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I was working more on the topic on my previous question when I have to know whether the following statement is true to circumvent the "exception" caused by division by singular matrices; again, long story short, the statement follows:

If two singular matrices $A, B$ exist s.t. the determinant of $EA-B$ is identically zero for all real matrices $E$, then either $A=YB$ or $B=ZA$, $Y$ and $Z$ being undetermined matrices.

Is it true (vacuously or not) in general?

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  • $\begingroup$ Should $C$ be $B$ in the above? $\endgroup$
    – J.J. Green
    Commented Feb 8, 2023 at 11:50
  • $\begingroup$ @J.J.Green Yes; fixed typo. $\endgroup$
    – Kanghun Kim
    Commented Feb 8, 2023 at 12:02

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No.

The first condition is satisfied if (and only if) there is some vector in the kernel of $A$ that is also in the kernel of $B$.

The second condition is satisfied (if and) only if the kernel of $A$ is contained in the kernel of $B$ or the kernel of $B$ is contained in the kernel of $A$.

To make a counterexample, we choose any two subspaces with nonempty intersection but neither contained in the other, and choose matrices with those kernels. This first occurs in dimension 3, with two 2-dimensional subspaces, giving Federico Poloni's answer.

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The statement is false.

Take $3\times 3$ matrices such that $A_{11}=B_{22}=1$ and all other entries are zero. Then $EA-hB$ has the third column equal to $0$, but the row spaces of $A$ and $B$ are disjoint and hence neither of $A=YB$ or $B=ZA$ holds.

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