Proving that a set is a subspace of $\mathbb{R}^{n}$.

Question

Let $A$ be a square matrix of order $n$, and let $\lambda$ be a scalar. Prove that the set $S = \{x : Ax = \lambda x\}$ is a subspace of $\mathbb{R}^{n}$.

Answer 1

We check the conditions on a subspace.

A set $U\subseteq V$ of a vector space $V$ is a subspace if

1. $\mathbf0\in U$(the null vector is in it)
2. If $v,w\in U$, then $v+w\in U$(closure under vector addition)
3. If $v\in U$, then $\mu v\in U$ f.a. scalars $\mu$(closure under scalar multiplication)

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Note, that the null vector is in $S$, $\mathbf0\in S$, as always $A\mathbf0=\mathbf0=\lambda\mathbf0$.

Secondly, it is closed under under addition, i.e. let $v,w\in S$. Then $Av=\lambda v$ and $Aw=\lambda w$ and followingly, we have

$$A(v+w)=Av+Aw=\lambda v+\lambda w=\lambda(v+W)$$

by distribution of linear maps over addition.

Lastly, it is closed under scalar multiplication, i.e. let $v\in S$. Then $Av=\lambda v$ and thus

$$A(\mu v)=\mu Av=\mu\lambda v=\lambda\mu v$$

as linear maps distribute over scalar multiplication.

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To alternatively check if a set $U\subseteq V$ of a vector space $V$ is a subspace, you may also just verify that $U\neq\varnothing$ and additionally that it is closed under linear combinations, i.e.

$$\text{If }v_1,\dots,v_n\in U\text{, }\lambda_1,\dots,\lambda_n\text{ scalars, then }\sum_{i=1}^n\lambda_iv_i\in U$$

It even suffices to check only for linear combinations of length two, i.e.

$$\text{If }v_1,v_2\in U\text{, }\lambda_1,\lambda_2\text{ scalars, then }\lambda_1v_1+\lambda_2v_2\in U$$

Can you see why? A good little exercise would be to prove these definitions equivalent to get a little grip on subspaces in general.