Find a matrix that diagonalizes the matrix A

[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?

 

[Hootenanny]

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.

 

[suspenc3]

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

 

[Hootenanny]

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 $\det(A-\lambda\cdot I) = 0$

 

[suspenc3]

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

 

[suspenc3]

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

 

[Hootenanny]

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.

 

[suspenc3]

ohhh, Ok, that makes sense,

Thanks

 

[Hootenanny]

ohhh, Ok, that makes sense,

Thanks

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