# Equalities between transforms of matrices that are extremely different

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.

• 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 $1$. – 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).
• @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. – Igor Khavkine May 13 '19 at 18:05