# Are these $4$ by $4$ matrices similar?

Are the matrices $$A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\0 & 0 & 0 & 1 \end{pmatrix}, B =\begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\0 & 0 & 0 & 1 \end{pmatrix}$$ similar over $\mathbb{R}$? Why or why not?

Since these two matrices have same characteristic polynomials and same eigenvalues, I think they are similar, but I am not sure about it. Could you help me to solve this problem? Thank you!

• Check their minimal polynomial – user251257 Mar 29 '16 at 4:33
• @user251257 for A I got $(x-1)^4$ and for B, $(x-1)^2$. So they are having different minimal polynomials. Does it mean they are not similar? – Nhay Mar 29 '16 at 5:29
• yes ${}{}{}{}{}{}{}$ – user251257 Mar 29 '16 at 7:40

The geometric multiplicity $\DeclareMathOperator{gmult}{gmult}\gmult_X(\lambda)$ of an eigenvalue $\lambda$ of a matrix $X$ is the smallest $n$ such that $(X-\lambda I)^n=0$.
Fact. If $Y$ is similar to $X$, then $\gmult_X(\lambda)=\gmult_Y(\lambda)$.
Now, note that $\lambda=1$ is an eigenvalue of both $A$ and $B$. The computations \begin{align*} (A-I)^1 &= \left[\begin{array}{rrrr} 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] & (B-I)^1 &= \left[\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \\ (A-I)^2&= \left[\begin{array}{rrrr} 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] & (B-I)^2&= \left[\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \\ (A-I)^3 &= \left[\begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \\ (A-I)^4 &= \left[\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{align*} show that $\gmult_A(1)=4$ while $\gmult_B(1)=2$. This proves that $A$ and $B$ are not similar.
• I got the ranks of them are 3 and 2. But I don't know how it is related to the $A= PBP^-1$ – Nhay Mar 29 '16 at 5:30
• Because $A - I = PBP^{-1} - PP^{-1} = P(B - I)P^{-1}$ and $P$ is invertible, we see that $A - I$ and $B - I$ must have the same rank. (Giving a contradiction). More conceptually, our rank calculationn showed there were two eigenvectors of B with eigenvalue 1 and only one eigenvector of A with eigenvalue 1. – Frederic Koehler Apr 5 '16 at 22:50