Vtyjkyuk

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 $mathbbR^n$.

We check the conditions on a subspace.

A set $Usubseteq V$ of a vector space $V$ is a subspace if

1. $mathbf0in U$(the null vector is in it)


2. If $v,win U$, then $v+win U$(closure under vector addition)


3. If $vin U$, then $mu vin U$ f.a. scalars $mu$(closure under scalar multiplication)

Note, that the null vector is in $S$, $mathbf0in S$, as always $Amathbf0=mathbf0=lambdamathbf0$.

Secondly, it is closed under under addition, i.e. let $v,win S$. 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 $vin S$. Then $Av=lambda v$ and thus

$$A(mu v)=mu Av=mulambda v=lambdamu v$$

as linear maps distribute over scalar multiplication.

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

$$textIf v_1,dots,v_nin Utext, lambda_1,dots,lambda_ntext scalars, then sum_i=1^nlambda_iv_iin U$$

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

$$textIf v_1,v_2in Utext, lambda_1,lambda_2text scalars, then lambda_1v_1+lambda_2v_2in U$$

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