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## Homework Statement

Prove if set A is a subspace of R4, A = {[x, 0, y, -5x], x,y E ℝ}

## Homework Equations

## The Attempt at a Solution

Now I know for it to be in subspace it needs to satisfy 3 conditions which are:

1) zero vector is in A

2) for each vector u in A and each vector v in A, u+v is also in A

3) for each vector u in A and each scalar a in ℝ, a*vector u is also in A

Now the problem I have is proving it...

Do I need to prove the 3 conditions above by substituting values for x and y?

Example:

1) letting x and y = 0, we get a zero vector, therefore vector 0 is in A (check)

2) let vector u have x = 1, y = 2 and vector v have x = 3, y =4

u + v = [1, 0, 2, -5] + [3, 0, 4, -15] = [4, 0, 6, -20] (which is also in A; check)

3) let a = 2, and vector u having x = 1, y = 3, so 2[1, 0, 3, -5] = [2, 0, 6, -10] (which is also in A; check)

The reason I am confused is because I have seen methods where the values are multiplied by the entire vector within the set, like...

1) 0*vector u = vector 0

2) u = b[itex]_{1}[/itex]u[itex]_{1}+...+[/itex]b[itex]_{k}[/itex]*u[itex]_{k}[/itex] and

and then you would have v = c[itex]_{1}[/itex]v[itex]_{1}+...+[/itex]c[itex]_{k}[/itex]*v[itex]_{k}[/itex], then u + v = (b[itex]_{1}[/itex]+c[itex]_{1}[/itex])u[itex]_{1}[/itex]+...+(b[itex]_{k}[/itex]+c[itex]_{k}[/itex])u[itex]_{k}[/itex]

3) and etc.

But the second method only applies to spans? Which can only be used if you know set A is a subspace of R4?

Anyways, any clarification would be appreciated, thanks

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