Show that if $A \in M_{2 \times 2}(\mathbb{R})$ then $A, A^T$ are similar.

Question

• Show that if $A \in M_{2 \times 2}(\mathbb{R})$ then $A, A^T$ are similar.
  
  We say that two matrices $A, B \in M_{n}$ are similar if there exists an invertible matrix $P \in M_{n}$ s.t $A = P^{-1}BP$

I was asked to decide wether this is true or false during an exam at algebra. Thing is, this exam was for algebra 1 and we still haven't started to learn about eigenvalues (so the notion of similar matrices was introduced very briefly so far).

I know that this is true for $n$, but I probably was expected to have enough knowledge to solve it for the case where $n=2$. I'm struggling to see how this can be proved with basic knowledge about similarity.

I hope to get some ideas, thanks

Answer 1

Let $K$ be a field and $A = \bigl(\begin{smallmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{smallmatrix}\bigr) \in M_2(K)$. We need to find an invertible matrix $V = \bigl(\begin{smallmatrix} v_{11} & v_{12} \\ v_{21} & v_{22} \end{smallmatrix}\bigr) \in \mathrm{GL}_2(K)$ with $A^T = V A V^{-1}$, meaning $$V A = A^{T} V. \quad (\ast)$$ But this is just a system of linear of equations in $v_{11},v_{12},v_{21},v_{22}$ which we can solve.

When $A$ is a diagonal matrix, nothing has to be done, since then $A = A^T$. Otherwise, we have $a_{12} \neq 0$ or $a_{21} \neq 0$. Let's say $a_{21} \neq 0$ without loss of generality. Then we are able to eliminate $v_{21}$ and $v_{22}$. Let me only present the result here (if necessary, I will add the calculation):

• $v_{11}$ and $v_{12}$ are arbitrary
• $v_{21} = v_{12}$
• $v_{22} = a_{21}^{-1} (a_{12} v_{11} + (a_{22} - a_{11}) v_{12})$

Now we need to find $v_{11},v_{12}$ so that $\det(V) \neq 0$. This evaluates to $$\det(V) = a_{21}^{-1} a_{12} v_{11}^2 + a_{21}^{-1} (a_{22} - a_{11}) v_{11} v_{12} - v_{12}^2$$ This is non-zero when $v_{11}=0$ and $v_{12}=1$, for example. So we may in fact choose $$V = \begin{pmatrix} 0 & 1 \\ 1 & a_{21}^{-1}(a_{22}-a_{11}) \end{pmatrix}.$$ It is also possible to verify directly now that this matrix satisfies $(\ast)$.

Answer 2

I thought about this problem for quite a long time now. At least for $n=2$, there is an easy solution.

The solution here looks simple, but there is a long story how I obtained this. Here's the full account of how I obtained this method: https://www.csun.edu/~sungjin/Linear%20Algebra.pdf

Let $A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$.

Case 1: $b\neq 0$, we can take $X=\begin{pmatrix} \frac{a-d}b & 1 \\ 1 & \ \end{pmatrix}$. Then $XA=A^T X$.

Case 2: $b=0$, $c\neq 0$, then we take $X=\begin{pmatrix} \frac{a-d}c & 1 \\ 1 & \ \end{pmatrix}$. Then $XA^T = AX$.

Case 3: $b=c=0$, then $A$ is diagonal, and $X=I$ works.