Solution to this problem is incorrect about a diagonalizable matrix A?

Question

I have this matrix A

$$ \begin{bmatrix} 1 & c & 0 & 0 & 0 \\ 0 & 1 & c & 0 & 0 \\ 0 & 0 & 1 & a & 0\\ 0 & 0 & 0 & b & c\\ 0 & 0 & 0 & 0 & b \end{bmatrix} $$

(d) (4 points) Find all values of a, b, c, if any, such that A is diagonalizable but not orthogonally diagonalizable. Justify your answer.

Solution: By spectral theorem, we know that A is orthogonally diagonalizable iff A is symmetric. So we know that at least one of a or c must be nonzero so that A is not symmetric. If b !=1, we know by part (c) that c=0 for the matrix to be diagonalizable. Therefore, a != 0. If b = 1, then the matrix is only diagonalizable iff a = c = 0 which is a contradiction as then A will be symmetric. Therefore, the requirements are the b != 1, c = 0, and a != 0.

However, b must be 0. For example, you cannot have b = 2. If b = 2, then the dimension of the kernel of this matrix is only $1$, so the eigenvalue $1$ has $gemu(1) < almu(1)$, so this matrix would not be diagonalizable. Is there anything wrong with my reasoning?

Answer 1

Why do you think that $b$ must be $0$? That's not true. Take, for instance this case:$$A=\begin{pmatrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2\end{pmatrix}.$$It is diagonalizable: for instance, the vectors $(1,0,0,0,0)$, $(0,1,0,0,0)$, and $(0,0,1,0,0)$ are eigenvalues with eigenvalue $1$ and $(0,0,1,1,0)$ and $(0,0,0,0,1)$ are eigenvectors with eigenvalue $2$.

Answer 2

With the given constraints on $a$ and $c$, $\det A=b^2$, so its kernel is always trivial for $b\ne0$. If instead by the first “this matrix” you mean $$A-bI = \begin{bmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&a&0\\0&0&0&0&0\\0&0&0&0&0\end{bmatrix},$$ its null space is quite clearly two-dimensional regardless of the value of $a$.