Given two subspaces of $R^4$

$W = \operatorname{Span}( (1,1,0,-1), (1,2,3,0) )$

$U = \operatorname{Span}( (1, 2, 2, -2), (2, 3, 2, -3) )$

Find the cartesian equations of the intersection.

I would solve this by putting all the vectors in a matrix as columns and the coordinates x,y,z,t on the side and doing gaussian elimination (this follows straight from the definition).

Instead I read a very strange (to me) solution to this problem.

Given that

$$ \det \left(\begin{bmatrix}1&1&1\\1&2&2\\0&3&2\end{bmatrix}\right) = -1 \neq 0$$

and that the last vector is the sum of the first and the third, a base of the sum is given by the first 3 vectors, so we can find cartesian equations for the sum by setting:

$$ \det \left(\begin{bmatrix}x&y&z&t\\1&1&1&-1\\1&2&2&0\\0&3&2&-2\end{bmatrix}\right) = 0.$$

No more explanation is given could you please explain me how this procedure works?

  • $\begingroup$ My guess: you found the the vectors are linearly independent and therefore span $\mathbb{R}^3$ (those they are a basis) for $x,y,z,t$ to be in the span we want them to be linear depdenet in the there vectors, asking for $0$ determinant P.S you take only 3 elements of the coordinates of the vector and not all the given 4? $\endgroup$
    – gbox
    Jan 29, 2018 at 14:32
  • $\begingroup$ @gbox that is the problem I don't understand why given solution does not take all 4 coordinates $\endgroup$ Jan 29, 2018 at 15:13
  • 1
    $\begingroup$ Interesting I will keep follow too $\endgroup$
    – gbox
    Jan 29, 2018 at 15:18
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    $\begingroup$ As it says, the fourth vector is a linear combination of two of the others, so if you were to include it with them in any sort of determinant, you’d always get zero. $\endgroup$
    – amd
    Jan 30, 2018 at 2:17
  • $\begingroup$ You essentially want to solve for the null space. So, rref(A) when $A=augment((1,1,0,-1)a, (1,2,3,0)b, -(1,2,2,-2)c, -(2,3,2,-3)d)=\vec{0}$, and find your values that way. $\endgroup$
    – W. G.
    Feb 1, 2018 at 14:37

1 Answer 1


The proposed solution is simply wrong. As both $W$ and $U$ are two-dimensional, $W\cap U$ is at most two-dimensional. However, the determinantal equation in the proposed solution is a single linear equation in four unknowns. So, what it specifies is a three-dimensional subspace, which cannot possibly be equal to $W\cap U$.

In fact, since the first three vectors in the specifications of $W$ and $U$ are linearly independent and the fourth vector is a linear combination of the first and the third, we know right away that $W\cap U$ is the span of the first vector $(1,1,0,-1)^T$ (and you need a system of three linear equations to specify this span, such as $x=y,\,x=-t$ and $z=0$). However, $(1,1,0,-1)$ is not even a solution to the determinantal equation. So, you see how wrong that proposed solution is.

Edit. I've just noticed that there is a mismatch between your question title (which is about the sum $W+U$) and the question body (which is about the intersection $W\cap U$). Anyway, the proposed solution is still wrong when $W+U$ is concerned. We have $(1,1,0,-1)\in W\subseteq W+U$, but the LHS of the determinantal equation (which amounts to $2x+4y-5z+t=0$) is $5$ when $(x,y,z,w)=(1,1,0,-1)$.

  • $\begingroup$ I quess that however it is sum (like in the tittle) not intersection ... $\endgroup$
    – Widawensen
    Feb 2, 2018 at 9:41

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