In the vector space $\mathbb{R}^3$, $S$ is a subspace of all the vectors with the first coordinate $0$, and $T$ is a subspace generated by the vectors $(1,1,1)$ and $(2,3,0)$. Find the subspace $S \cap T$.
Can someone help me with this problem? I have an exam tomorrow and I can't quite understand this problem.
On
Hint:
A generic element from the second subspace can be written as $$\vec v=t(1,1,1)+u(2,3,0)$$
Find for which values of $t$ and $u$ the first coordinate vanishes.
On
Here a general method: set $S = \mathrm{Span}\{(0,1,0),(0,0,1)\}$ and $T = \mathrm{Span}\{(1,1,1),(2,3,0)\}$. A vector in the intersection can be written as a linear combination in two ways as follows: $$a(0,1,0)+b(0,0,1) = c(1,1,1)+d(2,3,0).$$ This equality implies $$ \begin{cases} -c-2d = 0 \\ a-c-3d = 0 \\ b-c = 0, \end{cases} $$ namely $b = c, d = -\frac{c}{2}$ and thus $a = -\frac{c}{2}$. So the intersection is given by a one-dimensional space containing vectors of the form $$a(0,1,0)+b(0,0,1) = -\frac{c}{2}(0,1,0)+c(0,0,1) = c\left(0,-\frac{1}{2},1\right), c \in \mathbb{R}.$$
Take basis of both subspaces:
$$S=\text{Span}_{\Bbb R}\left\{\,\begin{pmatrix}0\\1\\0\end{pmatrix}\;,\;\;\begin{pmatrix}0\\0\\1\end{pmatrix}\right\}\;,\;\;T=\text{Span}_{\Bbb R}\left\{\,\begin{pmatrix}1\\1\\1\end{pmatrix}\;,\;\;\begin{pmatrix}2\\3\\0\end{pmatrix}\;\,\right\}$$
and now form the $\;3\times4\;$ corresponding matrix and reduce it by tows:
$$\begin{pmatrix} 1&1&1\\ 2&3&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\longrightarrow\begin{pmatrix} 1&1&1\\ 0&1&\!-2\\ 0&1&0\\ 0&0&1\end{pmatrix}\longrightarrow\begin{pmatrix} 1&1&1\\ 0&1&\!-2\\ 0&0&2\\ 0&0&1\end{pmatrix}\longrightarrow\begin{pmatrix} 1&1&1\\ 0&1&\!-2\\ 0&0&2\\ 0&0&0\end{pmatrix}$$
From the above we get $\;\dim \left(S+T\right)=3\;$, and thus $\;\dim(S\cap T)=1\;$ Why? Be sure you can explain/justify all this).
Observe also that
$$\begin{pmatrix}2\\3\\0\end{pmatrix}-2\begin{pmatrix}1\\1\\1\end{pmatrix}\in S\ldots$$
Try to take it from here...