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So I know that

$A = T^{-1}AT \implies T \text{ is a similarity transformation matrix}$.

Say $A = \begin{pmatrix}9 & 13 \\ -3 & -3\end{pmatrix}$, then how would I go about finding T without using eigenvalues? By sheer luck, I managed to find $T = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ by picking it at random, plugging in and seeing it satisfy the conditions. So I thought to myself that ANY invertible 2x2 matrix would work because if you multiply $A$ by $T$ and then by $T$'s inverse, you would end up where you started (this is obviously one of my conceptual issues).

So I picked $T$ to be $\begin{pmatrix}2 & 1 \\ 3 & 4\end{pmatrix}$, but I found that it didn't work for some reason.

Why is this? I must be missing something fundamental here. I often find myself struggling with the abstractness of Linear Algebra and I'm not sure what it is.

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    $\begingroup$ But $\;A=T^{-1}AT\iff TA+AT\iff T4\;$ commutes with $\;A\;$ , and there are lots of possible matrices like, among others the trivials ones: scalar matrices of the form $\;k\cdot I_2\;,\;\;k\in\Bbb F=$ our definition field $\endgroup$ – Timbuc Apr 10 '15 at 16:58
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    $\begingroup$ Look more closely at your definitions. The idea of similarity is to establish a relation between matrices $A$ and $B$ (rather than only $A$ and itself) that are connected by conjugation: $A = T^{-1}BT$. When matrices $A$ and $B$ are similar, they represent the same transformation with respect to different bases. The matrix $T$ transforms one basis into the other. $\endgroup$ – Sammy Black Apr 10 '15 at 17:01
  • $\begingroup$ @SammyBlack So how would I algorithmically go about finding T in this case? I only found ((1,0)(0,1)) by accident. $\endgroup$ – aidandeno Apr 10 '15 at 17:06
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    $\begingroup$ I'm actually not sure what question you're asking now. This? Given a matrix $A$ find a similar matrix $D$ that is diagonal; that is, find diagonal $D$ and invertible $T$ such that $D = T^{-1}AT$. $\endgroup$ – Sammy Black Apr 10 '15 at 17:16
  • $\begingroup$ @SammyBlack: If $A = \begin{pmatrix}9&13,-3&-3 \end{pmatrix}$, find a $T$ such that $A=T^{-1}AT$ $\endgroup$ – aidandeno Apr 10 '15 at 17:19
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Finding solutions $T$ to $A = T^{-1}AT$ is, to my knowledge, not a very fun procedure. Left-multiplying by $T$, we get the equivalent equation that $TA = AT$, so that $A$ and $T$ commute.

If, as in your example, $$A = \begin{pmatrix}9 & 13 \\ -3 & -3\end{pmatrix},$$ then we're looking for a matrix $T = \begin{pmatrix}a & b \\ c & d\end{pmatrix}$ so that

$$TA = \begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}9 & 13 \\ -3 & -3\end{pmatrix} = \begin{pmatrix}9a - 3b&13a - 3b\\9c - 3d&13c-3d\end{pmatrix}$$ while

$$AT = \begin{pmatrix}9 & 13 \\ -3 & -3\end{pmatrix}\begin{pmatrix}a & b \\ c & d\end{pmatrix} = \begin{pmatrix}9a + 13c&9b + 13d\\-3a-3c&-3b-3d\end{pmatrix}.$$

Now, as we want $TA = AT$, we can get a system of $4$ equations in $4$ unknowns:

\begin{align*} 9a - 3b &= 9a + 13c \\ 13a - 3b &= 9b + 13d \\ 9c - 3d &= -3a-3c \\ 13c-3d &= -3b-3d, \end{align*}

and consulting WolframAlpha, we get solutions perameterized by $c = \dfrac{-3b}{13}$ and $d = a - \dfrac{-12b}{13}$. (Note that the matrix you found is obtained when $b = 0$ and $a = 1$).

As you can see, it's not a very fun process! This is the most elementary way I can think of to answer the question. There are potentially more efficient methods known, especially if $A$ has some 'nice' properties. Generally, I think, it's still not a question whose concrete answer is nicely computable.

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  • $\begingroup$ You're right, that's not fun at all. But it made perfect sense. Thank you. $\endgroup$ – aidandeno Apr 10 '15 at 17:50
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    $\begingroup$ You're very welcome! I really should make it clear that this is one way to simply list the matrices $T$ for which $A = T^{-1}AT$; it certainly doesn't shed much light on why that relationship would hold, which is probably the right way to think about the situation. Unfortunately, I can't really comment on the right way to think about it. $\endgroup$ – pjs36 Apr 10 '15 at 18:45
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If you multiply $T^{-1}AT$, then multiplication of matrices is not commutative so you won't necessarily get $A$.The set of matrices you CAN get from such a product form what's called the conjugacy class of $A$, if you consider all possible $T$.

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