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Let $A \in M_n$ and $B,C \in M_m$. Prove that if

$$H= \begin{bmatrix} A&0 \\ 0 & B \end{bmatrix}$$

is similar to

$$K = \begin{bmatrix} A&0 \\ 0 & C \end{bmatrix}$$

then $B$ is similar to $C$.

I am not sure how I would do this I know that if $H$ is similar to $K$ then for some non-singular matrix $S$ then $S^{-1} H S=K$.

Rodrigo de Azevedo
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Fernando Martinez
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  • I'd write $A,B,C$ and $S$ explicitly. A proper submatrix of $S$ to show that $B$ and $C$ are similar should become apparent. – Git Gud Nov 13 '16 at 18:59

2 Answers2

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Using Jordan form: there exists matrices $P,Q,R$ such that $P^{-1}AP = J_A$, $Q^{-1}BQ = J_B$, and $RCR^{-1} = J_C$ are all in Jordan normal form.

We then note that $$ \pmatrix{P\\&Q}^{-1}\pmatrix{A\\&B} \pmatrix{P\\&Q} = \pmatrix{J_A\\& J_B}\\ \pmatrix{P\\&R}^{-1}\pmatrix{A\\&C} \pmatrix{P\\&R} = \pmatrix{J_A\\& J_C} $$ By the uniqueness of Jordan form (up to permutations of blocks), the two matrices on the right can only be similar if $J_B$ is similar to $J_C$, which is to say that $B$ is similar to $C$ as desired.

Eugene Zhang
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Ben Grossmann
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Since $H$ and $K$ are similar, $B$ and $C$ have same set of distinct eigenvalues. Also $B$ and $C$ have same generalized eigenspaces (same number of independent vectors) for distinct eigenvalues, i.e. $$ \operatorname{Rank}(\lambda_iI-B)^n=\operatorname{Rank}(\lambda_iI-C)^n $$ where $\lambda_i$ are distinct eigenvalues of $B$ and $C$. Hence $B$ and $C$ are similar.

Ben Grossmann
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Eugene Zhang
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  • is the minimum polynomial the same thing as the charecteritstic polynoial. – Fernando Martinez Nov 13 '16 at 19:20
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    It is not true that matrices with the same minimal polynomial are necessarily similar – Ben Grossmann Nov 13 '16 at 19:39
  • That true it only say if they are similar they have the minimal polynomial but how would I solve this – Fernando Martinez Nov 13 '16 at 19:41
  • Instead of minimal polynomials, we could use either Jordan form or Frobenius normal form. Do you know what either of those are? – Ben Grossmann Nov 13 '16 at 19:43
  • Why was this accepted? – copper.hat Nov 13 '16 at 19:44
  • Actually, as the comment below your question indicates, theres another approach – Ben Grossmann Nov 13 '16 at 19:46
  • I know the jordan form – Fernando Martinez Nov 13 '16 at 19:48
  • @User1006 it's still wrong. Matrices can have the same minimal polynomials, the same eigenvalues up to multiplicity, the same eigenspaces, and could still fail to be similar. I don't understand why you decided to use anything other than Jordan form in this context. – Ben Grossmann Nov 13 '16 at 19:55
  • @User1006 That isn't sufficient. There's is no way to answer this question using eigenvalues/eigenspaces alone, even if you also use the minimal polynomial. – Ben Grossmann Nov 13 '16 at 19:57
  • So maybe I am supposed to use jordan form because I covered it in class maybe but I am not sure how to apply it here – Fernando Martinez Nov 13 '16 at 19:59
  • @User1006 That's an interesting conjecture you have. I'll post a question about it. Feel free to try to prove it, and I'll try to construct a counterexample. In the mean time, I am fairly confident that you are wrong. – Ben Grossmann Nov 13 '16 at 20:06
  • @User1006 here's the question – Ben Grossmann Nov 13 '16 at 20:12
  • @User1006 I've posted the question and my answer in the link above, and so I am quite confident that you are wrong. I encourage you to find the flaw in my argument, if there is one. – Ben Grossmann Nov 13 '16 at 20:22
  • @Omnomnomnom, it should be $Rank(\lambda_iI-B)^n=Rank(\lambda_iI-C)^n$. So both have same number of independent eigenvectors – Eugene Zhang Nov 13 '16 at 21:19
  • @User1006 I agree that with the $n$ there (for arbitrary integers $n$), having that is enough; it is well known that these values can be used to "encode" the Jordan form. I don't know what you're saying about "independent eigenvectors", though. – Ben Grossmann Nov 13 '16 at 21:42
  • (The sequence of these values is known as the "Weyr characteristic") – Ben Grossmann Nov 13 '16 at 21:45
  • "$B$ and $C$ have same number of independent vectors for eigenspace of distinct eigenvalues" is not a correct interpretation of the statement $$ Rank(\lambda_iI-B)^n=Rank(\lambda_iI-C)^n \quad n=1,2,3,\dots $$ In particular, we would need to say something about generalized eigenspaces in order to capture what's going on. – Ben Grossmann Nov 14 '16 at 12:27
  • Also, now that you've effectively proven that $B$ and $C$ have the same Jordan form, your earlier statement about minimal polynomials is redundant. – Ben Grossmann Nov 15 '16 at 17:02