This problem involves finding the basis of the union and intersection of two subspaces.

We have $V$ and $U$ which are subspaces of $\mathbb R^4$

$$V = \begin{Bmatrix} {(x_1, x_2, x_3, x_4) : x_1 + x_2 = x_3 + x_4}\end{Bmatrix}$$ $$U = span \{ \begin{bmatrix}0\\0 \\1 \\1 \\ \end{bmatrix}, \begin{bmatrix}3\\0 \\1 \\1 \\ \end{bmatrix}, \begin{bmatrix}0\\-1 \\2 \\-1 \\ \end{bmatrix} , \begin{bmatrix} 0\\ 1 \\ 0 \\ 3 \\ \end{bmatrix} \} \\ $$

We want to find the dimension and basis for:

$a)\text{ } U + V$
$b)\text{ } U \cap V$

My attempt: Let me first try to find the column space of U and a basis for V The $$rref(U) = \begin{bmatrix} 1 & 0 & 0 & 2\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & -1\\ 0 & 0 & 0 & 0\\ \end{bmatrix}$$ Since only the first three columns have pivot elements, only the first three rows of $U$ make up the column space: $\begin{bmatrix}0\\0 \\1 \\1 \end{bmatrix}, \begin{bmatrix}3\\0 \\1 \\1 \end{bmatrix}, \begin{bmatrix}0\\-1 \\2 \\-1\end{bmatrix}$

Now let's find a basis for $V = \begin{Bmatrix} {(x_1, x_2, x_3, x_4) : x_1 + x_2 = x_3 + x_4}\end{Bmatrix}$. I did this by just making vectors that satisfied the constraint until I couldn't anymore. If this is the wrong way to do so, please let me know!

The vectors I found were: $\begin{bmatrix}1\\1 \\2 \\0 \end{bmatrix}, \begin{bmatrix}1\\1 \\1 \\1 \end{bmatrix}, \begin{bmatrix}1\\1 \\0 \\2\end{bmatrix}$.

So I know we want to find the the dimension and basis for $U + V$. I think what I should do now is find the linearly independent vectors of all seven vectors above. The number of vectors is going to be the dimension and the basis is just the independent vectors. Correct?

How should I go about doing the same for $U \cap V$. I'm stumped by the "intersect" symbol. Do they just want me to list the basis of vectors that the two have in common?

  • $\begingroup$ Alternatively for $V$ if we use column vectors with $x=(x_1,x_2,x_3,x_4)$ then $(1,1,-1,-1)^Tx=0$ so this is the halfspace orthogonal to $(1,1,-1,-1)$. Since it has a one dimensional nullspace you should expect the halfspace to be three dimensional. This shows that once you find three linearly independent vectors in that space, using guess and test or other means, you're done. $\endgroup$ Oct 5, 2020 at 1:21

1 Answer 1

  • One nitpick: one usually says "a basis" rather than "the basis," since there is not a unique basis for a given subspace.
  • Your approach for $U+V$ will work.
  • For the intersection: note that any member of $U$ can be written as a linear combination of the three basis elements of $U$ that you found. So, every element of $U$ is of the form $(3b, -c, a+b+2c, a+b-c)$ where $a,b,c$ are scalars. If this also belongs to $V$, then we must have $3b-c = 2a+2b+c$ i.e. $b = 2(a+c)$. So elements of $U \cap V$ are of the form $(6a+6c, -c, 3a + 4c, 3a + c)$. Can you take it from here?
  • $\begingroup$ I think so... I have a question about the subspace of $V = \begin{Bmatrix} {(x_1, x_2, x_3, x_4) : x_1 + x_2 = x_3 + x_4}\end{Bmatrix}$. How would I go about finding a basis for V? I think my technique worked, but it's flawed in that I don't know when I'm done if I don't have $n$ vectors in $R^n$ For the case of V, I wasn't sure if I had a full basis because I didn't have 4 vectors. $\endgroup$ Oct 5, 2020 at 1:16
  • $\begingroup$ I think I have something for the intersection. As you mentioned, elements of $U \cap V$ are of the form $(6a+6c, -c, 3a + 4c, 3a + c)$. If we put this into a $4 \times 3$ matrix and row reduce. We'll find that the matrix only has two independent columns. Those columns form the basis of $U \cap V$. Correct? $\endgroup$ Oct 5, 2020 at 1:53
  • $\begingroup$ @MatthewEngelstein $V$ is $3$-dimensional (the one linear constraint $x_1 + x_2 = x_3 + x_4$ "takes away" one dimension from $4$), so I think your basis is fine. And I think the approach in your second comment is correct. $\endgroup$
    – angryavian
    Oct 5, 2020 at 2:49
  • $\begingroup$ How do we know that the constraint "takes away" one dimension from four? Is it because we can rewrite the equation isolating one of the variables? $\endgroup$ Oct 5, 2020 at 2:56
  • $\begingroup$ @MatthewEngelstein $V$ can be described as "vectors orthogonal to $(1,1,-1, -1)$." So $V$ and $\text{span}\{(1,1,-1,-1)\}$ are orthogonal complements of each other, and their dimensions add up to $4$. $\endgroup$
    – angryavian
    Oct 5, 2020 at 4:00

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