# Proof of a matrix non-invertibility

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

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

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.