#Type/Definition #Topic/Linear_Algebra #calc

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Properties: 12.8 Solution Set to Square Linear-Systems Problem
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The Square Linear-Systems Problem

Let $n \in N$ . Let $A \in R^{n \times n}$ be a "given" square, nonsingular matrix. Let $\vec{b} \in R^{n}$ be a "given" vector. Then, the square linear-systems problem (AKA the NLSP) is to find an unknown vector $\vec{x} \in R^{n}$ such that

A \cdot \vec{x} = \vec{b}

Note that this is a backwards problem.

Remark. For now, we can think of a nonsingular matrix as having linearly independent columns. In other words, the individual columns of $A$ are unable to be created from linear combinations of the other columns.
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Relation to Matrix-Vector Multiplication Problem

Let the function $f : R^{n} \mapsto R^{N}$ such that $f (\vec{x}) = A \cdot \vec{x}$ .

\begin{aligned} Domain (f) & = R^{n} \\ Codomain (f) & = R^{n} \\ Rng (f) & = {\vec{b} \in R^{n} : \vec{b} = A \cdot \vec{x} for some \vec{x} \in R^{n}} \end{aligned}

Recall that an equivalent for the range of our 10 Matrix-Vector Multiplication function $f$ can be also be given by

Rng (f) = Span {A (:, k)}_{k = 1}^{n}

Tldr

We say $\vec{b}$ is in the range of $f$ iff we can write $\vec{b}$ as a linear combination of the columns of matrix $A$ .

In contrast, when solving the square linear-systems problem, we need to create our square matrix $A$ and a vector $\vec{b}$ . We then need to work backwards to calculate all possible vectors $\vec{x}$ such that

A \vec{x} = \vec{b}

or conclude that no $\vec{x}$ exists. In other words, the linear-systems problem is the inverse of the matrix-vector multiplication problem.