# Looking for a motivated proof of a very simple similarity relation between $2 \times 2$ matrices

I am reading a textbook which casually mentions that the matrices

$$A = \begin{pmatrix} a & b \\ c & a + b - c \end{pmatrix}$$ and

$$B = \begin{pmatrix} a+b & - b \\ 0 & a-c \end{pmatrix}$$ are similar because

$$B = \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} A \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix}^{-1},$$ which allows one to conclude that the eigenvalues of $A$ are $a+b,a-c$.

Its easy to verify by direct calculation that this is true, but it feels pulled out of a hat and I prefer to have proofs which "make sense" to me, i.e., proofs where I understand where each step "came from." I'm wondering if anyone can explain this similarity relation to me on that level.

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I can't say I have much experience coming up with shortcuts for producing similarity transformations, but I can speculate how they might look. You can think of building them up by doing a row operation on a matrix followed by the inverse column operation. e.g.

subtract the first row from the second, then add the second column to the first


(to see that this is right, look at the corresponding elementary matrices)

I could easily imagine a person with much practice in such calculations would notice the relationship between the four terms, and the possibility that you could combine three of them to cancel the fourth. Or, they might be armed with a general-purpose algorithm for canceling the bottom-left entry in any 2x2 square using simultaneous elementary operations as I described above, and it just happened to have a very simple form in this case.

Armed with this idea, the first thing I thought to try was actually the operation I described above in the block quote, which also is enough to reduce the matrix to upper-triangular. The difference is that I have a $b$ in the upper-right corner, rather than a $-b$.

As a meta-mathematical point, I think one should separate the idea of "a proof I understand" from the idea of "a proof I could have come up with". (and further, to separate the ideas of "a proof I could have come up with because I already knew some things about $A$ and $B$" versus "a proof I could have come up with if I saw $A$ without any context", which may or may not be relevant)

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Define matrix $P$,

$$P=\begin{pmatrix}a &b \\ c &d \end{pmatrix}$$

now, find $P^{-1}$, and solve:

$$B=PAP^{-1}$$

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What I'm looking for is along the following lines: just by looking at the matrix $A$, how could you one guess that its possible to so easily transform it into upper triangular form and that its eigenvalues are so simply expressed in terms of its elements? Apologies if I wasn't entirely clear. – robinson Apr 17 '12 at 22:38
@robinson It has to do with the fact that the sum of two rows are equal. For any such 2*2 matrix the eigenvalues are simple (linear expressions). But your question perhaps can be phrased as "is there a theorem for the eigenvalues of a matrix similar to rational root theorem of polynomials ?" – Maesumi Apr 18 '12 at 0:05

You can start with any parametric matrix $B$, especially one that is triangular with easy-to-find eigenvalues, and apply any similarity transformation to get another matrix A, and ask the same question.

Perhaps the question should be: "How can I tell if two given matrices are similar and how to find the similarity transformation?" If $X$ is the similarity transformation then $B X=X A$. This will give you an $n^2 \times n^2$ homogeneous system to solve. In this case a $4\times 4$ system. The system has to be solved for a non-singular answer that is valid for all values of parameters. In this case it is easy to solve the system and find that there is indeed one solution (up to a multiplication).

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First, some basics regarding similarities.

1. $PIP^{-1} = I\quad$ This is useful here because:$$P(A-aI)P^{-1} = PAP^{-1} - aI$$ What this means in this problem is that you need only look at $$A - aI = \pmatrix{0 & b \\ c & b-c}$$ and whatever result is obtained needs only the $aI$ restored to the diagonal, thus any diagonalization of it works. So that removes one variable $a$ from consideration here.

2. Any elementary row operation need be followed with the column operation that restores identity: $$\pmatrix{1 & 0 \\ 0 & 1}\overset{R2\leftarrow R2-R1}{\rightarrow} \pmatrix{1 & 0 \\ -1 & 1}\overset{C1\leftarrow C1+C2}{\rightarrow} \pmatrix{1 & 0 \\ 0 & 1}$$ I use this particular example of subtraction first (as opposed to adding first) because it simply is the operation needed for the problem at hand: $$\pmatrix{0 & b \\ c & b-c}\overset{R2\leftarrow R2-R1}{\rightarrow} \pmatrix{0 & b \\ c & -c}\overset{C1\leftarrow C1+C2}{\rightarrow} \pmatrix{b & b \\ 0 & -c}$$

3. Even though eigenvalues are established here and thus for your problem there is no real need to continue, I note here that scaling is also an elementary row operation:$$\pmatrix{1 & 0 \\ 0 & 1}\overset{R1\leftarrow (-1)R1}{\rightarrow} \pmatrix{-1 & 0 \\ 0 & 1}\overset{C1\leftarrow (-1)C1}{\rightarrow} \pmatrix{1 & 0 \\ 0 & 1}$$ This then helps complete the similarity to B in the problem $$\pmatrix{b & b \\ 0 & -c}\overset{R1\leftarrow (-1)R1}{\rightarrow} \pmatrix{-b & -b \\ 0 & -c}\overset{C1\leftarrow (-1)C1}{\rightarrow} \pmatrix{b & -b \\ 0 & -c}$$

These basics are important enough to remember if you want to be at all proficient in recognizing such similarities.

To summarize: the $a$ diagonal values are ignored, then the elementary similarity that removes the $c$ value is performed. From there if you want full similarity to $B$, an elementary row scaling similarity is performed to negate the off diagonal element $b$.

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