# Find a matrix that diagonalizes the matrix A

1. Dec 5, 2006

### 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?

2. Dec 5, 2006

### Hootenanny

Staff Emeritus
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.

3. Dec 5, 2006

### suspenc3

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

4. Dec 5, 2006

### Hootenanny

Staff Emeritus

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

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: Dec 5, 2006
5. Dec 5, 2006

### suspenc3

Yes thats how I got it. And Im not sure why they must equal 0.

6. Dec 5, 2006

### suspenc3

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

7. Dec 6, 2006

### Hootenanny

Staff Emeritus
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.

8. Dec 6, 2006

### suspenc3

ohhh, Ok, that makes sense,

Thanks

9. Dec 6, 2006

### Hootenanny

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