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Homework Statement
Let
[itex](**)[/itex]
\begin{matrix}
a_{11}x_1 & + & \ldots & + & a_{1n}x_n & = 0 \\
\vdots & & & & \vdots\\
a_{m1}x_1 & + & \ldots & + & a_{mn}x_n & = 0 \\
\end{matrix}
be a system of [itex]m[/itex] linear equations in [itex]n[/itex] unknowns, and assume that
[itex]n > m[/itex]. Then the system has a non-trivial solution.
[itex]Proof:[/itex]
Consider first the case of one equation in [itex]n[/itex] unknowns, [itex]n > 1[/itex]:
[itex]a_{1}x_1 + \ldots + a_{n}x_n = 0[/itex]
If all coefficients [itex]a_1, \ldots, a_n[/itex] are equal to [itex]0[/itex], then any value of the variables
will be a solution, and a non-trivial solution certainly exists. Suppose
that some coefficient [itex]a_i \neq 0[/itex]. After renumbering the variables and the
coefficients, we may assume that it is [itex]a_i[/itex] Then we give [itex]x_2, \ldots, x_n[/itex] arbitrary
values, for instance we let [itex]x_2, \ldots, x_n = 1[/itex] and solve for [itex]x_1[/itex] letting
[itex]x_1 = \frac {-1}{a_1} (a_2 + \ldots + a_n)[/itex]
In that manner, we obtain a non-trivial solution for our system of equations.
Let us now assume that our theorem is true for a system of [itex]m - 1[/itex]
equations in more than [itex]m - 1[/itex] unknowns. We shall prove that it is true
for [itex]m[/itex] equations in [itex]n[/itex] unknowns when [itex]n > m[/itex]. We consider the system
[itex](**)[/itex].
If all coefficients [itex](a_{ij})[/itex] are equal to [itex]0[/itex], we can give any non-zero value
to our variables to get a solution. If some coefficient is not equal to [itex]0[/itex],
then after renumbering the equations and the variables, we may assume
that it is [itex]a_{11}[/itex]. We shall subtract a multiple of the first equation from the
others to eliminate [itex]x_1[/itex]. Namely, we consider the system of equations
[itex] (A_2 - \frac {a_{21}}{a_{11}} A_1) \cdot X [/itex]
[itex] (A_m - \frac {a_{m1}}{a_{11}} A_1) \cdot X [/itex]
Which can be written also in the form
[itex](***)[/itex]
[itex]A_2 \cdot X - \frac {a_{21}}{a_{11}} A_1 \cdot X = 0[/itex]
[itex]\vdots[/itex]
[itex] A_m \cdot X - \frac {a_{m1}}{a_{11}} A_1 \cdot X = 0[/itex]
In this system, the coefficient of [itex]x_1[/itex] is equal to [itex]0[/itex]. Hence we may view
[itex](***)[/itex]as a system of [itex]m - 1[/itex] equations in [itex]n - 1[/itex] unknowns, and we have
[itex]n-1 > m-1[/itex].
According to our assumption, we can find a non-trivial solution
[itex](x_2, ... ,x_n)[/itex] for this system. We can then solve for [itex]x_1[/itex] in the first equation,
namely
[itex]x_1 = \frac {-1}{a_{11}} (a_{12}x_2 + \ldots + a_{1n}x_n)[/itex].
In that way, we find a solution of [itex]A_1 \cdot X = 0[/itex]. But according to [itex](***)[/itex], we
have
[itex]A_i \cdot X = \frac {a_{i1}}{a_{11}} A_1 \cdot X[/itex]
for [itex]i = 2, \ldots ,m[/itex]. Hence [itex]A_i \cdot X = 0[/itex] for [itex]i = 2, \ldots ,m[/itex], and therefore we have found a non-trivial solution to our original system [itex](**)[/itex]. The argument we have just given allows us to proceed stepwise from
one equation to two equations, then from two to three, and so forth.
This concludes the proof.
Homework Equations
The Attempt at a Solution
"We shall subtract a multiple of the first equation from the others to eliminate [itex]x_1[/itex]."
Why do we eliminate [itex]x_1[/itex] here? I mean we do that when solving non-general systems of equations. But here it looks like we could obtain the non-trivial solutions even with [itex](A_2 - A_1) \cdot X = 0[/itex]. Is it because subtracting the first equation from the second without eliminating any variables would just be a random, arbitrary operation?
Also, when solving the general system of equations, why do we eliminate only 1 variable, namely, [itex]x_1[/itex], as opposed to non-general cases where we can usually eliminate more than 1 variable? I can see how eliminating only [itex]x_1[/itex] brings us to the solution in the proof, but is that the only reason?
Thanks.