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I want to show that these two matrices are similar

$$\begin{pmatrix} 6 & 6 & -15 \\ 1 & 5 & -5 \\ 1 & 2 & -2 \end{pmatrix} \ \ \ \ \ \ \begin{pmatrix} 37 & -20 & -4 \\ 34 & -17 & -4 \\ 119 & -70 & -11 \end{pmatrix}$$

It is from my homework. I have showed that their characteristic polynomial are the same, $f(x)=(x-3)^3$, but I know that it isn't sufficient. How do I proceed?

In class, we not studying yet the Jordan canonical form.

Thank you very much!

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2 Answers 2

up vote 1 down vote accepted

This is what you can do. The possible minimal polynomials are $(x-3)=0$, $(x-3)^2=0$, $(x-3)^3=0$ In the order I wrote them down, you check, for each matrix, if each one of these is satisfied, if it is, you stop, that is your minimal polynomial. (you substitute your matrix where $x$ is)

If the min polynomials are the same, then they must be similiar. Because in 3 dimensions, the possible jordon blocks can only be 1,1,1, 1,2 and 3. and they correspond to the cases the min poly is $(x-3)$, $(x-3)^2$, $(x-3)^3$. this method will fail to work in higher dimensions, generally.

EDIT:

Every factor in the characteristic polynomial must feature in the minimal polynomials.

Suppose your characterstic equation is $(x-a)^3(x-b)^2$, the minimal polynomial must be $(x-a)^u(x-b)^v$ for some $1\leq u\leq 3$ and $1\leq u\leq 2$. In addition, $u$ and $v$ gives the size of the BIGGEST Jordon block, it must have for $a$ and $b$ respectively. The Jordon block sizes should add up to the $n$ for an $n\times n$ matrix.

2 matrices are similiar if and only if they have same Jordon blocks sizes for each different eigenvalue. Normally for higher dimensions, minimal polynomial is insufficient to tell us what Jordon block is. Imagine if the problem is 4-dimensional and your minimal polynomial is $(x-3)^2$, then the Jordon blocks can be $2,2$ or $1,1,2$. However, in the case of 3 dimesions. If the min poly is $(x-3)$, then the block is $1,1,1$, if the min poly is $(x-3)^2$, then since the biggest block must have size 2, you are only left with 1, so the only possible combination of block sizes is 2,1. For min poly $(x-3)^3$, then it must only be 1 Jordon block of size 3. Hence, min poly uniquely determines the sizes of the Jordon blocks.

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Can you elaborate more about: "the possible jordon blocks can only be 1,1,1, 1,2 and 3. and they correspond to the cases the min poly is (x−3), (x−3)2, (x−3)3. this method will fail to" ? Thank you! –  17SI.34SA Apr 10 '13 at 15:28
    
@17SI.34SA okay, let me edit –  Lost1 Apr 10 '13 at 15:32

Without really explaining Jordan canonical forms, consider:

To be similar, $P$ must exist such that: $$A = P^{-1}BP$$

Let $u_i$ be the eigenvectors of $A$ and $v_i$ be the eigenvectors of $B$. Define two matrices $G$ and $H$ with those eigenvectors as columns:

$$ G = \left( \begin{array}{ccc} u_1 & u_2 & ... \end{array} \right),\ \ \ H = \left( \begin{array}{ccc} v_1 & v_2 & ... \end{array} \right) $$

Since the eigenvectors are linearly independent, we know both $G^{-1}$ and $H^{-1}$ exist. Use these facts to come up with two new matrices: $$ S_A = G^{-1}AG, \ \ \ S_B = H^{-1}BH $$

$S_A$ is similar to $A$ and $S_B$ is similar to $B$.

If $S_A = S_B$, then: $$ S_A = S_B\\ G^{-1}AG = H^{-1}BH\\ AG = GH^{-1}BH\\ A = GH^{-1}BHG^{-1} $$ With $P=HG^{-1}$ and $P^{-1}= GH^{-1}$, $A$ and $B$ are similar: $$A = P^{-1}BP$$ Therefore, you just need to construct $G$ and $H$ from the eigenvectors to find $P$ and show that they're similar.

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