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How can I show that there exists a nonzero vector $x$ such that, provided $(I - M)$ is not invertible, we have $(I - M)x = 0$?

I'm not sure about how to go about this. I think that I'll need to use the fact that $(I - M)$ is not invertible, but I have no idea where.

Any help is appreciated.

2 Answers2

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Assume that $I-M$ is not invertible. Then, $I-M$ is not an injection (aka not 1-to-1) (see here) and so there exists $u$ and $v$ with $u\neq v$ such that $(I-M)v = (I-M)u$. Thus $(I-M)(u-v) = 0$ and by assumption $u\neq v$ so $u-v\neq 0$.

Jürgen Sukumaran
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Let $M:\ell_{2}\rightarrow \ell_{2}$ be given by $Me_{i}=e_{i}-e_{i+1}$, where $e_{i}$ is the standard unit vector. Then $(I-M)$ is given by $(I-M)e_{i}=e_{i+1}$. Note that $(I-M)$ is not invertible as there is no $x\in\ell_{2}$ such that $(I-M)x=e_{1}$, but $(I-M)x\neq 0$ for all $x\neq 0$.

Floris Claassens
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  • Note that this kind of construction only works in infinite dimensional space, for finite dimensional spaces you can follow the suggestions in the comments on your question. – Floris Claassens Apr 08 '19 at 13:49