1. ## diagonalizable 2x2 matrix

I need to find all real x values such that:
(x,1
1,0) is diagonalizable. I believe that x can be any real number however I am not sure how to check this.

Anyone have an idea?

2. Did you try looking for eigenvalues first?

3. Originally Posted by hansard
I need to find all real x values such that:
(x,1
1,0) is diagonalizable. I believe that x can be any real number however I am not sure how to check this.

Anyone have an idea?
You need to identify eigenvalues that produce an eigenspace of 2.

The $det=-1$ and $tr=x$; therefore, your characteristic polynomial is $\lambda^2-x\lambda-1$

4. Originally Posted by hansard
I need to find all real x values such that:
(x,1
1,0) is diagonalizable. I believe that x can be any real number however I am not sure how to check this.

Anyone have an idea?
$\sqrt{b^2-4ac}\geq 0$ We know $a=1, c=-1$

$\sqrt{b^2+4}\geq 0$ So we are good in that aspect but are there values of x that will produce eigenvalues of multiplicity 2?

5. Originally Posted by dwsmith
You need to identify eigenvalues that produce an eigenspace of 2.

The $det=-1$ and $tr=x$; therefore, your characteristic polynomial is $\lambda^2-x\lambda-1$

every number that I've tried so far, (eg. 1, -1, 5, -10) produces a quadratic characteristic eqn. that factors into 2 eigenvalues, so the eigenspace is 2.

6. Originally Posted by hansard
every number that I've tried so far, (eg. 1, -1, 5, -10) produces a quadratic characteristic eqn. that factors into 2 eigenvalues, so the eigenspace is 2.
You can run into problems only if $\lambda_1=\lambda_2$ because then there is a chance of having only 1 eigenvector.

7. Originally Posted by dwsmith
$\sqrt{b^2-4ac}\geq 0$ We know $a=1, c=-1$

$\sqrt{b^2+4}\geq 0$ So we are good in that aspect but are there values of x that will produce eigenvalues of multiplicity 2?
I couldn't find any eigenvalues with multiplicity 2. I think x can be any real number but I'm still not sure how to prove it.

8. Originally Posted by hansard
I couldn't find any eigenvalues with multiplicity 2. I think x can be any real number but I'm still not sure how to prove it.
If $x\in\mathbb{R}$, then $\sqrt{b^2-4ac}\geq 0$.

Now show that the quadratic cant be of the form $(\lambda\pm y)^2$ where $y\in\mathbb{R}$.