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I tried the following : $x = - y - z + w $ so we can express every vector $(x, y, z, w)$ as $(x, y, z, w) = (-y -z + w, y, z, w) = y(-1, 1, 0, 0) + z(-1, 0, 1, 0) + w(1, 0, 0, 1)$

so the basis vectors are : $(-1, 1, 0, 0)^T$, $(-1, 0, 1, 0)^T$, $(1, 0, 0, 1)^T$

Does this make sense? Is that really the basis?

Mavi
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1 Answers1

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It makes sense. Right now you are saying that any vector on the hyperplane is a linear combination of these $3$ vectors. It is indeed the basis once you verify that they are linearly independent.

Although you will later realize that verifying linear independence in these kinds of questions is not required. Linear independence is implied because if we assume$$y(-1, 1, 0, 0) + z(-1, 0, 1, 0) + w(1, 0, 0, 1)=(0,0,0,0)$$then we know the LHS is $(x,y,z,w)$. Thus we directly get $(x,y,z,w)=(0,0,0,0)$ or $y=z=w=0$, implying linear independence.

Shubham Johri
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