I have a question from a test I solved (without that question.. =)
"If a matrix $A$ s.t $A$ is in $M(\mathbb{C})$ have only $1$ eigenvalue than $A$ is a diagonalizable matrix"
That is a false assumption since a ($n\times n$ matrix) a square matrix needs to have at least $n$ different eigenvalues (to make eigenvectors from) – but doesn't the identity matrix have only $1$ eigenvalue?…
Best Answer
No, they need not be different, as you yourself have noticed (the identity matrix example) and others have also pointed it out.
You can write each matrix in the Jordan normal form:
$$A = S J S^{-1}, \quad J = \operatorname{diag}(J_1, J_2, \dots, J_m),$$
where
$$J_k = \begin{bmatrix} \lambda_k & 1 \\ & \ddots & \ddots \\ & & \lambda_k & 1 \\ & & & \lambda_k \end{bmatrix}.$$
A matrix $A$ is diagonalizable if and only if all $J_k$ are of order $1$, i.e., $J_k = \begin{bmatrix} \lambda_k \end{bmatrix}$. In other words, $m = n$ and
$$J = \operatorname{diag}(\lambda_1, \lambda_2, \dots, \lambda_n).$$
Now, if $A$ has only one eigenvalue, that means that $\lambda := \lambda_1 = \lambda_2 = \cdots = \lambda_n$, so
$$J = \operatorname{diag}(\lambda, \lambda, \dots, \lambda) = \lambda \operatorname{diag}(1, 1, \dots, 1) = \lambda {\rm I}.$$
Now, let us get back to $A$:
$$A = S J S^{-1} = S (\lambda {\rm I}) S^{-1} = \lambda S S^{-1} = \lambda {\rm I}.$$
So, $A$ is diagonalizable with only one eigenvalue if and only if it is a scalar matrix.
However, not every matrix with only one eigenvalue is diagonalizable. Just put $\lambda := \lambda_1 = \lambda_2 = \cdots = \lambda_m$, but let some of $J_k$ be of order strictly bigger than $1$. For example,
$$B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$
is not diagonalizable.