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.

  • 4
    $\begingroup$ This is clearly possible if $a = 0$ and $b$ is non-zero with just one non-zero row, for example. And this is basically the only time it is possible: your condition implies $\operatorname{rank}(A^\prime-B^\prime) \leq 1$ so $\operatorname{rank}(A-B) \leq 1$, so one of $a$ or $b$ is $0$ and the other has rank $0$ or $1$. $\endgroup$ – Nathaniel Johnston May 13 '19 at 10:47

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).

  • $\begingroup$ Taking inspiration from the comment of @NathanielJohnston, maybe it is possible to simplify the proof removing the initial part with the calculation with blocks. $\endgroup$ – Doriano Brogioli May 13 '19 at 12:50
  • $\begingroup$ @DorianoBrogioli, I agree that it's essentially the same argument. The block calculation helped me see how to find the kernel of the $S$ matrix; that is all. $\endgroup$ – Igor Khavkine May 13 '19 at 18:05

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