# Finding vector from sum of vector subspaces

Let $B=\begin{Bmatrix}v_1, v_2, v_3, v_4\end{Bmatrix}$ base of a vector space $V$

And $S=\left \langle v_1 + 2v_2, v_2 + v_3 \right \rangle$ and $T=\left \langle v_3 + v_4, v_1 + v_2 + v_4 \right \rangle$

Find a vector $v \in S + T$ so that $v \notin S$ and $v \notin T$

I know $S + T = \langle v_1 + 2v_2, v_2 + v_3, v_3 + v_4, v_1 + v_2 + v_4 \rangle$, but I can't figure out how to find the vector the problem asks.

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Firstly, the vector is not unique. Secondly, what vectors have you looked at? –  mixedmath Sep 25 '12 at 15:53

There are many ways to solve this. I would try the following strategy:

1. Find a vector $a \in S, a \notin T$.
2. Find a vector $b \notin S, b \in T$.
3. Let $v = a + b$. Just by the properties above, we can be sure that $v \in S + T$, but $v \notin S$ and $v \notin T$ (why?).

Let's get to work on finding $a$ and $b$ then:

For $a$ we can take, for example, $a = v_2 + v_3$. We can see that this vector cannot be generated by the basis we have for $T$ (why? hint: what form do vectors in $T$ take, and what happens to that form if we force the coefficients of $v_2$ and $v_3$ to be $1$?).

For $b$ we can take, for example, $b = v_3 + v_4$. This time it's even easier to see why $b \notin S$; vectors in $S$ cannot have a nonzero coefficient for $v_4$ (look at your basis).

Thus, the vector $v = v_2 + 2v_3 + v_4$ is a possible solution to your question.

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Thanks Yoni! I've faced the problem this way and the answer was finally correct. –  Lucas Sep 29 '12 at 4:17

You can take for example the vector $$w=(v_{1}+2v_{2})+(v_{2}+v_{3})+(v_{3}+v_{4})+(v_{1}+v_{2}+v_{4})$$ Try wo write this vector as a linear combination of vector from S and then from T. You gonna see that this is impossible.

In fact, there is a lot of vectors satisfying this property. Draw a picture of the problem and you gonna understand it better.

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Thank you so much for your time Kaye! I'm gonna take into account this advice, I found out that writing everything really helps a lot! –  Lucas Sep 29 '12 at 4:19