Let \begin{align*} S&=\langle(1,0,k,k),(3,−2,−1,k)\rangle \\ H&=\{x \in \Bbb{R}^4 \mid 3x_1 + x_2 − 2x_3 − kx_4=0\}. \end{align*} Find all $k \in \Bbb{R}$ such that $S \subset H$.
Okay... I'm a bit confused with linear transformations, I feel like they don't teach them very well. There I have two Subspaces from what I see.
The first thing I did was try to find the basis of $H$..
I have $H\{\langle(1,-3,0,0),(0,2,1,0),(0,k,0,1)\rangle\}$
and now ? I don't know where I'm in the subspace.
I made a drawing so that you understand what I ask.
and sorry if I draw like a child, but I need to understand well why.
Finding a basis for $H$ is not what you want to be doing. Deciding if $\langle w_1, \ldots, w_m \rangle \subset \langle v_1, \ldots, v_n\rangle$ is not nearly as simple as testing to see if $\{w_1, \ldots, w_m \} \subset \{ v_1, \ldots, v_n\}$. Non-trivial finite-dimensional vector spaces have many bases (infinitely many when they're real or complex), and two bases for the same space might have no vectors in common.
What you should do is see when the two spanning vectors of $S$ belong to $H$. Let's call these vectors $v$ and $w$. If we can see that $v, w \in H$, then because $H$ is a subspace, we know that $a v + b w \in H$. That is, every element of $\langle v, w \rangle = S$ belongs to $H$, i.e. $S \subset H$. Conversely, of course, if either of the vectors don't belong to $H$, then neither does all of $S$.
How do we see when these vectors belong to $H$? We just have to check when they each satisfy the definition of $H$. In particular, $$(1, 0, k, k) \in H \iff 3(1) + (0) - 2(k) - k(k) = 0 \iff 3 - 2k - k^2 = 0.$$ Similarly, $$(3, -2, -1, k) \in H \iff 3(3) + (-2) - 2(-1) - k(k) = 0 \iff 9 - k^2 = 0.$$ So, $S \subset H$ if and only if $3 - 2k - k^2$ and $9 - k^2 = 0$ at the same time! If you solve both quadratics, you'll find this happens if and only if $k = -3$.