# Is $(I-A)v = Iv - Av = Iv - \mu v = (I- \mu)v$ for an eigenvector $v$ with eigenvalue $\mu$ of $A$?

This is perhaps a silly question, but it has caused me some confusion.

Let $$\mathbb{R}^{n,n}$$ denote the space of all real $$n\times n$$ matrices. Let $$A \in \mathbb{R}^{n,n}$$ have an eigenvalue-eigenvector pair $$\mu \in \mathbb{R}, v$$. Let $$I$$ denote the identity matrix on $$\mathbb{R}^{n,n}$$, i.e., $$X I = I X = X$$, for all $$X \in \mathbb{R}^{n,n}$$.

Compute $$(I-A)v = Iv - Av = Iv - \mu v = (I- \mu)v$$, using the distributive law of matrix-vector multiplication.

But now it seems that $$I - \mu$$ is not well-defined, as the dimensions disagree, since $$I \in \mathbb{R}^{n,n}$$ and $$\mu \in \mathbb{R}$$ (here I assume we may only subtract two matrices of equal dimension).

My question is: what, if anything, has gone wrong here? Also, I realize we can probably set $$Iv = 1\cdot v = v$$ and get the seemingly more correct factorization $$(1-c\mu)v$$, but shouldn't the final equality in the original computation above follow just from basic operations of linear algebra? Perhaps my confusion stems from trying to compute $$(I-\mu)$$, which may not make sense - however, the product $$(I- \mu)v$$ does make sense?

• I think your last step is almost correct. $(I - A)v = Iv - Av = v - \mu v = (1 - \mu)v$. Why do you think that doesn't "follow just from basic operations of linear algebra"? – roundsquare May 29 '20 at 14:57
• @roundsquare they wrote $(I - \mu)$ not $(1-\mu)$ in the last step which is where their confusion stems from, in which case yea its a meaningless expression. – Osama Ghani May 29 '20 at 14:59
• Got it, I"ll move my comment to an answer. – roundsquare May 29 '20 at 15:02

The error is in the step $$Iv - \mu v = (I - \mu)v$$. You're trying to use the distributive law, but the problem is that the distributive law could in this case refer to one of two things:

1. For matrices $$A, B$$ and vectors $$v$$, we have $$(A + B)v = Av + Bv$$. This is a statement about the operation of matrix multiplication.
2. For scalars $$\alpha, \beta$$ and vectors $$v$$, we have $$(\alpha + \beta)v = \alpha v + \beta v$$. This is a statement about the operation of scalar multiplication.

The key point is that in your argument, actually neither of these laws applies, because you don't have two scalars or two matrices; you have one of each, and in this case that's not good enough, because it doesn't satisfy the hypotheses for either version of the distributive law given above.

Obviously the argument can very easily be salvaged, as is noted in the other answers, but it seems to me that this is where the confusion creeps in. It's a very good question, and in my opinion a subtle point.

So, to summarise, you are trying to use a rule (the distributive law) that by definition only applies when your operators are of the same type, in a situation where the operators are not of the same type, rendering the rule null and void.

You're getting there, but you don't want to do something like a matrix minus a scalar. Instead, you can do it like this:

$$(I-A)v=Iv-Av=v-\mu v = (1-\mu)v$$

The trick is to use both $$Iv$$ and $$Av$$ on their own to get a scalar times a vector and then combine.

Well, then just write "in the correct world":

$$(I-A)v=Iv-Av=Iv-\mu I v=(I-\mu I)v$$

or else

$$(I-A)v=Iv-\mu v=v-\mu v=(1-\mu)v$$