# If $\mathbf{V}$ and $\mathbf{W}$ are subspaces of $\mathbb{R}^{n}$, prove that so is $\mathbf{V} \cap \mathbf{W}$

The first step is clear -- the zero vector is in both, so it will be in the intersection, too. Scalar multiplication is pretty obvious too -- if the vector is in the intersection, it can be represented as some linear combination of vectors in $V$ = some linear combination of vectors in $W$. Multiplying both by $k$ preserves the equality. How can we prove that it is closed under addition, too? I can think of a 'cheat' way, saying that the intersection of $V$ and $W$ is the kernal of the matrix with spanning vectors of $V$ and $W$, and by properties of kernal, etc. but I was looking for a more simply-formulated solution. Thanks!

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You are overcomplicating things. There is absolutely no need to think about linear combinations.

If $\mathbf{x}\in\mathbf{V}\cap\mathbf{W}$, then $\mathbf{x}\in\mathbf{V}$ and $\mathbf{x}\in\mathbf{W}$. If $k$ is a scalar, then $k\mathbf{x}\in\mathbf{V}$ (because $\mathbf{V}$ is a subspace, and $\mathbf{x}$ is in $\mathbf{V}$), and likewise $k\mathbf{x}\in\mathbf{W}$ because $\mathbf{x}\in\mathbf{W}$ and $\mathbf{W}$ is a subspace. Since $k\mathbf{x}$ is in both $\mathbf{V}$ and $\mathbf{W}$, then it is in $\mathbf{V}\cap\mathbf{W}$.

For sums, do the same thing: take $\mathbf{x}$ and $\mathbf{y}$ which are both in $\mathbf{V}\cap\mathbf{W}$. Then $\mathbf{x}$ and $\mathbf{y}$ are both in each of $\mathbf{V}$ and in $\mathbf{W}$. Since $\mathbf{V}$ is a subspace, and $\mathbf{x}$ and $\mathbf{y}$ are both in $\mathbf{V}$, then so is $\mathbf{x}+\mathbf{y}$. Now show that $\mathbf{x}+\mathbf{y}$ is also in $\mathbf{W}$ to conclude that it must be in $\mathbf{V}\cap\mathbf{W}$, which is what you want.

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I was overcomplicating indeed, thanks! –  LinAlgStudent Feb 5 '11 at 23:16

Let $a,b \in V \cap W$. Then $a+b \in V$ and $a+b \in W$. Hence $a+b \in V \cap W$.

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PEV: Should one understand that a subset of a vector space is a vector subspace as soon as it is stable by addition? I wonder if you realize that this is exactly what your answer hints at. –  Did May 28 '11 at 19:27