Show that $A$ is similar to a diagonal matrix iff $b=c=d=e=f=g=0$

Question

Show that $A$ is similar to a diagonal matrix iff $$b=c=d=e=f=g=0$$ $$A= \left(\begin{array}{cccc}a & b & c & d \\ 0 & a & e & f\\ 0 & 0 & a & g\\ 0 & 0 & 0 & a \end{array}\right)$$

Attempt: If $$b=c=d=e=f=g=0$$, then clearly $A$ is similar to a diagonal matrix. $$A = I^{-1}AI$$ Conversely, if $A$ is similar to a diagonal matrix, then $$A=P^{-1}DP$$ where $P$ is non-singular. Suppose $$b,c,d,e,f,g \neq 0$$ We know that the only eigenvalue of $A$ is $a$. To be diagonalizable, we know that $A$ must have 4 eigenvectors. With the assumption of $$b,c,d,e,f,g \neq 0$$ we get that the only eigenvector w.r.t. the eigenvalue $$\lambda=a$$ is $$(1,0,0,0)$$ This contradicts the fact that $A$ is diagonalizable. Hence $$b=c=d=e=f=g=0$$

Is the above logic wrong? Is there any other way I can prove this?

Answer 1

There is an easier way to prove this: Suppose $A$ is similar to a diagonal matrix $D$, i.e. for some invertible $P$, $A=PDP^{-1}$. We know that $A$ and $D$ have the same eigenvalues (which are exactly the diagonal entries of $D$), but the only eigenvalue of $A$ is $a$. Hence $D=aI$ and $A=P(aI)P^{-1}=aPIP^{-1}=aI$.

Answer 2

If $b,c,d,e,f,g$ are not all $0$, then the rank of $A-a I$ is at least 1 (because it is not the $0$ matrix). Therefore the eigenspace for the eigenvalue $a$ has at most dimension three and $A$ is not diagonalizable.

Answer 3

For a simple proof this a Hint:

A scalar matrix is only similar to itself.