Determine complex bases for S, T, S ∩ T and S + T Assuming that S and T are subspaces of $V_3(\Bbb C)$ generated by
$\langle(1, 1, 0),(i, 1 + i, 1),(1 + i, 1 + i, 0)\rangle  $
and $\langle(1, 0, 1),(i, −i, 0),(0, i, i)\rangle$, respectively.   
Determine $\Bbb C$-bases for $S, T, S \cap T$ and $S + T$.
Anyone seen any examples like this online anywhere? Not too sure where to start and struggling to find sources online. 
 A: It would probably be easiest to start off finding bases for $S$ and $T$. In vector spaces, any generating set contains a basis. There may just be some elements that are linearly dependent on the rest, but after they are all removed, we should be able to work down to a linearly independent set.
You can achieve this by using row reduction on a matrix whose rows are the generating set. Let me show you for $S$:
$\begin{bmatrix}1&1&0\\i&1+i&1\\1+i&1+i&0\end{bmatrix}$
By subtracting $i+1$ times the first row from the last row, we get
$\begin{bmatrix}1&1&0\\i&1+i&1\\0&0&0\end{bmatrix}$
Then by subtracting $i$ times the first row from the second, you're at
$\begin{bmatrix}1&1&0\\0&1&1\\0&0&0\end{bmatrix}$
Now this is at echelon form, and so we can see that the first two rows are linearly independent, and that the original third row was dependent on the first two. We can conclude that $S$ is two dimensional, and generated by the rows of this matrix. You see, by backtracking our row operations, we can express all three of the original generators in $S$ by these two rows of the reduced matrix. Try the same thing for the generators of $T$. I'm going to supply you with my answer (which may not match yours!) so that I can keep going below with the intersection of the two spaces.
The particular way I reduce the matrix for $T$, I get:
$\begin{bmatrix}1&0&1\\0&1&1\\0&0&0\end{bmatrix}$, so that $[1,0,1]$ and $[0,1,1]$ form a basis for $T$.
Doing $S+T$ will probably be the next easiest thing. Just lump the two bases you got for $S$ and $T$ together and perform the same analysis as we did above. You will get down to a basis for $S+T$. I'd like to leave this work for you to try.
$S\cap T$ might be the most challenging. One way to do it would be to take a look at what the elements of $S$ look like, and the elements of $T$ look like, and see what vectors satisfy both pictures. Using the basis I got for $S$ above, I know that everything in $S$ looks like $\alpha[1,1,0]+\beta[0,1,1]=[\alpha,\alpha+\beta,\beta]$. Using my basis for $T$ above, everything in $T$ looks like $\gamma[1,0,1]+\delta[0,1,1]=[\gamma,\delta,\gamma+\delta]$.
Something in $S\cap T$ would have to look like $[\alpha,\alpha+\beta,\beta]=[\gamma,\delta,\gamma+\delta]$. Then $\alpha=\gamma$, $\alpha+\beta=\delta$ and $\beta=\gamma+\delta$, and we can attempt to solve for $\alpha,\beta$.
You should find that $\beta=\delta-\alpha=\delta+\alpha$ implies $\alpha=0$, and so we are looking at $[0,\beta,\beta]$. This is clearly the $1$ dimensional space generated by $[0,1,1]$, which conveniently is in both of our bases for $S$ and $T$. This shows that the intersection is $1$ dimensional.
