How do I show that this vector is a linear combination of column vectors of matrix?

Question

Given some $Ax = y$, I want to show $y$ is a linear combination of column vectors of $A$ if the equation has a solution.

I'm thinking along the lines of finding eigenvectors but this might not be necessary.

If there is a solution, should i find $det(Ax - y)$ and show that it is non-zero? How should I proceed?

Answer 1

You can write $Ax$ as

$$ \begin{pmatrix} a_{11} & a_{12}\cdots & a_{1n} \\ \vdots & a_{22}\cdots & a_{2n} \\ a_{m1} & \cdots & a_{mn} \\ \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}=x_1\begin{pmatrix} a_{11} \\ \vdots \\ a_{1m} \end{pmatrix} + x_2\begin{pmatrix} a_{21} \\ \vdots \\ a_{2m} \end{pmatrix} + \cdots + x_n\begin{pmatrix} a_{n1} \\ \vdots \\ a_{nm} \end{pmatrix} $$

Therefore, if $Ax=y$, then $y$ is a linear combination of $A$'s column vectors.

Answer 2

experiment with $$ A = \left( \begin{array}{cc} 1 & 100 \\ 10 & 1000 \end{array} \right) $$ and calculate $Ax$ with, say, $$ x = \left( \begin{array}{c} 3 \\ 7 \end{array} \right) $$

Then calculate $Ax$ with $$ x = \left( \begin{array}{c} -13 \\ 4 \end{array} \right) $$

Answer 3

Consider the following examples,

$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = 1 \begin{bmatrix} a \\ b\end{bmatrix} + 2 \begin{bmatrix} c \\ d \end{bmatrix}$$

$$ \begin{bmatrix} a & b & c\\ d & e & f \\ g & h & i \end{bmatrix} \begin{bmatrix} 4 \\ -1 \\ 11 \end{bmatrix} = 4 \begin{bmatrix} a \\ d \\ g \end{bmatrix} -1 \begin{bmatrix} b \\ e \\ h \end{bmatrix} + 11 \begin{bmatrix} c \\ f \\ i \end{bmatrix}$$

Answer 4

Your statement holds by definition. Always $Ax$ is a linear combination of the columns of $A$. Now if $x$ is a solution. Then it holds that $Ax = y$.