# diagonalizable 2x2 matrix

#### 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?

#### roninpro

Did you try looking for eigenvalues first?

#### dwsmith

MHF Hall of Honor
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 $$\displaystyle det=-1$$ and $$\displaystyle tr=x$$; therefore, your characteristic polynomial is $$\displaystyle \lambda^2-x\lambda-1$$

#### dwsmith

MHF Hall of Honor
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?
$$\displaystyle \sqrt{b^2-4ac}\geq 0$$ We know $$\displaystyle a=1, c=-1$$

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

#### hansard

You need to identify eigenvalues that produce an eigenspace of 2.

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

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.

#### dwsmith

MHF Hall of Honor
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 $$\displaystyle \lambda_1=\lambda_2$$ because then there is a chance of having only 1 eigenvector.

#### hansard

$$\displaystyle \sqrt{b^2-4ac}\geq 0$$ We know $$\displaystyle a=1, c=-1$$

$$\displaystyle \sqrt{b^2+4}\geq 0$$ 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.

#### dwsmith

MHF Hall of Honor
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 $$\displaystyle x\in\mathbb{R}$$, then $$\displaystyle \sqrt{b^2-4ac}\geq 0$$.

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

Last edited:
hansard