12

First assume that $A$ and $B$ are $p \times p$ matrices and that $\lambda_1,\ldots , \lambda_p$ are distinct eigenvalues of $A$ and $B$. I want to show that $A$ and $B$ are similar.

Here is my approach: The goal is to show that there is a nonsingular $p \times p$ matrix $P$ such that $B = P^{-1}A P$. We know the following is satisfied for each eigenvalue $$ A \textbf x_j = \lambda_j \textbf x _j \text { where } j = 1,\ldots,p$$ and $\textbf x_j$ are the eigenvectors. Since $A$ and $B$ share the same eigenvectors, $B$ must satisfy the following $$ B \textbf x_j = \lambda_j \textbf x _j \text { where } j = 1,\ldots,p$$ Thus we have $$ B \textbf x_j = A \textbf x_j$$ We can take the $\textbf x_j$ as the columns of the matrix $P$ and get $$BP=AP$$ we know that the columns of $P $ are linearly independent so the matrix must be non singular. Unfortunately this means that $$A=B$$ but this is not what we wanted to show. Am I on the right track? Is there an obvious mistake I am making?

Martin Argerami
  • 205,756
rioneye
  • 953
  • 2
    We know that $A$ and $B$ share the same eigenvalues, but not necessarily the same eigenvectors – tylerc0816 Nov 07 '13 at 23:28
  • 1
    it would be if you can change the eigenvalues in the title to distinct eigenvalues to better reflect your question. – abel Feb 24 '15 at 12:58

4 Answers4

17

The idea in your approach almost works. But you cannot assume that $A$ and $B$ have the same eigenvectors! So you will have $$ Ax_j=\lambda_jx_j,\ \ \ By_j=\lambda_jy_j. $$ As $\lambda_1,\ldots,\lambda_p$ are distinct, $x_1,\ldots,x_p$ and $y_1,\ldots,y_p$ are each linearly independent, so they are bases. Let $P$ be the change of basis $y\to x$, i.e. $P$ is the matrix such that $Py_j=x_j$, $j=1,\ldots,n$. It is clearly nonsingular: if $P(c_1y_1+\cdots+c_py_p)=0$, then $$ 0=c_1P(y_1)+\cdots+c_pP(y_p)=c_1x_1+\cdots+c_px_p, $$ so $c_1=\cdots=c_p=0$. Now, for each $j$, $$ PBy_j=\lambda_j Py_j=\lambda_jx_j=Ax_j=APy_j. $$ As the $y$ are a basis, we get that $PB=AP$, i.e. $PBP^{-1}=A$.

Martin Argerami
  • 205,756
12

Another way to proceed is to note that there is an invertible $p \times p$ matrix $S$ such that $S^{-1}AS = {\rm diag}(\lambda_{1},\ldots,\lambda_{p})$ and there is an invertible $p \times p$ matrix $T$ such that $T^{-1}BT = {\rm diag}(\lambda_{1},\ldots,\lambda_{p}).$ Then $TS^{-1}AST^{-1} = B,$ and we can take $P = (ST^{-1}).$

Geoff Robinson
  • 24,035
4

Be careful! If A and B are $n\times n$ matrices having the same p distinct eigenvalues where $p<n$ this argument doesn't work. Because in this case the base change matrix P would be a $p\times p$ matrix and we cannot multiply an $n \times n$ matrix with a $p\times p$ matrix.

amWhy
  • 209,954
hande kul
  • 41
-1

Two matrices with same eigenvalues are similar if they are non-singular! If the matrices are singular they may or may not be similar!

eg: $$M_1= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}$$ $$ M_2= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

here $M_1$ & $M_2$ have the same set of eigenvalues! but they are not similar.

SANGEETH S J
  • 1
  • This is not correct mathematically. Suppose we take $M_1,M_2$, which have the same eigenvalues but are not similar (as you say), then add a positive multiple of the identity to make them both invertible, e.g. $M_1+I,M_2+I$ are both invertible and have the same "non-similarity" as your example. – hardmath Jun 15 '17 at 05:00
  • Note that the Question postulates that $A,B$ are $p\times p$ matrices which share the same $p$ distinct eigenvalues. This is sufficient to guarantee $A,B$ are similar (since both are diagonalizable to the same diagonal matrix $D$). – hardmath Jun 15 '17 at 12:58
  • @hardmath The "counter-example" provided in this answer seems to use the fact the zeroes are the only eigenvalues of a nilpotent matrix. And that two nilpotent matrices are similar iff they have same order. In this case, these two nilpotent matrices have different order and hence, not similar. However, I think this doesn't dispute the claim in question because that clearly says "distinct eigenvalues". – Nothing special Feb 18 '24 at 07:28
  • @Nothingspecial "two nilpotent matrices are similar iff they have same order." Only with the order of nilpotentcy being the order of the matrix (or more trivially, nilpotency order $1$). – hardmath Feb 20 '24 at 11:31
  • @hardmath I am sorry. I was ignorant when I made that comment. However, it's true that "if two nilpotent matrices are similar, they must have same order." See nilpotent matrices up to similarity. All nilpotent matrices have the same characteristic equation $x^n=0$ and eigenvalues all zeroes yet they are not similar based on the block partition of their Jordan normal form. – Nothing special Feb 20 '24 at 14:52
  • 1
    @hardmath what you have stated is also true for $n-1$ i.e., All $n\times n$ nilpotent matrices with order of nilpotence $n-1$ are similar. As you see there's only one partition with $n-1$ on the lead, it's $n=(n-1)+1$. I would say your "only" is now complete (we also have this besides the trivial and order $n$ case). – Nothing special Feb 20 '24 at 15:00
  • Anyways, my initial reaction was too fast. $M_1$ and $M_2$ are not nilpotent. They do share the same characteristic equation and hence, the same eigenvalues i.e., $0$, $0$, $1$. As the answerer noted, $M_1$ and $M_2$ are not similar. However, this does not dispute the claim in the original post; The criteria of distinct eigenvalues is not met. $0$ is a repeated eigenvalue of both $M_1$ and $M_2$. – Nothing special Feb 20 '24 at 17:33