I'm having a conceptual issue with similarity matrices 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.
 A: 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.
A: 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 $T$.
