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Stuck on this question:

Let $$A=\begin{pmatrix} 2&1\\ -1&-1 \end{pmatrix}$$$$B=\begin{pmatrix} -2&5\\ -1&3 \end{pmatrix}$$$$C=\begin{pmatrix} 5&2\\ 4&1 \end{pmatrix}$$ Show that A is similar to B, but that A is not similar to C.

I can do the second part of the question as $det(A)\neq\det(C)$, therefore as similar matrices have the same determinant $A\nsim C$. I also understand that I need to find $P$ such that $AP=PB$ for the first part but have no idea how I would go about finding it.Would anyone be able to provide an answer and explanation of the method used?

Thanks

Edit:Had a look at the solution from the textbook, it gives $$P=\begin{pmatrix} 2&3\\ 1&1 \end{pmatrix}$$

George1811
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2 Answers2

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In order to find your $P$, you can do as follows:

  • First, you find a diagonal matrix $D$ to which both $A$ and $B$ are equivalent. For this, you need to find the eigenvalues of both matrices and if they coincide, they are equivalent. (The general case would be more involved: but in yours, both matrices diagonalize easily.)
  • Then you have to find bases of eigenvectors for both matrices and form with them change of bases matrices $S$ and $T$ such that

$$ D = S^{-1}A S \qquad \text{and} \qquad D = T^{-1}BT \ . $$

  • Now you'll have

$$ S^{-1}A S = T^{-1}BT \qquad \text{and hence} \qquad AST^{-1} = ST^{-1}B \ . $$

So $ST^{-1}$ will be your matrix $P$.

Agustí Roig
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If we want to perform as little calculations as possible I suggest the following line of reasoning.

Assuming the eigenvalues of a matrix are distinct it is diagonalizable. If two matrices are both diagonalizable with the same distinct eigenvalues then they are similar. (To see this just write them out in diagonalized form and construct a change of basis from the eigenvector-basis-matrices)

So what does this mean? Well if two matrices have the same characteristic equation and all the roots of that characteristic equation are distinct then they must be similar!

Since the characteristic equation is of second order we don't even need to solve it in order to check that it has two distinct solutions. Just check the discriminant instead.

EDIT: Note of course that we don't obtain the actual similarity transform from this but the point is we don't need to find it. Only show that it exists.

John B
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Squid
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  • The characteristic polynomial being equivalent is only a necessary condition not a sufficient one. Take the matrix $A=\begin{pmatrix} 1 &0 \ 0 &0 \end{pmatrix}$ This has determinant 0. As does the null matrix. However they are not similar – George1811 Jan 03 '14 at 15:29
  • The characteristic equation and the determinant are not the same thing. The characteristic polynomial contains more information than the determinant but it is true that just because two matrices have the same characteristic polynomial they aren't necessarily similar because it all hinges on the eigenvalues being distinct. If the eigenvalues aren't distinct then their Jordan-forms might be different.

    In the case of your matrix and the nullmatrix they don't have the same characteristic equation. A has eigenvalue 1 whereas the null matrix does not.

     – Squid Jan 03 '14 at 15:35
  • Ah I see finding the distinct eigenvalues themselves gives us enough information to deduce they are similar – George1811 Jan 03 '14 at 15:37
  • Precisely. This method might not work to establish similarity because you could be unlucky and find that the eigenvalues are not distinct but since the method requires very little actual computation I feel it is worth trying out before resorting to more computation-heavy methods. – Squid Jan 03 '14 at 15:41
  • @Squid : "If two matrices are both diagonalizable with the same eigenvalues then they are similar" . In the best case this might be misleading or confusing, in the worst it is utterly wrong. The matrices $$\begin{pmatrix}1&0&0\0&1&0\0&0&2\end{pmatrix};,;;\begin{pmatrix}1&0&0\0&2&0\0&0&2\end{pmatrix}$$ are diagonalizable and "have the same eigenvalues" but they're obviously not similar. In the present question's case it is very important we've $;2\times 2;$ matrices... – DonAntonio Jan 03 '14 at 15:41
  • I think the condition that the eigenvalues were assumed to be distinct was pretty clear at this point but I'll edit it into the statement for clarity. Thanks for expressing your concern, it's a valid point. – Squid Jan 03 '14 at 15:45
  • I can't see how writing "distinct" addresses my last comment's concern, @squid... – DonAntonio Jan 03 '14 at 15:50
  • I'm sorry but then I guess I don't understand what your concern is. Could you please clarify. The eigenvalues of your counter-example aren't distinct (both have two of the same) so my reasoning wouldn't suggest them to be similar.

    Furthermore I don't find that my reasoning relies heavily on the fact that we have a 2D situation except for the suggested method for showing that the eigenvalues are distinct. Should apply to all dimensions even if the methods for sorting out eigenvalues have to be adjusted.

     – Squid Jan 03 '14 at 16:02
  • I don't follow you, @Squid : in my counterexample, both matrices have eigenvalues $;1,2;$ ...What has to be considered here, imo, is the multiplicity, perhaps. In the case of $;2\times 2;$ this consideration is moot. – DonAntonio Jan 03 '14 at 21:32
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    In the literature I've read, and the in the terminology I'm used to, saying that the eigenvalues of a matrix are distinct precisely means there is no multiplicity in the eigenvalues, not that there are more than 2. In that case I would say that a matrix has at least 2 distinct eigenvalues. What term would you use to communicate that an $n \times n$ matrix has $n$ different eigenvalues without having to specify the dimension $n$? Maby I can use that in case this convention is confusing or incorrect in other literature. – Squid Jan 03 '14 at 23:12