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?
Best Answer
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 $A=P(aI)P^{-1}=aPIP^{-1}=aI$.