Prove that if $S$ and $T$ are subspaces of $V$, then so is $S\cap T$

Question

I am asked to prove that

If $S$ and $T$ are subspaces of $V$, then so is $S\cap T$.

I have been thinking about this for a while but remain perplexed. I tried this:

Let $\{e_1, e_2, ..., e_k\}$ be a basis for $S$ and let $\{u_i, u_2, ..., u_k\}$ be a basis for $T$.
Let all $u_n=e_n$ be denoted $d_n$.
$\therefore$$\{d_1, d_2, ..., d_p\}$ is a basis for $S \cap T$ if $S\cap T$ is a linear space.
Now all elements $x$ in $S$ have form $x=\sum^{k}_{i=1}c_ie_i$ for some $c_i \in \Bbb R$ and all elements $y$ in $T$ have form $y=\sum^k_{i=1}c_iu_i$ for some $c_i \in \Bbb R$.
Now $S\cap T$ contains all elements of $S$ and $T$ such that $x=y$.
$\therefore \sum^k_{i=1}c_ie_i=\sum^k_{i=1}c_iu_i$.
$\therefore e_i=u_i$.
$\therefore S\cap T$ is a subspace with basis $\{d_1, d_2, ..., d_p\}$.

I am not sure that this proof is valid, nor that it even really proves anything? Is there any merit to it? Otherwise, what would be a better approach to proving this?

Answer 1

1) Since S, T are subspaces of V, the unique zero vector is contained in both, and so the zero vector is in the intersection.

2) Let $a,b \in S \cap T$. That means a, b are in both, S and T. Since $S,T$ are subspaces, they are closed under addition and so $a+b$ is in S and in T and so in the intersection.

3) Let c be an element of the reals. Let a be in the intersection of $S,T$. Since, $S,T$ are both subspaces, It follows that $ca$ is in both and therefore in the intersection.

Answer 2

You can't assume that for any given basis $B$ of $M$ and subspace $N$ of $M$ that there is a subset of $B$ that is a basis for $N$.

For example, let $V = \mathbf{R}^3$ and $S = \{(x,y,z) : x = 0\}$ and $T = \{(x,y,z) : y = 0\}$. Now I choose the bases

$$ \left\{ (0,1,1), (0,1,-1) \right\} \text{ and } \left\{ (1,0,1), (1,0,2) \right\} $$

for $S$ and $T$. Notice that I have chosen these bases in such a way that neither contains a basis for $S \cap T$.

For an example element of $S \cap T$, we have $(0,0,1)$ and we write this in each basis as

$$ \frac12(0,1,1) - \frac12(0,1,-1) = -(1,0,1) + (1,0,2) \tag{1}$$

this shows that your equation

$$ \sum^k_{i=1}c_ie_i=\sum^k_{i=1}c_iu_i $$

doesn't work because the coefficients on the left in $(1)$ are different than those on the right.