Is matrix $X$ similar to matrix $Y$ for all $w,x,y,z$?

Question

$$X = \left( \begin{array}{ccc} w & x \\ y & z \end{array} \right)$$

$$Y = \left( \begin{array}{ccc} w & y \\ x & z \end{array} \right)$$

I know that matrix $X$ is similar to matrix $Y$ when $Y=S^{-1}XS$ for some invertible matrix $S$. So consider:

$$S = \left( \begin{array}{ccc} a & b \\ c & d \end{array} \right)$$

$$S^{-1} = \frac{1}{ad-bc}\left( \begin{array}{ccc} d & -b \\ -c & a \end{array} \right)$$

I was thinking that we could compute $S^{-1}XS$ and set that equal to $X$, but I was wondering if there was an easier way to determine whether $X$ is similar to $Y$ for all $w,x,y,z$. When I tried to solve the multiplication by hand the expressions got very long, so I'm assuming there must be some trick I'm not seeing.

Answer 1

Note that $Y=X^t$.

Case 1: If there exist some matrix $S$ such that $S^{-1}XS=\left( \begin{array}{cc} \lambda_1& \\ &\lambda_2 \\ \end{array} \right)$, then $X\simeq S^{-1}XS=(S^{-1}XS)^t=S^t Y (S^t)^{-1}\simeq Y$.

Case 2: If there exist some matrix $S$ such that $X\simeq S^{-1}XS=\left( \begin{array}{cc} \lambda&1 \\ 0&\lambda \\ \end{array} \right)$, then $Y\simeq S^t Y (S^t)^{-1}=\left( \begin{array}{cc} \lambda&0 \\ 1&\lambda \\ \end{array} \right)$, and $\left( \begin{array}{cc} 0&1 \\ 1&0 \\ \end{array} \right)\left( \begin{array}{cc} \lambda&1 \\ 0&\lambda \\ \end{array} \right)\left( \begin{array}{cc} 0&1 \\ 1&0 \\ \end{array} \right)=\left( \begin{array}{cc} \lambda&0 \\ 1&\lambda \\ \end{array} \right)$, so $X\simeq Y$.

In case 1 and Case 2, the only problem is that the matrix may not in $M_2(\mathbb{R})$. If $X\in M_2(\mathbb{R})$, then you can find a matrix $S\in M_2(\mathbb{C})$ first, such that $Y=S^{-1}XS$ or $SY=XS$, write $S=U+iV, U,V\in M_2(\mathbb{R})$, then $UY=XU$ and $VY=XV$, the polynimial $\det(U+\lambda V)\neq0$, because $\det(U+iV)=\det(S)\neq0$, so there exist some $\lambda\in\mathbb{R}$, such that $U+\lambda V$ invertible and $(U+\lambda V)Y=X(U+\lambda V)$.

Moreover, $\forall X\in M_n(\mathbb{R})$, $X\simeq X^t$, you can prove this by induction on $n$.

Answer 2

Yes every matrix$~X$ is similar to its transpose (over its field of definition). However there is no easy formula that expresses the a valid base change matrix$~S$ (for which there is a lot of choice) in terms of$~X$. This means that you attempts to solve explicitly for$~S$ are not likely to lead to a satisfactory conclusion.

For the $2\times2$ case one can see that there is always a solution for $S$ by writing down the equation $XS=SY$ (where $X$ and $Y=X^t$ are fixed) which is a linear equation in that coordinates $a,b,c,d$ of $S$ (which are the unknowns here). The equations are trivial when $X$ is a multiple of the identity matrix (i.e., when $x$, $y$ and $w-z$ are all zero), so we can exclude that case henceforth. Then the equations force $b=c$ (in several ways) which means you need $S$ to be symmetric. Assuming that (and dropping the unknown$~c$) there one linear equation in terms of $a,b,d$ that remains, which is $ya+(z-w)b-xd=0$. This defines a $2$-dimensional subspace of matrices. One is not quite done, because one needs to show that the expression $ad-b^2$ for $\det S$, which needs to be nonzero so to have a true solution, does not vanish everywhere on the subspace. But the zeros of this quadratic form form a cone, which contains $1$-dimensional subspaces but no $2$-dimensional ones.