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Let H and K be subspaces of a vector space $V$. The intersection of $H$ and $K$, , is the set of $v$ in $V$ that belong to both $H$ and $K$. Show that the intersection of $H$ and $K$ is a subspace of $V$. Give an example in $R^2$ to show that the union of two subspaces is not, in general, a subspace. enter image description here

I know that in order to prove that a set is a subspace of a vector space, we need to prove that it contains the zero vector, closed under multiplication and addition. I know that the zero vector exists in the intersection from the graph above.

And it seems to me that the intersection of $H$ and $K$ is along the z-axis (i think...) so i am assuming that i can take any two vectors along the z-axis (0,0,1), (0,0,2)... and add it together to prove it is closed under addition, and the same logic applies to using any scalar and multiplying with any vector along the z-axis to prove it is closed under multiplication

Is this the right approach? or did i make a mistake somewhere. Any help will be appreciated.

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You should prove things in general, i.e. using the properties that $H$ and $K$ are subspaces to prove that $H \cap K$ is a subspace.

$0 \in H$ and $0 \in K$, hence $0 \in H \cap K$.

Now let $u \in H \cap K$ and $v \in H \cap K$, justfity that $u +v \in H \cap K$.

Similarly verify for scalar multiplication.

As for counter example for the union, just have to find one in $\mathbb{R}^2$, consider $2$ different straight lines that passes through the origin, what can you say about their union? which property fails?

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  • $\begingroup$ For scalar multiplication, if we let $x$ be any vector in $H \cap K$ and $c$ be any scalar, then $cx \in H \cap K$ because $H$ and $K$ are subspaces of $V$ and any scalar multiplication involving the vectors will always result in another vector in the subspaces. Is this a proper verification for scalar multiplication? $\endgroup$ – Soon_to_be_code_master Apr 5 '18 at 6:08
  • $\begingroup$ yes, perhaps slowing down to illustrate that you understand is good for beginner. $x \in H \cap K$, then $x \in H \land x \in K$, then we have $cx \in H \land cx in K$ due to closure of scalar multiplication of $H$ and $K$, hence $cx \in H \cap K$. $\endgroup$ – Siong Thye Goh Apr 5 '18 at 6:11
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The figure shows that the line passes through the origin and a line that passes through the origin is a subspace.

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