Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

up vote 6 down vote accepted

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.

share|improve this answer
    
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 $a+b \in V \cap W$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.