Hello friends, today it’s all about the characteristic polynomial by Faddeev. Have a look!!

### Characteristic polynomial by Faddeev

In one of my earlier posts on eigenvalues of matrices, I have already mentioned about the characteristic equations or characteristic polynomials.

Now if the matrix is very large, then the method of becomes very difficult to handle.

In that situation, the method of Faddeev works very well.

You can also check out: Eigenvalues of any matrix

Suppose any matrix is

According to the method of Faddeev, the characteristic polynomial of the matrix is

Here is the order of the matrix.

Thus for any matrix is 3.

Also is the related to the trace of the matrix

The formula for is

At the beginning,

That means the initial value of is .

For any value of more than 1, .

Also,

Here is the identity matrix.

So, now I combine all these information and get three equations.

These are

(1)

(2)

and

(3)

These three equations are the basis to get the Characteristic polynomial by Faddeev.

Now I’ll solve an example for you. Hope that will help!!

#### Example on the characteristic polynomial by Faddeev

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)* “Using the method of Faddeev, obtain the characteristic polynomial of the matrix

”

##### Solution

First of all, I’ll give this matrix a name, say

Therefore I can say,

As per the method of Faddeev, let the characteristic equation be

This is because the matrix has order 3. So the degree of the characteristic polynomial will also be 3.

Anyways, now I’ll start with

###### Step 1

From the equation (1), I can say for the value of is = trace of the matrix .

Now the trace of any matrix is the sum of its diagonal elements.

Thus

I can substitute this value of in equation (2) to get

Thus it will be

I can substitute this value of in equation (3) to get as

So now I’ll go to the next step, that is, to get the value of

###### Step 2

From the equation (1), I can say that for the value of is

Now the trace of any matrix is the sum of its diagonal elements.

Thus the trace of the matrix is

Thus

I can substitute this value of in equation (2) to get

Thus it will be

Again I substitute this value of in equation (3) to get as

So now I’ll go to the next step, that is, to get the value of

###### Step 3

From the equation (1), I can say that for the value of is

Thus the trace of the matrix is

Thus

Now I’ll stop my calculation here.

Thus the characteristic polynomial will be

Hence I can conclude that using the method of Faddeev, the characteristic polynomial of the given matrix is

This is the answer to this example.

Dear friends, this is the end of my 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!!

###### *Reference: G. James et al. (2011): Advanced modern engineering mathematics, Prentice Hall; 4th Edition (March 8, 2011), Chapter: Matrix analysis, Sec. 1.4: The eigenvalue problem, p. 23, Q. No. 5(a).

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