How to correctly convert vectors to a matrix

Question

I am confused about a sample solution to a problem (please read until the end, since I am not looking for a solution to the problem itself)

We were asked if the vector

$\begin{bmatrix} -3 \\ 4\\ 7 \end{bmatrix}$

can be written as a linear combination of

$\begin{bmatrix}1 \\ 2\\ 1 \end{bmatrix}$,$\begin{bmatrix}-2 \\ 1 \\ 3 \end{bmatrix}$, $\begin{bmatrix}4 \\ 3\\ -1 \end{bmatrix}$

Clearly this can be solved by a matrix $(A|b)$ The solution that was given to us solved it this way:

$\begin{bmatrix} 1 & -2 & 4 \\ 2 & 1 & 3 \\ 1 & 3 & -1 \end{bmatrix} * \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} -3 \\ 4\\ 7 \end{bmatrix}$

But I would say that this is the wrong way to convert these vectors to a matrix. I would have written it down like this:

My trial: $\begin{bmatrix} 1 & 2 & 1 \\ -2 & 1 & 3 \\ 4 & 3 & -1 \end{bmatrix} * \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} -3 \\ 4\\ 7 \end{bmatrix}$

So I am not looking for a solution for the problem itself - I just want to have cleared out why the sample solution uses a wrong ( so I suggest) built matrix. Is there a reason I haven't been awared of yet? Or is it just a mistake.

! I do not want to know if the starting vector can be written as a linear combination of the other ones. I am able to solve this on my own. I want to know which way to solve it is the correct one. Handed out a sample solution which I suggest is wrong is confusing me.

Answer 1

The sample solution is actually right. Let $\mathbb{F}$ be some field, and $V$ be a vector space over $\mathbb{F}$. When we say we want to find $x\in V$ such that it's a solution to $Ax=v$ for some $v\in V$, we actually think about $v$ as a vector in the Image, or Range of the linear transformation represented by the matrix $A$. The Image is actually spanned by the Column space of the matrix- That be, linear combinations of the columns, and not the rows- so if $v \in Im(A)$ it should be a linear combination of the columns of the matrix. This can be seen by writing the matrix-vector product explicitly for $c_1,...,c_n\in\mathbb{F}$ and $A\in M_{n\times m}(\mathbb{F})$

$$ A\left[\begin{array}{c}{c_{1}} \\ {\vdots} \\ {c_{n}}\end{array}\right]=\left[\begin{array}{ccc}{a_{11}} & {\cdots} & {a_{1 n}} \\ {\vdots} & {\ddots} & {\vdots} \\ {a_{m 1}} & {\cdots} & {a_{m n}}\end{array}\right]\left[\begin{array}{c}{c_{1}} \\ {\vdots} \\ {c_{n}}\end{array}\right]=\left[\begin{array}{ccc}{c_{1} a_{11}+} & {\cdots} & {+c_{n} a_{1 n}} \\ {\vdots} & {\vdots} & {\vdots} \\ {c_{1} a_{m 1}+} & {\cdots} & {+c_{n} a_{m n}}\end{array}\right]=c_{1}\left[\begin{array}{c}{a_{11}} \\ {\vdots} \\ {a_{m 1}}\end{array}\right]+\cdots+c_{n}\left[\begin{array}{c}{a_{1 n}} \\ {\vdots} \\ {a_{m n}}\end{array}\right] $$

Notice that after the second equation we have a vector, not a matrix. In your case it's something like

$$ \left[\begin{array}{ccc}{1} & {-2} & {4} \\ {2} & {1} & {3} \\ {1} & {3} & {-1}\end{array}\right]\left[\begin{array}{l}{a} \\ {b} \\ {c}\end{array}\right]=\left[\begin{array}{c}{1 a+(-2 b)+4 c} \\ {2 a+1 b+3 c} \\ {1 a+3 b+(-1) c}\end{array}\right]=a\left[\begin{array}{c}{1} \\ {2} \\ {1}\end{array}\right]+b\left[\begin{array}{c}{-2} \\ {1} \\ {3}\end{array}\right]+c\left[\begin{array}{c}{4} \\ {3} \\ {-1}\end{array}\right] $$

We get a linear combination of the columns, not the rows.

Answer 2

We are looking for existence of solutions of A·x=b.

We first decompose the matrix A into its columns: A = ($a_1$, . . . , $a_n$), where $a_1$, . . . , $a_n$ are column vectors. Then we rewrite the matrix-vector equation:

A·x=$($a_1$, . . . , $a_n$)·($x_1$,...,$x_n$)^T$= $x_1$·$a_1$+···+$x_n$·$a_n$ = b. Thus we have written b as a linear combination of the column vectors of A. (If you expand A·x of the solutions and collect all the a,b and c together you will see that it is indeed true).