# Find a matrix that diagonalizes the matrix A

• suspenc3
In summary, the conversation discusses finding a matrix that diagonalizes another matrix A by first finding eigenvalues and then eigenvectors. There is confusion about the number of eigenvectors needed to diagonalize an nxn matrix and the definition of an eigenvector. The conversation concludes with a clarification on the process of finding eigenvectors and solving the characteristic equation.
suspenc3
This could be a dumb question..but here goes:

Im trying to find a matrix that diagonalizes the matrix A. If I find my eigenvalues and then proceed to get my eigenvectors...after row reducing I get 1, 1, 0, 0.(2X2 matrix, two 1's on top).what is the eigenvector?

In this case you have a single eigenvector of value (1,0) and I say you would have made a mistake somewhere in the question since n eigenvectors are required to diagonalise an nxn matrix.

I got two eigenvectors, i just didnt know what to do with:$$\left(\begin{array}{cc}1&1\\0&0\end{array}\right)$$

Apologies, I misread your post. To find your eigenvector you need the following to be true;

$$\left(\begin{array}{cc}1&1\\0&0\end{array}\right)\cdot \left(\begin{array}{c}x\\y\end{array}\right) = 0$$

Do you know why?

P.S. I assume you have obtained the above matrix by applying the Characteristic equation, namely $\det(A-\lambda\cdot I) = 0$

Last edited:
Yes that's how I got it. And I am not sure why they must equal 0.

No one cares that your an english major..have you pointed that out in every post?

suspenc3 said:
Yes that's how I got it. And I am not sure why they must equal 0.
Okay, so you've got your matrix A, solved the characteristic equation and obtained your eigenvalues. You've then applied your eigenvalues to your characteristic equation of A and obtained the following;

for $\lambda =1$

$$( A - (1)I ) = \left(\begin{array}{cc}1&1\\0&0\end{array}\right)$$

Now the definition of an eigenvector (v) of matrix A is a non-zero vector such that $A\underline{v}=\lambda\underline{v}$, where $\lambda$ is your eigenvalue. We can rewrite the above relationship as

$$A\underline{v} - \lambda\underline{v} = 0$$

Using the fact the $I\underline{v} = \underline{v}$ (I being the Identity matrix) we obatin;

$$A\underline{v} - \lambda I \underline{v} = 0$$

$$\therefore \; (A-\lambda I)\underline{v} = 0$$

Does that make sense? In R2 (as in your case);

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

which is where the equation I posted comes from. In fact the characteristic equation follows directly from this relationship.

ohhh, Ok, that makes sense,

Thanks

suspenc3 said:
ohhh, Ok, that makes sense,

Thanks
No problem, so what do you obtain for your eigenvector corresponding to $\lambda=1$?

## What does it mean to "diagonalize" a matrix?

Diagonalizing a matrix means finding a new matrix that has the same eigenvalues as the original matrix, but with all non-diagonal elements equal to zero. This makes solving equations involving the matrix much simpler.

## Why is it important to diagonalize a matrix?

Diagonalizing a matrix can make it easier to perform calculations and solve equations involving the matrix. It also reveals important information about the matrix, such as its eigenvalues and eigenvectors.

## How do you find a matrix that diagonalizes a given matrix?

To find a matrix that diagonalizes a given matrix, you need to find the eigenvectors and eigenvalues of the original matrix. Then, create a new matrix using the eigenvectors as columns and the corresponding eigenvalues on the diagonal. This new matrix will be the diagonalizing matrix.

## Can any matrix be diagonalized?

Not all matrices can be diagonalized. A matrix can only be diagonalized if it has n linearly independent eigenvectors, where n is the dimension of the matrix. If the matrix does not have enough linearly independent eigenvectors, it cannot be diagonalized.

## What are the benefits of diagonalizing a matrix?

Diagonalizing a matrix can make it easier to solve equations and perform calculations involving the matrix. It also allows for a better understanding of the matrix's properties and can reveal important information about its behavior and relationships with other matrices.

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