I have the following HW problem:

Suppose $A$ and $B$ are $n\times n$ matrices with $AC=CB$ for an invertible matrix $C$. Also let $B-\lambda I_n$ be non-invertible for a real number $\lambda$. Prove that $A-\lambda I_n$ is also non-invertible.

I know that $\det{A} = 0 \iff$ the system $A\vec{x} = \vec{0}$ has a non-zero solution. So I let $(A-\lambda I_n)\vec{x}=\vec{0}$ and then rewrote this as $A\vec{x} = \lambda\vec{x}$. But I'm unsure what properties I can use from this point on to actually prove that this is the case (and also show $\vec{x} \ne \vec{0}$; that is, that the system has a non-zero solution).

Thank you!

Edit: Didn't cover similar matrices, characteristic polynomials, or eigenvalues yet

  • 2
    $\begingroup$ We have $B=C^{-1}AC$, so $A$ and $C$ are similar, i.e. they have the same characteristic polynomial. $\endgroup$ Nov 10, 2021 at 18:58
  • $\begingroup$ We didnt learn that material yet @DietrichBurde . I'll edit the question to clarify this $\endgroup$ Nov 10, 2021 at 18:58
  • 3
    $\begingroup$ It is not necessary to "cover" similar matrices. It is just $B=C^{-1}AC$, which we have. Clearly then the determinants are equal. $\endgroup$ Nov 10, 2021 at 19:00
  • $\begingroup$ If $v\neq 0$ is such that $Bv=\lambda v$ show that $w=Cv$ satisfies $Aw=\lambda w$. Why can we say $w\neq 0$? $\endgroup$
    – Matthew H.
    Nov 10, 2021 at 19:01
  • $\begingroup$ $B-\lambda I=C^{-1}AC-C^{-1}\lambda I C$ $\endgroup$
    – Widawensen
    Nov 11, 2021 at 13:42

1 Answer 1


Because $B-\lambda I_n$ is not invertible, then there exists $X \neq 0$ such that $(B-\lambda I_n)X=0$, i.e. such that $BX=\lambda X$.

Hence $ACX = CBX = \lambda CX$, so $(A-\lambda I_n)(CX)=0$. But because $C$ is invertible and $X \neq 0$, then $CX \neq 0$, and you deduce that $A-\lambda I_n$ is not invertible.


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