# Finding alternate transformation matrix for similarity transformation

A pair of square matrices $$X$$ and $$Y$$ are called similar if there exists a nonsingular matrix $$T$$ such that $$T^{-1}XT=Y$$ holds. It is known that the transformation matrix $$T$$ is not unique for given $$X$$ and $$Y$$. I'm just wondering whether those non-unique transformation matrices would have any relation among themselves, like having column vectors with same directions...

What I want to mean is: Given $$X$$ and $$Y$$ a pair of similar matrices, if $$S$$ and $$T$$ are two possible transformation matrices satisfying $$S^{-1}XS = T^{-1}XT = Y$$, is there any generic (apart from scaling) relation between $$T$$ and $$S$$ (e.g., direction of column vectors)?

For a specific example, consider $$X = \begin{bmatrix} A & BK\\ C & 0 \end{bmatrix}$$ and $$Y = \begin{bmatrix} A+A^{-1}BKC & -A^{-1}BKCA^{-1}B\\ KC & -KCA^{-1}B \end{bmatrix}$$. Assuming $$K$$ to be invertible it can be shown that $$X$$ and $$Y$$ are similar with transformation matrix $$T = \begin{bmatrix} I & -A^{-1}B\\ 0 & K^{-1} \end{bmatrix}$$. Can there be any other matrix $$S$$ which will be independent of $$K$$, and would result $$S^{-1}XS=Y$$?

You can always use $$S=\alpha\,T$$ with $$\alpha$$ some non-zero scalar not equal to one. Another way of finding more solutions is by using the Kronecker product, which allows you to write the transformation as
\begin{align} X\,T &= T\,Y \\ (I \otimes X)\,\text{vec}\,T &= (Y^\top \otimes I)\,\text{vec}\,T \\ (I \otimes X - Y^\top \otimes I)\,\text{vec}\,T &= 0 \end{align}
Using any (non-zero) vector which lies in the null space of $$I \otimes X - Y^\top \otimes I$$ for $$\text{vec}\,T$$ will satisfy the equation (however I am not sure if every $$T$$ would be invertible). If the dimension of the null space is one then you can only do the scalar multiplication. If the dimension is bigger then one you can also take a linear combinations of linear independent solutions (this both holds for $$\text{vec}\,T$$ and $$T$$, since the vectorization transformation and its inverse are also linear).
• @neelarnab I suspect that you will have multiple solutions if $X$ and $Y$ have repeated eigenvalues/Jordan blocks. – Kwin van der Veen Oct 23 '18 at 15:55
• are you saying that it would be a necessary condition for having multiple transformation matrices (excluding the scaling)? Given a transformation matrix $X$, can we then find an alternate transformation matrix $Y$, like in the case of the mentioned example? – neelarnab Oct 24 '18 at 10:41
• @neelarnab I did some testing and it seems that when $X,Y,T\in\mathbb{R}^{n \times n}$ you always seem to have that the dimension of the null space is equal or greater then $n$. But I haven't found an intuitive explanation for this. – Kwin van der Veen Oct 24 '18 at 11:18