Canonical form. Hello friends, today it’s all about how to transform the quadratic form of an expression to a canonical form. Have a look!!

### Canonical form in the matrix analysis

Suppose I have a quadratic form

And I want to transfer it to a canonical form.

First of all, I’ll write it in a matrix form as

Next, I’ll get its eigenvalues, say, and . Then the conic section of will be

After that, I have to get the normalised eigenvector corresponding to each eigenvalue. Then I can write in terms of .

Now I’ll give you an example to explain it in a better way.

#### A solved example of how to transform quadratic form to canonical form

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 Kreyszig (2005)*, “What kind of conic section (or pair of straight lines) is given by the quadratic form? Transform it to principal axes. Express in terms of the new coordinate vector .

”

**Solution**

Now here I have the quadratic form as

First of all, I’ll check the conic section of it.

**Step 1**

As per the standard formula, I have

Also, the matrices and are

Now I’ll get the * eigenvalues of the matrix* . If is the eigenvalue of the matrix , the characteristic equation is the determinant of the matrix

Therefore the characteristic equation is

Next, I’ll solve it to get the values of . Thus it will be

Then I’ll simplify it to get

Now I’ll factorize it. So that gives

Thus I can say that are the two eigenvalues of the matrix .

Hence the quadratic form becomes

Also, it’s given that . Therefore becomes

which means

Hence I can conclude that gives an equation of a hyperbola.

Next, I’ll transform it into principal axes.

**Step 2**

Now, in order to transform the quadratic form to the canonical form or principal axes, I have to get the normalized eigenvectors. And for the normalized eigenvectors, I should get the * eigenvectors corresponding to each eigenvalue*. So I’ll start with that.

Now I’ll get the normalized eigenvector for the eigenvalue .

Thus for , the matrix becomes

So let’s say is the corresponding eigenvector for the eigenvalue .

Thus I can say

And this means

Now the first equation gives

And the second equation gives

Since these two equations are identical, So I can choose any of these.

Therefore let’s choose which gives

Now, let’s say, . Then the value of in terms of will be

Thus the eigenvector corresponding to the eigenvalue is

Hence I can conclude that for , the corresponding eigenvector is .

Now I’ll get its normalized eigenvector. Since , I can say that the normalized eigenvector is .

Next, I’ll get the normalized eigenvector for the eigenvalue .

**Step 3**

First of all, I’ll get the eigenvector corresponding to the eigenvalue .

Thus for , the matrix becomes

So let’s say is the corresponding eigenvector for the eigenvalue .

Thus I can say

And this means

Now the first equation gives

And the second equation gives

Since these two equations are identical, So I can choose any of these.

Therefore for ,

Now, let’s say, . Then the value of in terms of will be

Thus the eigenvector corresponding to the eigenvalue is

Hence I can conclude that for , the corresponding eigenvector is .

Now I’ll get its normalized eigenvector. Since , I can say that the normalized eigenvector is .

Next, I’ll express in terms of the new coordinate vector .

**Step 4**

Since I have already got two normalized eigenvectors as

will be

So this gives

Hence I can say that the values of in terms of the new coordinate vector will be

And these are the answers to the given 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!!

## Leave a Reply