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We have $T=span\bigg\{\begin{bmatrix}2&0\\ 1&3\end{bmatrix}\begin{bmatrix}1&9\\ 6&2\end{bmatrix}\begin{bmatrix}2&4\\ 1&1\end{bmatrix}\bigg\}$

And $S=span\bigg\{\begin{bmatrix}-7&1\\ 2&-6\end{bmatrix}\begin{bmatrix}3&3\\ -4&-2\end{bmatrix}\begin{bmatrix}2&-8\\ 1&7\end{bmatrix}\bigg\}$

My task is to find the dimension and basis of $S \cap T$ and $S+T$

Rewriting T as $T=\begin{bmatrix}2&1&2\\ 0&9&4\\ 1&6&1\\ 3&2&1\end{bmatrix}$ and calculating the rank of that matrix we get that the dimension

of T is 3, also doing the same with S we get that the dimension of S is 3 as well.

Now instead of doing any calculation, can I just conclude since both $T$ and $S$ span the same vector space, that $S=T$ and $S+T=S=T$, thus also giving me the intersection as $S \cap T=S=T$

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    equality of dimensions does not guarantee that the spaces are same. For example in $\mathbb{R}^2$, take $S=\{(x,0)\}$ and $T=\{(0,y)\}$. Both have dimension $1$ but spaces are very different. Moreover $S \cap T=\{(0,0)\}$ and $S+T=\mathbb{R}^2$. – 2017-01-08
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    I see, but here it turned out to be the case, so I came to the wrong conclusion. EDIT: Also is my conclusion that S and T span the same vector space wrong? Since that's what I based by answer on. – 2017-01-08

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The formula $\\ \dim(S + T) = \dim(S) + \dim(T) - \dim(S \cap T)$ is whatcha wanna use here. Since ya already calculated the dimension of S and T, all ya gotta do is calculate the dimension of $S + T$, and from the above formula this will tell you for the dimension of the intersection.

Now to compute the dimension of $S + T$, just recognize $S + T$ is the span of the 6 vectors that span S and T together. So using this, just compute the rank of the standard 4 x 6 matrix you get from those 6 vectors to get the dimension of $S + T$.

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    That's what I have done... but in this case my question was since both S and T span the same vector space, could I have skipped the calculation ? – 2017-01-08
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    If you know that S = T, then yes, you can say that $S + T = S \cap T = S = T.$ However, just because they have the same dimension does NOT tell ya they are the same SPACE. To verify they're the same space, you have to check that each vector that is spanned by S (or T) is a linear combination of the vectors that span T (or S respectively). – 2017-01-08