Suppose $A=\begin{pmatrix}λ&a\\0&λ\end{pmatrix},B=\begin{pmatrix}λ&b\\0&λ\end{pmatrix} \in \Bbb C^{2\times 2}$ with $m_A(x)=m_B(x)=(x-λ)^2$ $(1)$.

I'm asked to prove that $A,B$ are similar.

From $(1)$ we get that $ab\ne 0 \implies$ both $A$ and $B$ are not diagonalizable. Also, we can see that they have the same corresponding eigenvectors. How do I continue from this part?

  • 2
    $\begingroup$ What about just finding an invertible matrix $M$ such that $MA = BM$? $\endgroup$ – Michael Biro Mar 26 '17 at 19:20
  • $\begingroup$ @MichaelBiro Yes, what matrix do you suggest? $\endgroup$ – ZeroPancakes Mar 26 '17 at 19:35
  • $\begingroup$ For a first guess, I would try a matrix of the form $\begin{bmatrix} c & d \\0 & e\end{bmatrix}$ $\endgroup$ – Michael Biro Mar 26 '17 at 19:49

Let $\{ e_1, e_2 \}$ denote the basis vectors with respect to which these matrices are written.

Try to show that, if we define a new basis, $\{ f_1 = a e_1, f_2 = e_2\}$, then in this new basis, the matrix for $A$ is $$ A = \left( \begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array}\right).$$

Can you find another basis in which the matrix for $B$ takes the form $$ B = \left( \begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array}\right)?$$

If so, then you've succeeded in showing that $A$ and $B$ are similar!

Finally, for $\{ f_1 = a e_1, f_2 = e_2 \}$ to actually be a basis, we need $f_1$ and $f_2$ to be linearly independent, i.e. we need $a \neq 0$. Otherwise none of this will work!

So you need to use the fact that the minimum polynomial is $(x - \lambda)^2$ (as opposed to $x - \lambda$) to show that $a$ cannot be zero - but you appear to have already done that!

By the way, ANY two-by-two matrix with minimal polynomial $(x- \lambda)^2$ has Jordan canonical form $\left( \begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array} \right)$.

  • $\begingroup$ I'm not sure I understood what you did there $\endgroup$ – ZeroPancakes Mar 26 '17 at 19:38
  • $\begingroup$ Do you understand what it means to write a matrix with respect to another basis? $\endgroup$ – Kenny Wong Mar 26 '17 at 19:38
  • $\begingroup$ Yes, but shouldn't the new basis be $\{e_1,ae_2\}$? $\endgroup$ – ZeroPancakes Mar 26 '17 at 19:39
  • $\begingroup$ No, for example, $A(f_1) = A(ae_1) = ae_1 = f_1 $ and $A(f_2) = A(e_2) = a e_1 + e_2 = f_1 + f_2$. So the entries in the matrix are correct. $\endgroup$ – Kenny Wong Mar 26 '17 at 19:41
  • $\begingroup$ Okay, so to show that two matrices are similar instead of showing that there exists an invertible matrix $P$ such that $B=PAP^{-1}$. we can show that there are exist two bases such that $(A; \hat a_1)=(B; \hat a_2)$ (if there exists such notation for matrices) $\endgroup$ – ZeroPancakes Mar 26 '17 at 19:48

The minimal polynomials of $A,B$ are of degree$~2$, so they are both equal to the respective characteristic polynomials (both are $(X-\lambda)^2$). It is then a general fact that there exists a cyclic vector$~v$, one such that $v,Av,\ldots,A^{n-1}v$ (respectively $v,Bv,\ldots,B^{n-1}v$) are linearly independent, and therefore form a basis of you space$~\Bbb C^n$ (here of course with $n=2$). After base change to those bases, the matrices become companion matrices for the minimal (or characteristic) polynomial, here $\left(\begin{smallmatrix}0&-\lambda^2\\1&2\lambda\end{smallmatrix}\right)$. In particular $A$ and $B$ a similar to each other (and similar to that companion matrix).

Concretely here any vector$~v$ not in the ($1$-dimensional) eigenspace for$~\lambda$ is a cyclic vector. So you can start with taking the same such vector for $A$ and for $B$ (the second standard basis vector will do fine). In the above approach the bases will differ by choosing the second vector differently (namely $Av$ versus $Bv$). But one could alternatively complete with $(A-\lambda I)v$ respectively $(B-\lambda I)v$, again different though this time both do lie in the eigenspace for$~\lambda$, and if you then swap the order of the vectors, you get a Jordan block $\left(\begin{smallmatrix}\lambda&1\\0&\lambda\end{smallmatrix}\right)$ in both cases after the change of basis.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.