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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?

Zero Pancakes
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2 Answers2

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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 $\left( \begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array} \right)$.

Kenny Wong
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  • I'm not sure I understood what you did there – Zero Pancakes Mar 26 '17 at 19:38
  • Do you understand what it means to write a matrix with respect to another basis? – Kenny Wong Mar 26 '17 at 19:38
  • Yes, but shouldn't the new basis be ${e_1,ae_2}$? – Zero Pancakes Mar 26 '17 at 19:39
  • 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. – Kenny Wong Mar 26 '17 at 19:41
  • 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) – Zero Pancakes Mar 26 '17 at 19:48
  • Exactly. Though when you change basis, the new matrix $A'$ is given in terms of your old matrix $A$ by a formula of the kind $A' = PAP^{-1}$, where the columns of $P$ are just the old basis vectors written with respect to the new basis. So the two things are the same. – Kenny Wong Mar 26 '17 at 19:52
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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.

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Marc van Leeuwen
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