Characteristic polynomial in $\mathbb{C}^2$

Question

Consider a finite-dimensional vector space $V$ over a field $\mathbb{F}$.Let $a,b\in\mathbb{F}$. State whether the following is true or not: If $\begin{pmatrix} b & a \\ 0 & b \end{pmatrix}$ is diagonalizable, then $a=0$.

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To my confusion, the solution says that the statement is true, even though I thought I had found a counterexample:

Consider $V=\mathbb{C}^2$. Then the characteristic polynomial of the map given by the above matrix is $(X-b)^2+a=((X-b)+ai)((X-b)-ai)$ which gives us two distinct eigenvalues and hence $a$ can be an arbitrary complex number. Where have I made a mistake?

Answer 1

That matrix has a double eigenvalue, which is $b$. So, if it was diagonalizable, it would be similar to $\left[\begin{smallmatrix}b&0\\0&b\end{smallmatrix}\right]$. But that's a scalar matrix and the only matrix similar to a scalar matrix is that matrix itself. So, if your matrix is diagonalizable, $a=0$.

Answer 2

Let's call your matrix $A$ and suppose $A$ is diagonalizable and $a \neq 0$. Since the characteristic polynomial is $(X-b)^2$ this means that $A$ is similar to the diagonal matrix $$D=\begin{pmatrix} b & 0 \\ 0 & b \end{pmatrix}$$ i.e. there exists an invertible matrix $P$ such that $A=PDP^{-1}$ or equivalently $AP=PD$. Let now $$P=\begin{pmatrix} x & y \\ z & w \end{pmatrix}$$ with $\mathrm{det \ } P=xw-zy \neq 0$. Now $$PA=\begin{pmatrix} x & y \\ z & w \end{pmatrix}\begin{pmatrix} b & a \\ 0 & b \end{pmatrix}=\begin{pmatrix} xb & xa+yb \\ zb & za+wb \end{pmatrix}$$ and $$DP=\begin{pmatrix} b & 0 \\ 0 & b \end{pmatrix}\begin{pmatrix} x & y \\ z & w \end{pmatrix}=\begin{pmatrix} bx & by \\ bz & bw \end{pmatrix}$$ So $PA=DP$ implies that $xa=za=0$. Since $a \neq 0$, this implies $x=z=0$. But then $$P=\begin{pmatrix} 0 & y \\ 0 & w \end{pmatrix}$$ and this is a contradiction, since $P$ needs to be invertible.

Answer 3

Say $A=\begin{pmatrix}b&a\\0&b\end{pmatrix}$, $D$ is diagonal and $D=P^{-1}AP$. Then $A$ and $D$ have the same eigenvalues. So $D$ is a diagonal matrix and the only eigenvalue of $D$ is $b$. The eigenvalues of $D$ are just the entries on the diagonal; hence $D=bI$, so $A=PDP^{-1}=bI$.

Answer 4

$2\times 2$ matrix is diagonalizable iff it has a $2-$dimensional eigenspace.

$b$ is the only eigenvalue (double). The matrix $A-bI$ is \begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix} The eigenspace is $2-$dimensional only if $a=0.$