Finding the basis from two subspaces

Question

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?

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?