Matrix Diagonalizable Question

[lubricarret]

Homework Statement

1. Let A =
[-8 k
0 -8]
Then A is diagonalizable exactly for the following values of k

2. Let B =
[-8 k
0 1]
Then B is diagonalizable exactly for the following values of k

Homework Equations

-Equations for eigenvalues, eigenvectors... and D=PA(P^-1)
-A matrix is invertible if and only if its determinant does not equal 0
-If matrix P created from the eigenvectors [X_1, X_2] is invertible, then matrix A is diagonalizable.

The Attempt at a Solution

I seem to only get half of this answer correct, and am not sure why. I tried solving for the eigenvectors and then checking if the matrix P (from the eigenvectors) has a determinant = to zero.

For the first part, I got the eigenvalue -8 with an algebraic multiplicity of 2; then when solving for the eigenvectors, I get:
[0 -k
0 0]
and the eigenvector t[-k,0] and the matrix P =
[-k -k
0 0]

Therefore, I say that the matrix A can never be diagonalizable for any value of k, as the number of parameters will always be less than the algebraic multiplicity, and the determinant will always be = 0, regardless of the k value.

For the second part, I get eigenvalues -8 and 1.

I then get the eigenvectors t[1,0] and t[(-k/9),1], and the matrix P =
[1 (-k/9)
0 1]

Therefore, I said this matrix will be diagonalizable for all values of k, since the determinant will always equal 1, and is therefore always diagonalizable, regardless of the k value.


Any ideas on what I am doing wrong; and if my logic is messed up?

Thanks!

 

[Dick]

You aren't that messed up. The second one is diagonalizable for any value of k. It has two distinct eigenvalues, hence two distinct eigenvectors. I don't think [-k/9,1] is an eigenvector though. A sign doesn't seem right. You are only a little bit wrong on the first one. If k=0 it's is diagonalizable. In fact, it's already diagonal.

 

[lubricarret]

Hi Dick,

Thanks a lot for the help!

The question was only asking for which k values, so your explanation of k=0 for the first one solved my mistake. But, if you could just clear something up for me that I don't really understand:

As you mention the first one is diagonalizable if k=0, to make it basically the identity matrix. But, isn't there a theorem that says "A matrix A is diagonalizable if and only if the multiplicity of every eigenvalue lambda of A equals the number of basic eigenvectors corresponding to lambda , which is the number of parameters in the solution of (lambda*I - A)X = 0." In other words, I got:
[0 -k
0 0]
and the eigenvector t[k,0]
for the first part; so there is only 1 parameter for the eigenvalue with a multiplicity of 2... what am I not understanding here?

Also, that was a silly mistake that I had (-k/9); I just forgot to change the sign, as it as negative within my augmented matrix. Is that the mistake you were referring to?

Thanks again!

 

[HallsofIvy]

Hi Dick,

Thanks a lot for the help!

The question was only asking for which k values, so your explanation of k=0 for the first one solved my mistake. But, if you could just clear something up for me that I don't really understand:

As you mention the first one is diagonalizable if k=0, to make it basically the identity matrix.

No, it makes it -8 times the identity matrix.

But, isn't there a theorem that says "A matrix A is diagonalizable if and only if the multiplicity of every eigenvalue lambda of A equals the number of basic eigenvectors corresponding to lambda , which is the number of parameters in the solution of (lambda*I - A)X = 0." In other words, I got:
[0 -k
0 0]
and the eigenvector t[k,0]
for the first part; so there is only 1 parameter for the eigenvalue with a multiplicity of 2... what am I not understanding here?

How do you arrive at the idea that there is only one parameter? "k" is NOT the parameter meant if that is what you are thinking. With k= 0, "lambda*I- A" is the 0 matrix. The solutions to the equation 0X= 0 are all vectors in R² and that has two parameters: any vector <x, y>. x and y are the two parameters.

Also, that was a silly mistake that I had (-k/9); I just forgot to change the sign, as it as negative within my augmented matrix. Is that the mistake you were referring to?

Thanks again!

 

[Dick]

Hi Dick,

Thanks a lot for the help!

The question was only asking for which k values, so your explanation of k=0 for the first one solved my mistake. But, if you could just clear something up for me that I don't really understand:

As you mention the first one is diagonalizable if k=0, to make it basically the identity matrix. But, isn't there a theorem that says "A matrix A is diagonalizable if and only if the multiplicity of every eigenvalue lambda of A equals the number of basic eigenvectors corresponding to lambda , which is the number of parameters in the solution of (lambda*I - A)X = 0." In other words, I got:
[0 -k
0 0]
and the eigenvector t[k,0]
for the first part; so there is only 1 parameter for the eigenvalue with a multiplicity of 2... what am I not understanding here?

Also, that was a silly mistake that I had (-k/9); I just forgot to change the sign, as it as negative within my augmented matrix. Is that the mistake you were referring to?

Thanks again!

Yeah, that was the silly mistake I was referring to. And, sure, like Halls (and you) said, if k is not zero you only have the eigenvector [1,0]. If k=0 then [0,1] (and every other vector) also becomes an eigenvector. So you don't just have one anymore.

 

[lubricarret]

Ahhh, Okay. I don't know why I didn't see that before... obviously then, there are two parameters for the 0 matrix, which would make it invertible... the parameters there would be t[1,0] and s[0,1] (or as you said any vector <x, y>.
And thanks Dick, I see that when you have a k value other than 0, then you get:
[0 1
0 0]
which would create only one parameter. And
[0 0
0 0]
would create two.


Thanks guys!