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Could you please check my solution?

Q. State all conditions that make A and B similar.

$$A= \begin{pmatrix} 0 & 4 \\ a & 4 \\ \end{pmatrix} $$

$$B= \begin{pmatrix} 2 & b \\ 0 & c \\ \end{pmatrix} $$

My solution : Since similar matrices have the same eigenvalues, trA=4=trB. so C=2.
DetB=4=DetA so a=-1
Since both the sum and the product of the two eigenvalues are 4, the eigenvalues of A and B are 2,2 (multiplicity 2)
$$B-2I= \begin{pmatrix} 0 & b \\ 0 & 0 \\ \end{pmatrix} $$ $$(B-2I)v= \begin{pmatrix} 0 & b \\ 0 & 0 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ \end{pmatrix}=\begin{pmatrix} by \\ 0\\ \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ \end{pmatrix} $$, where $$v=\begin{pmatrix} x \\ y \\ \end{pmatrix}$$ is an eigenvector of B
The value of $by$ must always be 0. So b is 0.
So the answer is a=-1 and b=0 and c=2

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    $\begingroup$ Ah, true: the OP took $\;(B-2I)=0\;$ as truth instead of the always true in this case $(B-2I)^2=0\; $ . The upvote remains though because the OP at least showed a real effort and self work. $\endgroup$ – DonAntonio Jun 3 '17 at 9:59
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Your arguments for $c=2$ and $a=-1$ look convincing, but $b=0$ can't be right. That would lead to $B=2I$, which is not similar to anything except itself.

Your mistake is that you're assuming that the geometric multiplicity of the eigenvalue $2$ is $2$ (such that any $(^x_y)$ would be an eigenvector) whereas this is actually one of the cases where the geometric multiplicity is smaller than the algebraic multiplicity that you've correctly found to be $2$.

Actually $b$ can take any nonzero value, but specifically can't be $0$.

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  • $\begingroup$ Could you please let me know how to reach to "actually $b$ can take any nonzero value"? $\endgroup$ – Mathman2017 Jun 3 '17 at 10:30
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    $\begingroup$ @Mathman2017: I've explained how $b$ cannot be zero. On the other hand you can show explicitly that if $b\ne 0$ then the matrix is similar to the one with $b=1$ -- by a coordinate transformation of the form $({}^1_0\;{}^0_K)$. $\endgroup$ – Henning Makholm Jun 3 '17 at 10:33
  • $\begingroup$ Are identity matrices scalar-multiplied by a real number similar to only themselves? $\endgroup$ – Mathman2017 Jun 3 '17 at 10:41
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    $\begingroup$ @Mathman2017 Yes. $P^{-1}(cI)P = cI$ for any invertible $P$. $\endgroup$ – Christian Sykes Jun 3 '17 at 10:44

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