Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

3 Answers

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:

$$\begin{bmatrix}3&-5\\-2&6\\1&1\end{bmatrix}\cdot\begin{bmatrix}\lambda\\\mu\end{bmatrix}=\begin{bmatrix}0\\4\\4\end{bmatrix}$$

Hint: This system has a unique solution.

share|improve this answer
add comment

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.)

share|improve this answer
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.