Hello,

I want to,

Show that the matrices $\begin{pmatrix} 1&1&1\\1&1&1\\1&1&1 \end{pmatrix}$ and $\begin{pmatrix} 3&0&0\\0&0&0\\0&0&0 \end{pmatrix}$ are similar.

So I find the characteristic polynomial of each transformation and get, $\lambda^2(3-\lambda)$. So each transformation has eigenvalues 0 and 3 with algebraic multiplicity 2 and 1 respectively. So the Jordan normal form can either be $$\begin{pmatrix}0&0&0\\0&0&0\\0&0&3\end{pmatrix}, $$ or $$\begin{pmatrix} 0&1&0\\0&0&0\\0&0&3 \end{pmatrix}. $$ Since I can either have a block of size 2 then a block of size 1 or three blocks of size one.

So now I am stuck. I have a procedure we outlined in class but it is at this point I have the cryptic comments

"Study $(A-\lambda I)$ on the $\lambda$-eigenspace. The nullity tells you the number of zeros on the superdiagonal, and the index of nilpotence tells you the largest block size."

And I do not know how to decode this.

For $\begin{pmatrix} 1&1&1\\1&1&1\\1&1&1 \end{pmatrix}$, $(A-\lambda I)=\begin{pmatrix} 1-\lambda&1&1\\1&1-\lambda&1\\1&1&1-\lambda \end{pmatrix}$ which does not appear to be nilpotent. So my procedure seems too vague to follow.

I have some other text which tells me to use the minimal polynomial. I have a condition to find such a polynomial: if $m_A$ is the minimal polynomial then $m_A(A)=0$ and $m_A$ must divide the characteristic polynomial of $A$. So I know that in this case I have two options, either the minimal polynomial is $(3-\lambda)\lambda$ or $(3-\lambda)\lambda^2$. Is the rule then that the multiplicity of a root of the minimal polynomial is the geometric multiplicity of the eigenvalue? (Where geometric multiplicity is I think the dimension of the $\lambda$-eigenspace associated with a specific root of said minimal polynomial)

I should be able to show that both matrices have the same minimal polynomial. I used a calculator to find $(3-A)(A)$ and $(3-A)(A)^2$ for $A$ is equal to the first and then the second matrix and both were the zero transformation in the $(3-A)(A)$ case. So then once I find some justification I should be good.