Determining values of a parameter for which a matrix is complexly diagonalizable

Question

Given is the $3 \times 3$ matrix $A$ below. I want to know for which values of $c$ the matrix is complex diagonalizable. Based on my understanding, a matrix is complex diagonalizable if and only if there exist 3 independent eigenvectors.

A =$ \begin{bmatrix} i & c & 1 \\ 0 & i & 2 \: i \\ 0 & 0 & 1 \end{bmatrix} $

The first step seems to me to be determining eigenvalues, and when I do that I come up with: $\lambda_1 = i \: \land \: \lambda_2 = 1$. When I then determine the general solution sets for the eigenvectors I get:

$\{ x_3 \cdot\begin{pmatrix} \frac{-2c+1+i}{2} \\ -1+i \\ 1 \end{pmatrix} \}$ for $\lambda_2$

and

$\{ x_1 \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \}$ for $\lambda_1$

I further do not see how this matrix for a single value for $c$ would not be complex diagonalizable? The solution manual says this is only true for $c=0$.

Answer 1

For the matrix to be diagonalizable you need $3$ linearly independent eigenvectors. Since $i$ is the only eigenvalue with multiplicity greater than $1$ (equal to $2$) it's enough to find $2$ independent eigenvectors for $i$. Now let $X = [x, y, z]'$ then your eigenvectors are solutions to equations $(A - i) X = 0$ which is set of equations $cy + z = 0, 2iz = 0, (1-i)z = 0$

This gives $z = 0$ and if $c$ was nonzero it would imply $y = 0$ leaving you with only one eigenvector namely $[x, 0, 0]'$ for arbitrary $x$.

Answer 2

It is clear that $1$ is a single root of the characteristic polynomial of $A$, whereas $i$ is a double root. So, $A$ is diagonalizable if and only if there are two linearly independent eigenvectors corresponding to the eigenvalue $i$. If $(x,y,z)\in\Bbb C^3\setminus\{(0,0,0)\}$, then $(x,y,z)$ is an eigenvectors corresponding to the eigenvalue $i$ if and only if $A.(x,y,z)=i(x,y,z)$. But\begin{align}A.(x,y,z)=i(x,y,z)&\iff\left\{\begin{array}{l}ix+cy+z=ix\\iy+2iz=iy\\z=iz\end{array}\right.\\&\iff\left\{\begin{array}{l}cy+z=x\\2iz=0\\(1-i)z=0.\end{array}\right.\end{align}It follows from the second and the third equation that $z=0$. Now, the first equation becomes $cy=0$, which is equivalente to $y=0$ if and only if $c\ne0$. So, if $c\ne0$, both $(1,0,0)$ and $(0,1,0)$ are eigenvectors corresponding to the eigenvalue $i$. But if $c\ne0$, every eigenvectors corresponding to the eigenvalue $i$ are multiples of $(1,0,0)$.