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Problem 73: Prove that the subset $W = \{(x,2x,3x)|x \in \R\}$ of $\R^3$ is a subspace of $\R^3$.

Problem 73: Prove that the subset $W = \{(x,2x,3x)|x \in \R\}$ of $\R^3$ is a subspace of $\R^3$.

Solution:
$W = \{(x,2x,3x)|x \in \R\}$.
zero check:
If we take $x = 0$, we see that $(0,0,0) \in W$ , so W is non-empty.
Now let $u = (x,2x,3x)$ and $v = (y,2y,3y)$ be any two elements of W for $x, y \in \R$.Then
Closure under Addition:
$u+v = (x,2x,3x) + (y,2y,3y) = (x+y,2x+2y,3x+3y)$
$((x+y),2(x+y),3(x+y)) = (z,2z,3z) \in W$
if $z=x+y$
So, $u+v \in W$
Hence,W is closed under addition.
Closure under Scalar Multiplication:
Let $k \in \R$ and $u=(x,2x,3x) \in W$
Then, $ku = (kx,2kx,3kx)$
which shows that $ku \in W$
Therefore W is a subspace of $\R^3$

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