Find a normal matrix with a given characteristic polynomial Find a normal matrix with characteristic polynomial $t^2 + 4$ and eigenspace
$E_{2i} = span  {(1 \ 3i)^t}$
Since any vector in different eignespaces are perpendicular to each other, so i compute $E_{-2i} = span  {(-3 \ i)^t}$. Then I conclude that the matrix is \begin{matrix}
        1 & 3i  \\
        -3 & i\\
        \end{matrix}
Then when I try to calculate back the char. polynomial, I didn't get back the same ans.
What's wrong with my ans?
Thank you!
 A: You have just put the eigenvectors into a matrix (in rows). However, your matrix $A$ is really defined by $A=PDP^{-1}$, with
$$D=\left(\begin{matrix}2i & 0 \\ 0 &-2i\end{matrix}\right)$$
And
$$P=\left(\begin{matrix}1 & -3 \\ 3i & i\end{matrix}\right)$$
Then you find
$$A=-\frac{2}{5}\left(\begin{matrix}4i & -3 \\ 3 & -4i\end{matrix}\right)$$
Then $AA^*=A^*A=\left(\begin{matrix}4 & 0 \\ 0 &4\end{matrix}\right)$ so $A$ is normal, and by construction it has the stated eigenvalues.

The spectral theorem tells us a normal matrix is diagonalizable by a unitary matrix, hence you can normalize the colums of $P$ to make it unitary:
$$P=\frac{1}{\sqrt{10}}\left(\begin{matrix}1 & -3 \\ 3i & i\end{matrix}\right)$$
Then $PP^*=I$.
Note also that your statement that eigenspaces are orthogonal to each other is wrong in general. Here it's true because $A$ is supposed to be normal.
The converse is easy to check: when $A$ is diagonalizable by a unitary matrix, that is $A=PDP^*$, then you get $AA^*=(PDP^*)(PD^*P^*)=PDD^*P^*=PD^*DP^*=(PD^*P^*)(PDP^*)=A^*A$.
