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Subspace $S_1$ is generated by vectors $v_1=(1,0,4,1)$, $v_2=(0,1,1,1)$, $v_3=(1,3,7,4)$;

Subspace $S_2$ is generated by vectors $w_1=(1,1,3,4)$, $w_2=(1,0,0,-2)$, $w_3=(3,1,3,0)$.

Find the sum and the intersection of these two subspaces.

Solving attempt:

$S_1$: let $a,b,c,x,y,z,t$ be real numbers, so that $a(1,0,4,1) + b(0,1,1,1) + c(1,3,7,4)= (x,y,z,t)$.

This leads to the following system: $$ \begin{cases} a+c=x \\ b+3c=y \\ 4a+b+7c=z \\ a+b+4c=t \end{cases} $$ which I tried to solve by adding and substracting the equations: $$ \begin{cases} z-t-3x=0 \\ z-y-4x=0 \\ x+y-t=0 \\ \end{cases} $$

which gives the following solution: $$ \begin{cases} x=x \\ y=t-x \\ z=t+3x \\ t=t \end{cases} $$

This results in $S1=\{(x,t-x,t+3x,t) \mid x,t \in \mathbb R \}$

Same for $S_2$. I obtained: $S_2=\{(x,y,3y,t) \mid x,y,t \in \mathbb R\}$.

I am not sure about the results below:

\begin{align} S_1+S_2& = \{ (2x,t-x+y,t+3x+3y,2t) \mid x,y,t \in \mathbb R\}\\ S_1\cap S_2& = \{ (x,0,0,t) \mid x,t \in \mathbb R\} \end{align}

Is it correct? Very few information available online on this.

Thank you!

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  • $\begingroup$ It looks to me like you made a mistake in computing $S_2$. Note that $w_3=w_1+2w_2$, so you can simplify the system that you’re solving for that subspace by considering only two of the generators. $\endgroup$ – amd Dec 23 '15 at 19:30
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First, note that $v_3=v_1+3v_2$ and $w_3=w_1+2w_2$, so you can ignore these two vectors. Carrying them along in the computations won’t change the answer, but why make things more complicated than necessary?

$S_1$ consists of all linear combinations of $v_1$ and $v_2$ and $S_2$ is all linear combinations of $w_1$ and $w_2$, so their sum will consist of all possible linear combinations of these four vectors. In general, you can find this space by computing the column space of the matrix with these vectors as its columns, which in this case will give you the identity matrix. In this instance, though, you can take a shortcut: $w_1$ and $w_2$ are linearly independent of $v_1$ and $v_2$, which is easily seen by inspection, so the sum of their spans must be the entire vector space $\mathbb R^4$.

The intersection of $S_1$ and $S_2$ consists of all vectors that can be expressed both as a linear combination of $v_1,v_2$ and of $w_1,w_2$. Since we’ve already shown that these pairs are mutually linearly independent, the only vector that’s in both spaces is $\mathbf 0$, i.e., $S_1\cap S_2=\{\mathbf 0\}$. More generally, you would need to find scalars $a_1$, $a_2$, $b_1$, $b_2$ such that $a_1v_1+a_2v_2=b_1w_1+b_2w_2$, which would lead to another system of linear equations to solve.

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  • $\begingroup$ This makes perfect sense. Thank you! $\endgroup$ – Michael Dec 23 '15 at 20:30

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