diagonalizable 2x2 matrix

May 2010
5
0
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?
 
Nov 2009
485
184
Did you try looking for eigenvalues first?
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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
Mar 2010
3,093
582
Florida
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?
 
May 2010
5
0
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
Mar 2010
3,093
582
Florida
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.
 
May 2010
5
0
\(\displaystyle \sqrt{b^2-4ac}\geq 0\) 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
Mar 2010
3,093
582
Florida
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 \(\displaystyle y\in\mathbb{R}\).
 
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