If we have a set of vectors $(v_1, \ldots, v_n)$ that are linearly independent, then it means the only solution to $a_1v_1 + \ldots a_nv_n = 0$ is the trivial solution of $a_1 = \ldots = a_n = 0$.
For linear independence of equations, (which is inhomogenous) like the one below,
\begin{align}
x_1 \begin{bmatrix}a_{11}\\a_{21}\\ \vdots \\a_{m1}\end{bmatrix} +
x_2 \begin{bmatrix}a_{12}\\a_{22}\\ \vdots \\a_{m2}\end{bmatrix} +
\cdots +
x_n \begin{bmatrix}a_{1n}\\a_{2n}\\ \vdots \\a_{mn}\end{bmatrix}
=
\begin{bmatrix}b_1\\b_2\\ \vdots \\b_m\end{bmatrix}
\end{align}
we have linear independence if none of the equations can be written in terms of others. My questions is what the connection is between linear independence of vectors versus equations. It seems that for vectors, it is just the special case of a homogeneous system of equations since the constant terms are zero. Is there more to it? Thanks.
Best Answer
Let $A$ be the matrix
$$ A=\left(\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\\ \end{array}\right) $$
and let $b$ the vector
$$ b=\left(\begin{array}{c} b_1\\ b_2\\ \vdots\\ b_n \end{array}\right). $$
Trivially the space of solutions of your problem is the space
$$ S = \left\{x\in\mathbb{R}^n| A\cdot x=b\right\} $$
the system as a unique solution if and only if $det(A)\neq 0$ which is true if and only if the vector columns of $A$ are linearly independent which is equivalent to say that none of the equations of the system can be written in terms of others. For the case $b=0$ since $A$ is linear and then $A\cdot 0=0$, if the solution is unique, it can be only the zero vector, so $x_1=0,...,x_n=0$.