# Thread:

1. ## 2 proofs regarding linear transformations and operators and subspaces

1.) Prove that $l^1(\mathbb{R}^w) = \{(\beta\subscript_1, \beta\subscript_2, \beta\subscript_3.......) \in \mathbb{R}^w : \sum\limits_{n = 0}^\infty |\beta\subscript_n| < \infty \}$ is a subspace of $\mathbb{R}^w$

2.) Prove that if $\phi: V -> W$ is a linear transformation, the the set $Null(\phi) = \{v \in V: \phi(v) = 0\}$ is a subspace of V.

Okay, so I really know very little about these things. (I'm also currently enrolled in a non proof base linear algebra class)

I know (see read) that W is a subspace of V iff:
1.) W is nonempty
2.) $\alpha, \beta \in \mathbb{R}$ and $w\subscript_1, w\subscript_2 \in W$ always implies $\alpha w\subscript_1 \oplus \beta w\subscript_2 \in W$

I also get the definition of a linear operator (but latex takes along time for me) I can try to type it out if it helps anyone.

Is the Null function significant here? I've never seen it. Any other hints or words of advice will be greatly appreciated. I hope the class goes back to differential equations soon, although I'm guessing this all will be significant.

Thanks!

2. You're definitely on the right track. As you said, just show that elements of $\ell^1$ are closed under addition and scalar multiplication. This will imply that zero is in there (as long as one element is there--then add it to itself multiplied by -1).

For the other one, just use the property of linearity: if f is linear, then f(a*x)=a*f(x) and f(x+y)=f(x)+f(y). That will show that the null set (usually called the kernel) is closed under addition and scalar multiplication. The kernel of a linear map is always a subspace.
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