How do you represent constants in a spanning set of vectors?

Question

Let's say I have linear system A$\vec{x}=\vec{b}$ where $$ A = \begin{pmatrix} 1 & 3 \\ 0 & 0 \\ 0 & 0 \\ \end{pmatrix} $$ and $$ \vec{b} = \begin{pmatrix} 4 \\ 0 \\ 0 \\ \end{pmatrix}. $$ I then determine the solution set of this system to be: $$ \begin{pmatrix} x \\ y \\ \end{pmatrix} = \left\{ \left( \begin{array} {c} 4-3y \\ y \end{array} \right) % \begin{array} {|c} \\ \\ \\ \end{array} % \begin{array} {c} y \in \mathbb{R} \end{array} \right\} $$ where x is a dependent variable and y is a free variable.

If I also wanted to represent this solution set as a spanning set, I first tried: $$ =span \left\{ \begin{pmatrix} 4 \\ 0 \\ \end{pmatrix} , \begin{pmatrix} -3 \\ 1 \\ \end{pmatrix} \right\} $$ however this implies that you can do a linear combination of these vectors with a scalar, let's say $\alpha\begin{pmatrix}x \\y \\\end{pmatrix}+\beta\begin{pmatrix}x \\y \\\end{pmatrix}$ but what I really want is the first vector to be constant and the second vector to be scaled by some arbitrary scalar.

I thought of doing $\alpha\begin{pmatrix}1 \\0 \\\end{pmatrix}$ where $\alpha = 4$ but I really don't want to state something like this and it doesn't really make too much sense to do so in the context of spanning.

What is the proper way to represent this information? A constant vector amidst a span of vectors you do intend to combine linearly.

Answer 1

Span of a set of vectors forms a subspace, hence it must contain the zero vector.

However, the zero vector is clearly not a solution to the system of equation, hence the solution set can't be written as a span of a set of vectors.

It is however, an affine subspace,

$$\begin{pmatrix} 4 \\ 0\end{pmatrix} + \operatorname{Span} \left\{ \begin{pmatrix} -3 \\ 1\end{pmatrix}\right\}.$$

Answer 2

Whenever you have a linear system of equations $A\vec{x} = \vec b$, and you have found

• a particular solution $\vec{x}_p$, which means $A \vec{x}_p = \vec b$ (in your case $\vec{x}_p = \begin{pmatrix} 4 \\ 0\end{pmatrix})$ and
• a basis $B =\{\vec b_1,\ldots , \vec b_r\}$ of the solution set of $A\vec{x} = \vec{0}$ (in your case this is only one vector $\vec b_1 = \begin{pmatrix} -3 \\ 1\end{pmatrix}$),

then you can write the solution of $A\vec{x} = \vec b$ as

• $\vec{x}_p + span B$ (in your case $\begin{pmatrix} 4 \\ 0\end{pmatrix} + span\left\{\begin{pmatrix} -3 \\ 1\end{pmatrix} \right\}$)