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Let $u$ be the $3\times 1$ matrix $$\begin{bmatrix} 0\\ 4\\ 4 \end{bmatrix}$$

and $A$ the $3\times2$ matrix: $$\begin{bmatrix} 3 & -5\\ -2 & 6\\ 1 & 1 \end{bmatrix}.$$

Here's where the title comes in: is $u$ in the plane $\mathbb R^3$ spanned by the columns of $A$? Why or why not? I know that it is, I just don't know why. This is the chapter before we learn about linear dependence and independence, so I doubt it has to do with either of these, beyond that I haven't a clue.

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I tried $5/3a_1 + a_2$. Then I tried $5a_1 + 3a_2$. It became obvious. Alternatively, see the answer below where you actually solve a system to find the linear combination. A nice alternative is to find the linear combination of $a_1, a_2$ that gives you $(0, 1, 0)$ and the linear combination that gives you $(0, 0, 1)$. – user2468 Sep 4 '12 at 21:53

Assuming that the column vectors of your matrix $A$ can be written with $v=\begin{bmatrix}3&-2&1\end{bmatrix}^T$ and $w=\begin{bmatrix}-5&6&1\end{bmatrix}^T$ where $A=[v\quad w]$ then your (linear independant) vectors $v$ and $w$ span obviously one plane $P=\operatorname{span}\{v,w\}\subset\mathbb{R}^3$ with $\dim(P) = 2$.

Assume that there is one vector $x\in P$. $x$ has then the following form $$x=\lambda v+\mu w\in P$$ and therefore can be written as a linear combination of the two vectors $v,w$ that span $P$. If you want to check whether $u=\begin{bmatrix}0&4&4\end{bmatrix}^T$ is in $P$ you have to find $\lambda,\mu$ such that the combination of the vectors $v,w$ equals $u$. If there is no solution, then the vector is not part of the plane.

Basically you solve the following system of equations:


Hint: This system has a unique solution.

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You should try to write $u$ as $u = Ax$ when $x$ is a 2d vector. If you can find $x$, then the answer is yes. (What is more difficult is to show that it's impossible when it is actually impossible.)

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simply compute the rank of the the augmented matrix [A u]. It tells is the number of linearly independent columns. If rank is 2, then u is in the plane spanned by columns of A.

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You also need to show that columns of $A$ are linearly independent, but that's kind of obvious I guess. – Tunococ Sep 4 '12 at 21:47

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