None of these examples are mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here is the example.

**Example **

According to James et al. (2011)* “Obtain the eigenvalues and corresponding eigenvectors of the matrix

”

**Solution**

Let the given matrix is .

So I can say that the given matrix is

To start with, first of all, I’ll get the characteristic equation of the matrix.

**Step 1**

Now the characteristic equation of the matrix is

Thus is

Therefore the characteristic polynomial is

Next I simplify it. And the simplified version is

Therefore the characteristic equation is

If you want, you can also try it as an exercise.

Now I’ll get the eigenvalues.

**Step 2**

In order to get the eigenvalues of the matrix , I’ll solve the characteristic equation to get the values of .

I can rewrite the characteristic equation as Then I’ll solve it.

Thus it will be

Now I can see that the matrix has one repeated eigenvalue 1. It has another separate eigenvalue 2 as well.

I have already discussed how to get the eigenvector of an eigenvalue. I’ll not go into much detail.

So, now comes the next step.

**Step 3**

I’ll start with the separate separate eigenvalue 2, that is,

Thus for the matrix is

Suppose is the corresponding eigenvector for the eigenvalue

Thus I can say

Now I’ll use Gaussian elimination method to simplify these equations.

See also: Gaussian elimination method in 3 × 3 matrices

Thus at the end, I get the system of equations in matrix form as

So I get two equations as

(1)

and

(2)

From equation (2), I can say

Let . Therefore

Now I put these two values in equation (1). Thus it will be

Now I’ll solve this equation to get the value of in terms of .

So it will look like

Thus the eigenvector is

Therefore I can say for , the corresponding eigenvector is

**Step 4**

Now is the next step.

Here I will get the eigenvectors of repeated eigenvalues.

Thus for the matrix is

Suppose is the corresponding eigenvector for the eigenvalue

Therefore I can say

Now this gives the set of equations as

As you can see, here all these three equations are identical.

So the only equation I can get is

(3)

Since the eigenvalue comes twice, I should also get two eigenvectors related to this eigenvalue.

Now in equation (3), I have two options.

In the first option, I choose as zero. In the second option, I will choose as zero.

**Step 5**

So here is the next step.

Here I choose as zero.

Thus for , equation (3) becomes

This means

Let . Therefore .

Thus the eigenvector is

Therefore I can say for , one corresponding eigenvector is

**Step 6**

Now for the second option, I have .

Thus for , equation (3) becomes

This means

Let . Therefore .

Thus the eigenvector is

Therefore I can say that for , the other corresponding eigenvector is

Hence I can conclude that are the eigenvalues of the given matrix.

For the corresponding eigenvectors are . For the corresponding eigenvector is .

This is the answer to this example.

Dear friends, this is the end of today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

Hari Krishnan says

Hi its a wonderful article , i was wandering out in search of this topic.

My question here is from equation 3 there is a possibility of taking x2 as 0 that would give us another set of eigen vectors wouldn’t it ?

is my logic correct ?

Dr. Aspriha Peters says

Hi Hari,

Yes, it’s also possible. But it has to be either both x1 and x2 as 0 or both x2 and x3 as zero. All three variables x1, x2 and x3 can’t be 0 at the same time.

Jalaiah says

Mam..should we have to use gaussian elimination method to solve this example or nor?

(Or) we can do it directly to solve this??

Dr. Aspriha Peters says

Hi Jalaiah,

First of all, there is no such hard-and-fast rule that it must be the Gaussian elimination method. You can try any other method as well. But generally, it’s one of the easiest methods. Also, the direct solution makes life harder and more complicated. Sometimes it doesn’t work even properly. So I always prefer to stick to the Gaussian elimination method.

Manas Mohapatra says

Hi Aspriha,

Since the equation 3 satisfy a plane equation so any value of X, Y, Z would be considered to make eigen vector. Infact there would be more possibilities of eigen vectors. Is this correct?

One more doubt I have. Basically when we diagonalize a matrix through similarity transformation, we must have constraint that the simillarity matrix should not be singular. SO that way we can diagonalise. I think I am correct.

Somewhere I have seen that according to make a diagonalisable matrix there must be a relation that satisfy AS=SD. D for Diagonalied matrix through S. My question is S depends upon our choice. As far as I know it would be difficult to make a matrix S by hand by putting random numbers. SO how do we do tho this to make a diagonalisable matrix except the similarity transformation?? Pl do make a reply. Thank you