I have two $2N\times 2N$ matrices, defined by blocks:


$$ A = \begin{bmatrix} a & 0 \\ 0 & 0 \end{bmatrix} $$
$$ B = \begin{bmatrix} 0 & 0 \\ 0 & b \end{bmatrix} $$


where $a$ and $b$ are $N\times N$ matrices. I assume that $A\ne B$, i.e. they are not trivially 0.


Then I have a coordinate
transformation defined by a generic invertible matrix $T$. Through this transformation $A$ becomes $A'$ and $B$ becomes $B'$:


$$ A' = T^{-1} A T $$
$$ B' = T^{-1} B T $$


Of course, it is not possible that $A'=B'$ (proof: this would imply $A=B$)


Instead my question is if it is possible that the rows of $A'$ and $B'$ are equal, except one:


$$ \left[A'\right]\_{i,j} = \left[B'\right]\_{i,j} \tag{1} $$


for $i=1\dots 2N-1$ and $j=1\dots 2N$ (here $i$ the subscript of the row and
$j$ of the column).


I guess that this cannot happen, unless $A$ and $B$ are trivially 0. But I could not prove it, although it does not seem too difficult.



Equalities between transforms of matrices that are extremely different
Let $S = T^{-1} = \begin{bmatrix} S\_{00} & S\_{01} \\ S\_{10} & S\_{11} \end{bmatrix}$. If all rows but the last of the matrices $A'$ and $B'$ are equal, then the same is true for $SA = A'T^{-1}$ and $SB = B'T^{-1}$ (right multiplication mixes columns but not rows). Computing the products
\begin{align\*}
 SA &= \begin{bmatrix} S\_{00} a & 0 \\ S\_{10} a & 0 \end{bmatrix}, \\
 SB &= \begin{bmatrix} 0 & S\_{01} b \\ 0 & S\_{11} b \end{bmatrix} ,
\end{align\*}
you can see that all but the last row are equal only when $S\_{00} a = 0 = S\_{01} b$ and the first $N-1$ rows of both $S\_{10}a$ and $S\_{11} b$ are zero. If $\alpha$ and $\beta$ are vectors such that $a\alpha \ne 0 \ne b\beta$ (otherwise, $a$ and $b$ are identically zero), the two vectors $S [\begin{smallmatrix} a\alpha \\ 0 \end{smallmatrix}]$ and $S[\begin{smallmatrix} 0 \\ b\beta \end{smallmatrix}]$ will have non-zero components only in the very last row and hence be linearly dependent. But that implies that $S$ is not invertible, contrary to what was assumed.


Thus, for the equality of all but the last rows of $SA$ and $SB$ to be realized, at least one of $a$ and $b$ must be identically zero (in which case the situation is obvious).


