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$W = \{(x_{1}, x_{2}, x_{3}, x_4)\in R^4: x_1+x_3=x_4\}\\$ $$$$ I'm a bit confused about how to get the basis. I worked it out in terms of $x_4$ and got $$ x_4\left[\begin{array}{c}1 \\ 0 \\ 1 \\ 1 \end{array}\right] - x_3\left[\begin{array}{c}1 \\ 0 \\ 0 \\ 0 \end{array}\right] + x_2\left[\begin{array}{c}0 \\ 1 \\ 0 \\ 0 \end{array}\right] - x_1\left[\begin{array}{c}0 \\ 0 \\ 1 \\ 0 \end{array}\right]$$

So would the basis just be $$\left[\begin{array}{c}1 \\ 0 \\ 1 \\ 1 \end{array}\right] \left[\begin{array}{c}1 \\ 0 \\ 0 \\ 0 \end{array}\right] \left[\begin{array}{c}0 \\ 1 \\ 0 \\ 0 \end{array}\right] \left[\begin{array}{c}0 \\ 0 \\ 1 \\ 0 \end{array}\right]$$

samir91
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2 Answers2

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The subspace $W$ consists of all vectors $\mathbf{x}$ that satisfy: $$ x_{1} +x_{3} = x_{4} \Leftrightarrow x_{1} +x_{3} -x_{4}= 0 \Leftrightarrow \begin{bmatrix} 1& 0 & 1 & -1 \end{bmatrix} \mathbf{x} = 0. $$ That is, $W$ contains all vectors that are in the nullspace of $\begin{bmatrix} 1& 0 & 1 & -1 \end{bmatrix}$. All you have to do now, is find a basis for the nullspace: Finding the basis of a null space

Community
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megas
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Well, you know that solutions are of the form $(x_1, x_2, x_3, x_4)$, conditioned to

$$x_1 + x_3 = x_4$$

Hence, you'd know that solutions are of the form $(x_1, x_2, x_3, x_1 + x_3)$, which means you can rewrite it as

$$(x_1, x_2, x_3, x_1 + x_3) = x_1(1, 0, 0, 1) + x_2(0, 1, 0, 0) + x_3(0, 0, 1, 1)$$

Hence, notice that any vector that is a solution is a linear combination of

$$\langle(1, 0, 0, 1), (0, 1, 0, 0), (0, 0, 1, 1)\rangle$$

Notice that such vectors are linearly independent, which means that it is also a basis for $W$

Misguided
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