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Consider the set $W$ of all vectors $(x_1, x_2, x_3, x_4)\in \Bbb R^4$ satisfying:

$3x_1 − 2x_2 − x_3 − 4x_4 = 0$

$x_1 + x_2 − 2x_3 − 3x_4 = 0$

  1. Show that $W \subset \Bbb R$
  2. Find a basis for $W$.

I'm really struggling to grasp this. I know what you need to show to prove a set is a subspace. But I'm having issues showing that it's closed under Vector Addition and Scalar Multiplication.

And I don't really know how to find a basis, I know that it should span the set W and be Linearly Independent, but how do I find it.

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  • $\begingroup$ I'm able to prove that W is not empty, further that I'm having issues. $\endgroup$ Jul 12, 2017 at 12:22
  • $\begingroup$ And if anyone could give me some guidelines on how to exactly find bases, that would be appreciated. $\endgroup$ Jul 12, 2017 at 12:22
  • $\begingroup$ Can you find the solution set for the above system of equations? $\endgroup$
    – Naive
    Jul 12, 2017 at 12:23
  • $\begingroup$ About the closure, note that if we have a two solutions of the system, then the linear combinations of the solutions are again a solution of the system. $\endgroup$
    – Our
    Jul 12, 2017 at 12:28
  • $\begingroup$ I can solve the system of equations yes. $\endgroup$ Jul 12, 2017 at 12:33

2 Answers 2

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I'm really struggling to grasp this. I know what you need to show to prove a set is a subspace. But I'm having issues showing that it's closed under Vector Addition and Scalar Multiplication.

It's not hard, but it's a matter of realizing what exactly you have to show and then writing it up nicely. Since the zero vector clearly satisfies both (homogeneous) equations, $W$ is not empty. Suppose you have two elements $(a_1,a_2,a_3,a_4) \in W$ and $(b_1,b_2,b_3,b_4) \in W$; then verify:

  • $(a_1,a_2,a_3,a_4)+(b_1,b_2,b_3,b_4) \in W$
  • $(ka_1,ka_2,ka_3,ka_4)\in W$ for $k \in \mathbb{R}$ arbitrary.

Because the two equations are linear and homogeneous, this should be easy. Now for the basis:

I've solved the system yes and gotten the various elements in terms of others (given that we have 2 equations and 4 unknowns, some elements will be written in terms of others)

Depending on which variables you chose (to solve for, in terms of the others), you should have a solution set of the form (choosing $x_1,x_2 \in \mathbb{R}$): $$\left\{ \begin{array}{l} x_3 = -x_1+2x_2 \\ x_4 = x_1-x_2 \end{array}\right.$$ This means that any element of $W$ can be written in the form: $$\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\x_4 \end{pmatrix} = \begin{pmatrix}x_1 \\ x_2 \\ -x_1+2x_2 \\ x_1-x_2 \end{pmatrix} = x_1\color{blue}{\begin{pmatrix}1 \\ 0 \\ -1 \\ 1 \end{pmatrix}} + x_2\color{blue}{\begin{pmatrix}0 \\1 \\ 2 \\-1 \end{pmatrix}}$$ So any element of $W$ can be written as a linear combination of the two vectors in blue. Note (and check) that they are linearly independent.

Can you take it from here?

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  • $\begingroup$ Thank you very much. This clears up much of the confusion. $\endgroup$ Jul 12, 2017 at 14:19
  • $\begingroup$ @ArshadMonsterAbrahams Alright, you're welcome! $\endgroup$
    – StackTD
    Jul 12, 2017 at 14:49
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Since there are two equation and there are independent you should be able to observe that then dimension of $W$ is 2.
Put $x=x_1$ and $y=x_2$ (this is only for better understanding). So we have two equations, $$ 3x-2y=x_3+4x_4$$ $$ x+y=2x_3+3x_4$$ From these two equations solve for $x_3$ and $x_4$ in terms of $x$ and $y$ we get $$ x_3=2y-x$$ $$ x_4=x-y$$ Hence we can see that $(x_1,x_2,x_3,x_4)=x*(1,0,-1,1)+y*(0,1,2,-1)$.
Therefore $\{(1,0,-1,1),(0,1,2,-1)\}$ form a basis of $W$

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