Prove that $W_\lambda$ is a subspace of $R^n$ Let $A$ be an $n \times n$ matrix, and let $\lambda$ be an eigenvalue for $A$. The eigenspace of $A$ corresponding to $\lambda$ is the set
$$W_\lambda = \{x \in R^n : Ax = \lambda x\}$$
Prove that $W_\lambda$ is a subspace of $R^n$.
I am not sure how to prove that. Any known theorems I can make use of?
 A: The idea is to work straight from the definition of subspace.  All we have to do is show that
$W_\lambda = \{x \in \Bbb R^n : Ax = \lambda x \}$ satisfies the vector space axioms; we already know $W_\lambda \subset \Bbb R^n$, so if we show that it is a vector space in and of itself, we are done.  So, if $\alpha, \beta \in \Bbb R$ and $v, w \in W_\lambda$, then
$A(\alpha v + \beta w) = \alpha A v + \beta A w = \alpha \lambda v + \beta \lambda w = \lambda (\alpha v + \beta w), \tag{1}$
showing that $\alpha v + \beta w \in W_\lambda$; the other definitive properties of a vector space, commutativity of addition, existence of additive identity, properties of scalar multiplication, etc.,  apply to $W_\lambda$ via inheritance from $\Bbb R^n$; being closed under linear combination, as we have just seen,  this implies $W_\lambda$ is a subspace of $\Bbb R^n$.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: Hint:
Let $x,y\in W_\lambda$, then $x+y\in W_\lambda$ and $\alpha x\in W_\lambda$, for scalar $\alpha\in R$. Also $0 \in W_\lambda$.
