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I came across this problem which says:

Let $A$ be a $2 \times 2$ matrix such that only $A$ is similar to itself.Then show that A is a scalar matrix, that is $A$ is of the form \begin{pmatrix} a &0 \\ 0 & a \end{pmatrix} ? My attempts: Since $A$ is similar to itself,there exists an invertible matrix P such that A= $P^{-1}AP$. Then I tried to solve it by choosing A and P of the form \begin{pmatrix} a &b \\ c & d \end{pmatrix} and \begin{pmatrix} x &y \\ z & w \end{pmatrix} respectively. But I could not get the desired result. Please help. Thanks everyone in advance for your time.

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Stop me if I'm wrong, but isn't every square matrix similar to itself?! (take the identity matrix for $P$) May I suggest that you check the problem ? –  Siméon Dec 5 '12 at 11:23
    
@JulienB. Yes.But here the question says that A is only similar to itself. –  learner Dec 5 '12 at 11:29

3 Answers 3

up vote 5 down vote accepted

HINT:

The matrix $A$ is similar only to itself. Thus $PAP^{-1}=A $ for all invertible $P.$
Use as $P$ the matrices: $$P=\begin{pmatrix}1 &1 \\ 0 &1 \end{pmatrix} \ \ \text{and} \ \ P=\begin{pmatrix}1 &0 \\ 1 &1 \end{pmatrix}.$$

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Is there any way to find P in two different forms ? –  learner Dec 5 '12 at 11:59
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@learner: The answer means write out what this means when one takes for $P$ the first matrix, then (independently) write out what it means when one takes for $P$ the second matrix. The hypothesis is that it holds for all invertible matrices, but it suffices to use that hypothesis just for those two to reach the desired conclusion. –  Marc van Leeuwen Dec 10 '13 at 13:02

Do you know that every $2\times2$ matrix is similar either to a matrix of the form $$A=\pmatrix{a&0\cr0&b\cr}$$ or to a matrix of the form $$B=\pmatrix{a&1\cr0&a\cr}?$$ If so, then all you have to show is that $A$ is similar to $$\pmatrix{b&0\cr0&a\cr}$$ and $B$ is similar to $$\pmatrix{a&0\cr1&a\cr}$$ (and the latter follows from the theorem that every matrix is similar to its transpose).

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That's using quite a lot of structure theory for a question that can be answered very basically. Also it is only valid for matrices over algebraically closed fields, which is not implied in the question. –  Marc van Leeuwen Dec 10 '13 at 13:06

Alternatively, that $A$ is only similar to itself means $PA=AP$ for all invertible matrix $P$. Therefore $$ \begin{pmatrix}a+cx&b+dx\\ cy&dy\end{pmatrix}= \underbrace{\begin{pmatrix}1&x\\0&y\end{pmatrix}}_{P} \underbrace{\begin{pmatrix}a&b\\c&d\end{pmatrix}}_{A}= \begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}1&x\\0&y\end{pmatrix}= \begin{pmatrix}a&ax+by\\c&cx+dy\end{pmatrix} $$ for all $y\not=0$ and all $x$.

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