"Determine the values for a for which the matrix A is diagonizable. "
$$ A = \begin{bmatrix}1&1\\a&1\end{bmatrix}$$
My first attempt solving this problem was to find the characteristic polynomial equation for lamda. Which got me to.. :
$$ λ = 1 ±\sqrt{(1)^2 +a}$$
Now my solution sheet states that the correct answer is for values $a > 0$
I have hard time understanding why that is that way. How do they know that it's exactly over $0$ There are obviously infinitely many numbers$$0<a$$ that would render two real eigenvalues. Would that not make it possible for the matrix to be diagonizable for those values? Why does it have to be $a>0$?
So what I was thinking was the answer was $a>-1$ Because if $a=1$ then there would only be one eigenvalue, which is $ λ=0$ and hence the matrix would be not diagonizable, because the eigenbasis that would be obtained from that particular eigenvalue would not span the space.
The characteristic polynomial of your matrix is$$x^2-2x+1-a=(x-1)^2-a,$$which has real roots if and only if $a>0$; these roots are $1\pm\sqrt a$. So, in order that your matrix is diagonalizable over $\mathbb R$, it is necessary that $a>0$ indeed. Can you prove that it is also sufficient?