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.
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 $\displaystyle \lambda^2-x\lambda-1$
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?
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.
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.