$A:=\begin{bmatrix}
2 & 6 & 0 \\
3 & 1 & 3 \\
1 & 0 & 0 \\
4 & 8 & 1
\end{bmatrix}$
I've found the basis for the column space by doing row reduction (i.e. basis is just the columns vectors of A in this case), and the null space only has the trivial solution.
Question
Find a basis for $B = \{b \in R^4 \ | \ Ax = b \ has \ no\ solution\}$.
Is there a quick way of doing this? I know I can augment A with $b = (b_1, b_2, b_3, b_4)^T$, do row reduction, and then look at the row of zeros, but that seems quite laborious?
Best Answer
When you multiply a matrix by a (column) vector, you get a vector that is a linear combination of the columns of the matrix, where the coefficients/weights come from the components of the column vector. I'm too lazy to illustrate with a $4\times 3$ example, so here's a $2\times 2$ example: $$Ax=\begin{pmatrix} \mathbf{A_1} & \mathbf{A_2}\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=x_1\mathbf{A_1}+x_2\mathbf{A_2}$$ where $\mathbf{A_1}$ is the first column of the matrix $A$ and $\mathbf{A_2}$ is the second column.
In other words, $Ax$ is a linear combination of the columns of $A$. But the column space of $A$, i.e. the image of $A$, is just the span of the columns of $A$. So the equation $Ax=b$, which says that $b$ is a linear combination of the columns of $A$, will have a solution if and only if $b$ is in the column space of $A$.