Similarity Transformation Matrix $AC=CB$

Question

I am trying to find a matrix such that $AC=CB$.

$A= \begin{pmatrix} 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0 \end{pmatrix}$

$B= \begin{pmatrix} 0 & 5 & 0 & 0\\ 5 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$

I am trying to solve to find an invertible similarity transformation matrix. I tried using Sylvestor's equations but ended up with a singular matrix so I was wondering if there is an easier and less tedious way to find such a matrix without having to find the eigenvalues.

Answer 1

Let $C =\begin{pmatrix} a_{1} & b_{1} & c_{1} & d_{1} \\ a_{2} & b_{2} & c_{2} & d_{2} \\ a_{3} & b_{3} & c_{3} & d_{3} \\ a_{4} & b_{4} & c_{4} & d_{4} \\ \end{pmatrix}$

Then if you compute $AC$ and $CB$, you obtain

$AC = \begin{pmatrix} a_{2} + a_{4} & b_{2} + b_{4} & c_{2} + c_{4} & d_{2} + d_{4} \\ a_{1} + a_{3} & b_{1} + b_{3} & c_{1} + c_{3} & d_{1} + d_{3} \\ a_{2} + a_{4} & b_{2} + b_{4} & c_{2} + c_{4} & d_{2} + d_{4} \\ a_{1} + a_{3} & b_{1} + b_{3} & c_{1} + c_{3} & d_{1} + d_{3} \\ \end{pmatrix}$

$CB = \begin{pmatrix} 5b_{1} & 5a_{1} & 0 & 0 \\ 5b_{2} & 5a_{2} & 0 & 0 \\ 5b_{3} & 5a_{3} & 0 & 0 \\ 5b_{4} & 5a_{4} & 0 & 0 \\ \end{pmatrix}$

$ \ \ $

$a_{2} + a_{4} = 5b_{1}$ and $a_{2} + a_{4} = 5b_{3} \implies b_{1} = b_{3}$

$a_{1} + a_{3} = 5b_{2}$ and $a_{1} + a_{3} = 5b_{4} \implies b_{2} = b_{4}$

$b_{2} + b_{4} = 5a_{1}$ and $b_{2} + b_{4} = 5a_{3} \implies a_{1} = a_{3}$

$b_{1} + b_{3} = 5a_{2}$ and $b_{1} + b_{3} = 5a_{4} \implies a_{2} = a_{4}$

Thus $10a_{1} = 5(a_{1} + a_{3}) = 5(5b_{2}) = 25b_{2} \implies 20a_{1} = 50b_{2} = 25(b_{2} + b_{4}) = 25(5a_{1}) = 125a_{1} \implies a_{1} = 0$

Similar for $a_{2},a_{3},a_{4},b_{1},b_{2},b_{3}$ and $b_{4}$

Thus we have that $C =\begin{pmatrix} 0 & 0 & c_{1} & d_{1} \\ 0 & 0 & c_{2} & d_{2} \\ 0 & 0 & c_{3} & d_{3} \\ 0 & 0 & c_{4} & d_{4} \\ \end{pmatrix}$

Then $det(C) = 0$

There is not exist some no singular matrix $C$ such that $AC = CB$

Answer 2

A more methodical method:

Applying some row/column operations: \begin{align}Q_1^{-1}AQ_1&=\begin{pmatrix}1&0&0&0\\0&1&0&0\\-1&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&1&0\\0&1&0&1\\1&0&1&0\\0&1&0&1\end{pmatrix}\begin{pmatrix}1&0&-1&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\\ &=\begin{pmatrix}2&0&0&0\\0&1&0&1\\1&0&0&0\\0&1&0&1\end{pmatrix} \end{align} Applying further the matrices $Q_2= \begin{pmatrix}1&0&0&0\\0&1&0&-1\\0&0&1&0\\0&0&0&1\end{pmatrix}$, $Q_3= \begin{pmatrix}1&0&0&0\\0&1&0&0\\\frac{1}{2}&0&1&0\\0&0&0&1\end{pmatrix}$, $Q_4= \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&\frac{1}{2}&0&1\end{pmatrix}$ gives $$Q^{-1}AQ=Q_4^{-1}Q_3^{-1}Q_2^{-1}Q_1^{-1}AQ_1Q_2Q_3Q_4=\begin{pmatrix}2&0&0&0\\0&2&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$$

As can be seen, the eigenvalues of $A$ are $2,2,0,0$. These are not the eigenvalues of $B$, so the two are not similar to each other.