Diagonalization problems (eigenvalues and vectors)

Question

I am trying to diagonalize the following matrices:

$$A = \begin{pmatrix}0 & 1\\-1 & 2\end{pmatrix}\qquad B = \begin{pmatrix}1 & 2\\-1&-1\end{pmatrix}$$

For matrix $A$, I find an eigenvalue of $1$ with algebraic multiplicity of $2$. I find, though, that the dimension of its eigenspace consists of $1$ vector? Can I still construct a matrix to diagonalize with somehow?

For matrix B, I get the same scenario for eigenvalues (value of $1$ repeated twice), but am unsure if my eigenspace, consisting of $(1,0)$, is correct? Then again, I am left in the same situation.

Are those matrices un-diagonalizable? (spelling?) Can someone help me out?

Thank you.

Answer 1

Hints:

The first is not diagonalizable, but we can use the Jordan Normal Form. We get:

$$A = PJP^{-1} =\left( \begin{array}{cc} 1 & -1 \\ 1 & 0 \\ \end{array} \right) \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ -1 & 1 \\ \end{array} \right)$$

The second is diagonalizable and we get:

$$A = PJP^{-1} = \left( \begin{array}{cc} i-1 & -i-1 \\ 1 & 1 \\ \end{array} \right) \left( \begin{array}{cc} -i & 0 \\ 0 & i \\ \end{array} \right) \left( \begin{array}{cc} \frac{1}{2 i} & \frac{i+1}{2 i} \\ -\frac{1}{2 i} & \frac{i-1}{2 i} \\ \end{array} \right)$$

Answer 2

If you have an eigenvalue with multiplicity n, then you must have n linearly independent eigenvectors from that eigenvalue in order for the matrix to be diagonalizable.