Q:

Determine the values for a for which the matrix is diagonizable.

"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.

A:

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?

