Diagonalisability with $\lambda = 2,x$

Question

I have a $2 \times 2$ matrix, $Y(x)$, with $2$ eigenvalues: $\lambda = 2,x$ where $x\in \mathbb{R}$

Now $Y(x)$ is only diagonalisable if $\lambda_1,\lambda_2$ are distinct. Are $2,x$ distinct? I think not due to the fact that $x$ can take value $2$, is this intuition correct?

Would $Y(x)$ be diagonalisable $\forall x \ne 2$?

Answer 1

Your statement that $Y(x)$ is only diagonalisable if $\lambda_1,\lambda_2$ are distinct is not correct.

If the eigenvalues are distinct then the matrix is diagonalisable. So, if $x\ne2$ then $Y$ is diagonalisable.

But if there are repeated eigenvalues the matrix may still be diagonalisable. For example, if $$Y=\pmatrix{2&0\cr0&2\cr}$$ then the eigenvalues are $2,2$ and $Y$ is diagonalisable. On the other hand, if say $$Y=\pmatrix{2&1\cr0&2\cr}$$ then the eigenvalues are $2,2$ and $Y$ is not diagonalisable.

Answer 2

Actually, having two different eigenvalues is sufficient but not necessary for diagonalizability of a $2\times 2$ matrix. You need an eigenspace of dimension $2$ as a sufficient condition. So, yes, any $2\times 2$ matrix with two eigenvalues is diagonalizable. To illustrate this point while being lazy, let me ripoff David's example:

$$Y=\pmatrix{2&0\cr0&2\cr}$$

You can see that the eigenspace is the whole of $\mathbb R^2$, since for any vector $(x,y)^T$ we have $$Y(x,y)^T=(2x,2y)=2(x,y) $$

So the eigenspace has dimension 2 (which happens to be the full dimension of the space).