1. ## Help finding basis

Hi.

i have the following subspaces:

$W=Span\left \{ (0,2,1), (1,2,0) \right \}$

$U=Span\left \{ (5,4,-5),(0,1,0) \right \}$

i need to find the basis for $U+W$ and $U\cap W$ and to determine if $U+W$ is a direct sum of $\mathbb{R}$.

so, this is what i did to find the basis for $U+W$:
i set $(5,4,-5),(0,1,0),(0,2,1), (1,2,0)$ in one $3\times 4$ matrix and found its rank to be 3.
so i can just throw one of this vectors out, and it can be basis for $U+W$.
and obviously $U+W$ can't be direct sum of $\mathbb{R}$, since $U+W\neq \left \{ \vec{0} \right \}$ because $dim(U\cap W)=1>0$.

that all leaves me with 2 questions:

1. first of all: how do i find the basis for $U\cap W$?
2. is it enough to say that " $U+W$ can't be direct sum of $\mathbb{R}$, since $U+W\neq \left \{ \vec{0} \right \}$ because $dim(U\cap W)=1>0$" in order to prove that $U+W$ is not a direct sum of $\mathbb{R}$?

2. ## Re: Help finding basis

first of all, U+W is a direct sum if and only if: dim(U∩W) = 0 (which is the same thing as saying U∩V = {0}).

presumably you mean "direct sum decomposition of R3", but that is not what you wrote.

you cannot just arbitrarily "throw one vector out", you have to verify that the remaining 3 are linearly independent!

for example, {(1,0,0),(0,1,0),(0,0,1),(2,0,0)} is a linearly dependent set, but throwing out (0,1,0) won't fix this.

since you have deduced that dim(U∩V) = 1, we just need to find one non-zero vector in it, to use as a basis.

so suppose a(0,2,1) + b(1,2,0) = c(5,4,-5) + d(0,1,0), that is:

(b,2a+2b,a) = (5c,4c+d,-5c).

then b = 5c = -a, so:

2a+2b = 0 = 4c+d, so d = -4c = (4/5)(-5c) = (4/5)a. this means if we pick a, everything else falls into place. try a = -1.

3. ## Re: Help finding basis

Hi Deveno.

thanks for the great explanation.

Originally Posted by Deveno
you cannot just arbitrarily "throw one vector out", you have to verify that the remaining 3 are linearly independent!

for example, {(1,0,0),(0,1,0),(0,0,1),(2,0,0)} is a linearly dependent set, but throwing out (0,1,0) won't fix this.
would you say the best way to go is to throw away the vector that was eliminated during the gaussian elimination?
for example, in my case:

$\begin{bmatrix}v_1\\ v_2\\ v_3\\ v_4\end{bmatrix}\begin{bmatrix} 5 & 4 & -5\\ 1 & 2 & 0\\ 0 & 1 & 0\\ 0 & 2 & 1 \end{bmatrix}\rightarrow \begin{bmatrix} 5 & 4 & -5\\ 0 & -6 & -5\\ 0 & 1 & 0\\ 0 & 2 & 1 \end{bmatrix}\rightarrow \begin{bmatrix} 5 & 4 & -5\\ 0 & -6 & -5\\ 0 & 0 & -5\\ 0 & 0 & -2 \end{bmatrix}\rightarrow \begin{bmatrix} 5 & 4 & -5\\ 0 & -6 & -5\\ 0& 0 & -5\\ 0 & 0 & 0 \end{bmatrix}$

so $v_4(0,2,1)$ is the right vector to throw?

4. ## Re: Help finding basis

Originally Posted by Stormey
Hi Deveno.

thanks for the great explanation.

would you say the best way to go is to throw away the vector that was eliminated during the gaussian elimination?
for example, in my case:

$\begin{bmatrix}v_1\\ v_2\\ v_3\\ v_4\end{bmatrix}\begin{bmatrix} 5 & 4 & -5\\ 1 & 2 & 0\\ 0 & 1 & 0\\ 0 & 2 & 1 \end{bmatrix}\rightarrow \begin{bmatrix} 5 & 4 & -5\\ 0 & -6 & -5\\ 0 & 1 & 0\\ 0 & 2 & 1 \end{bmatrix}\rightarrow \begin{bmatrix} 5 & 4 & -5\\ 0 & -6 & -5\\ 0 & 0 & -5\\ 0 & 0 & -2 \end{bmatrix}\rightarrow \begin{bmatrix} 5 & 4 & -5\\ 0 & -6 & -5\\ 0& 0 & -5\\ 0 & 0 & 0 \end{bmatrix}$

so $v_4(0,2,1)$ is the right vector to throw?
We can easily see that the the 2 vectors spanning W are linerely independent, so taking one vector from U and verifying that this set is linearly independent, is akin to showing that
$\begin{bmatrix} 0 & 2 & 1 \\ 1 & 2 & 0 \\ 5 & 4 & -5 \end{bmatrix}$ has a non zero determinant, which is does. Also the other vector from U also works and would create an linearly independent set of vectors. But your way should also give the right answer

5. ## Re: Help finding basis

thanks jakncoke.