If I solve a system and get infinite many solutions, will it be linearly independent?

Question

irishboy123 · Answer

other way round:(

dumptybaby · Answer

So for it to be linearly independent it has to only have trivial solutions (nothing else) right?

irishboy123 · Answer

I wouldn't normally comment on stuff like this because I have a working rather than theoretical knowledge of it,......., but you do seem to be in bit of a pickle :(

Summary:


For nxn matrix A, and nx1 vectors $\mathbf x$ and $\mathbf b$ in system $A \mathbf x = \mathbf b$, then if A has a determinant $A^{-1}$, it follows that there is a non trivial solution: $A^{-1} A \mathbf x = A^{-1} \mathbf b , \implies \mathbf x = A^{-1} \mathbf b $

A hole bunch of other stuff follows too, especially the fact that the column vectors of A will be **linearly independent** and the row vectors of A will be **linearly independent**.  And that might be what you are interested in, maybe.

If A is non-invertible, then yes there is something wrong with your system.  When the determinant is zero you can an infinite number of, or no solutions.  Here is a simple and pretty crap 2x2 example.  

In the first, you are solving $y = 2x + 1, y = 2x + 1$, infinite solutions:

$\left( \begin{matrix} 2 & -1  \ 2 & -1 \end{matrix} \right) \left( \begin{matrix} x  \ y \end{matrix} \right) = \left( \begin{matrix} -1  \ -1 \end{matrix} \right)$

In the second, you are solving $y = 2x + 1, y = 2x + 2$, no solutions:
$\left( \begin{matrix} 2 & -1  \ 2 & -1 \end{matrix} \right) \left( \begin{matrix} x  \ y \end{matrix} \right) = \left( \begin{matrix} -1  \ -2 \end{matrix} \right)$

The determinant tells you something is wrong.  And look at the row and column vectors.


All that said, I am more familiar with **homogeneous** systems, $A \mathbf x = \mathbf 0 $, where you may want there to be no inverse as in the case of systems of linear differential equations.  Otherwise you could say the same thing, namely that $ \mathbf x = A^{-1} \mathbf 0 \implies \mathbf x= \mathbf O$ and you're stuffed with a trivial solution.  This is where eigenvalues kick in.  They avoid trivial solutions but denying the matrix an inverse.

irishboy123 · Answer

PS This is the back page of Gilbert Strang's book. He presents this as Lin Alg for nxn matrices "In a Nutshell". And if he thinks that, it is worth taking seriously :) |dw:1481722893589:dw| His online MIT course is legendary but I doubt you have the time to watch much of it as he goes through in splendid detail. and the URL is case that doesn't show up too well https://i.gyazo.com/a436f7c20d713c1a61b2f10a11ff4c36.png Good luck