Prove that $V$ is a subspace of $\Bbb R^n$ Let $V = \left\{x \in\Bbb R^n | Ax= \lambda x\right\}$ where $A$ is an $n \times n$ matrix and $\lambda\in\Bbb R$, together with the usual operations for vector addition and scalar multiplication from $\Bbb R^n$. Prove that $V$ is a subspace of $\Bbb R^n$.
I'm sure this isn't that difficult, I just have a hard time understanding subspaces and the notation is slightly confusing as well.
 A: We see that $0 \in V$, as $A*0 = \lambda * 0 = 0$. Therefore, V is non-empty.
Now, let $x,y \in V$. Let $\alpha \in \mathbb{R}$.
Thus, $Ax = \lambda x$ and $ Ay = \lambda y$
Because $A(x+y) = Ax + Ay = \lambda x + \lambda y = \lambda(x+y)$, we conclude that $x + y \in V$
Because $A(\alpha x) = \alpha(Ax) = \alpha (\lambda x) = \lambda (\alpha x)$, we conclude that $\alpha x \in V$
Therefore, $V$ is closed under scalar multipliction and vector addition. Hence, $V$ is a subspace of $\mathbb{R}^n$.
A: You need to show that $V$ is closed under addition and scalar multiplication.
For instance: Suppose $v,w \in V$. Then $Av = \lambda v$ and $Aw = \lambda w$. Therefore:
$$ A(v+w) = Av + Aw = \lambda v + \lambda w = \lambda (v+w).$$
So $V$ is closed under addition.
Now let $c \in \mathbb{R}$. The last step, which I'll leave to you, is to show that $cv \in V$.

Edit: As StackTD pointed out, we should verify that $V \neq \emptyset$. Note that $0 \in V$.
A: I suppose you have seen the following definition or equivalent characterization of subspace. A set $V \subset \mathbb{R}^n$ is a (linear) subspace of $\mathbb{R}^n$ if the following criteria are met:


*

*$V$ is non-empty (usually checked as: does $V$ contain the zero vector?);

*$V$ is closed under addition and scalar multiplication, i.e.:


*

*for arbitrary $\vec x, \vec y \in V$, you need $\vec x + \vec y \in V$;

*for an arbitrary $\vec x \in V$ and scalar $\alpha$, you need $\alpha\vec x\in V$.




Don't let the notation of the given set $V = \left\{ \vec x \in \mathbb{R}^n \vert A\vec x = \lambda \vec x\right\}$ confuse you:


*

*check whether $\vec 0 \in V$: you have $A\vec 0 = \vec 0$, so ...

*take $\vec x,\vec y \in V$: if $A\vec x = \lambda \vec x$ and $A\vec y = \lambda \vec y$, then $A\left( \vec x + \vec y \right) = \ldots$

*take $\vec x \in V$ and $\alpha \in \mathbb{R}$: if $A\vec x = \lambda \vec x$ , then $A\left( \alpha\vec x \right) = \ldots$


Can you fill in the gaps?

Referring to lhf's comment:

$V = \ker (A-\lambda I)$. Does that help?

You may have seen that the kernel is always a (linear) subspace and note that $V$ can be seen as the kernel of $A-\lambda I$, and thus... This could be an elegant way, but the intended approach may be the 'direct one' from above.
