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

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?

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Does $b,c,d,e,f,g\neq 0$ mean that at least one of them is not $0$? How do you go from that to 'we get that the only eigenvector w.r.t. the eigenvalue $a$ is $(1,0,0,0)$?' –  Git Gud Aug 25 '13 at 19:45
Yes, the above logic is wrong: '$b=c=d=e=f=g=0$ is false' is not equivalent to '$b,c,d,e,f,g\neq 0$', but it is equivalent to 'one of $b,c,d,e,f,g$ is $\neq 0$'. –  walcher Aug 25 '13 at 19:47
it should be a simple matter if you use Jordan form to show that there is a generalized eigenvector if any of the off diagonal is nonzero. Then it cannot be similar to diagonal because it has a Jordan block of size larger than 1. –  Evan Aug 25 '13 at 19:49

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

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Ah yes of course , this is very straightforward. –  Evan Aug 25 '13 at 22:11

For a simple proof this a Hint:

A scalar matrix is only similar to itself.

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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.

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This is good, but for completeness I think one should say that $a$ is the only possible eigenvalue (this is easy) –  Deven Ware Aug 25 '13 at 19:55