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 $$P=\begin{pmatrix} 2&3\\ 1&1 \end{pmatrix}$$
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$\begingroup$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 $P$.
$\endgroup$ $\begingroup$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.
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