Main Theorem

THEOREM 2.5.1   [Existence and Non-existence] Consider a linear system $A {\mathbf x}= {\mathbf b},$ where $A$ is a $m \times n$ matrix, and $\; {\mathbf x}, \; {\mathbf b}$ are vectors with orders $n \times 1,$ and $m \times 1,$ respectively. Suppose ${\mbox{rank }}(A) = r$ and ${\mbox{rank}} ([A \; \;{\mathbf b}]) = r_a.$ Then exactly one of the following statement holds:

1. if $\; r_a = r < n,$ the set of solutions of the linear system is an infinite set and has the form
   $$\{ {\mathbf u}_0 + k_1 {\mathbf u}_1 + k_2 {\mathbf u}_2 + \cdots... ...n-r} {\mathbf u}_{n-r} \; : \;\; k_i \in {\mathbb{R}}, \; 1 \leq i \leq n-r \},$$
   where ${\mathbf u}_0, {\mathbf u}_1, \ldots, {\mathbf u}_{n-r}$ are $n \times 1$ vectors satisfying $A {\mathbf u}_0 = {\mathbf b}$ and $A {\mathbf u}_i = {\mathbf 0}$ for $1 \leq i \leq n-r.$
2. if $\; r_a = r = n,$ the solution set of the linear system has a unique $n \times 1$ vector ${\mathbf x}_0$ satisfying $A {\mathbf x}_0 = {\mathbf b}.$
3. If $\; r < r_a,$ the linear system has no solution.

Remark 2.5.2   Let $A$ be an $m \times n$ matrix and consider the linear system $A {\mathbf x}= {\mathbf b}.$ Then by Theorem 2.5.1, we see that the linear system $A {\mathbf x}= {\mathbf b}$ is consistent if and only if

$${\mbox{rank }}(A) = {\mbox{rank}} ([A \; \;{\mathbf b}]).$$

The following corollary of Theorem 2.5.1 is a very important result about the homogeneous linear system $A {\mathbf x}= {\mathbf 0}.$

COROLLARY 2.5.3   Let $A$ be an $m \times n$ matrix. Then the homogeneous system $A {\mathbf x}= {\mathbf 0}$ has a non-trivial solution if and only if rank$(A) < n.$

Proof. Suppose the system $A {\mathbf x}= {\mathbf 0}$ has a non-trivial solution, ${\mathbf x}_0.$ That is, $A {\mathbf x}_0 = {\mathbf 0}$ and ${\mathbf x}_0 \neq {\mathbf 0}.$ Under this assumption, we need to show that ${\mbox{rank}} (A) < n.$ On the contrary, assume that rank$(A) = n.$ So,

$$n = {\mbox{rank}}(A) = {\mbox{rank}} \bigl([A \;\; {\mathbf 0}]\bigr) = r_a.$$

Also $A {\mathbf 0}= {\mathbf 0}$ implies that ${\mathbf 0}$ is a solution of the linear system $A {\mathbf x}= {\mathbf 0}.$ Hence, by the uniqueness of the solution under the condition $r = r_a = n$ (see Theorem 2.5.1), we get ${\mathbf x}_0 = {\mathbf 0}.$ A contradiction to the fact that ${\mathbf x}_0$ was a given non-trivial solution.

Now, let us assume that rank$(A) < n.$ Then

$$r_a = {\mbox{rank}} \bigl( [A \;\; {\mathbf 0}] \bigr) = {\mbox{rank}} (A) < n.$$

So, by Theorem 2.5.1, the solution set of the linear system $A {\mathbf x}= {\mathbf 0}$ has infinite number of vectors ${\mathbf x}$ satisfying $A {\mathbf x}= {\mathbf 0}.$ From this infinite set, we can choose any vector ${\mathbf x}_0$ that is different from ${\mathbf 0}.$ Thus, we have a solution ${\mathbf x}_0 \neq {\mathbf 0}.$ That is, we have obtained a non-trivial solution ${\mathbf x}_0.$ height6pt width 6pt depth 0pt

We now state another important result whose proof is immediate from Theorem 2.5.1 and Corollary 2.5.3.

PROPOSITION 2.5.4   Consider the linear system $A {\mathbf x}= {\mathbf b}.$ Then the two statements given below cannot hold together.

1. The system $A {\mathbf x}= {\mathbf b}$ has a unique solution for every ${\mathbf b}.$
2. The system $A {\mathbf x}= {\mathbf 0}$ has a non-trivial solution.

Remark 2.5.5  

1. Suppose ${\mathbf x}_1, {\mathbf x}_2$ are two solutions of $A {\mathbf x}= {\mathbf 0}.$ Then $k_1 {\mathbf x}_1 + k_2 {\mathbf x}_2$ is also a solution of $A {\mathbf x}= {\mathbf 0}$ for any $k_1, k_2 \in {\mathbb{R}}.$
2. If ${\mathbf u}, {\mathbf v}$ are two solutions of $A {\mathbf x}= {\mathbf b}$ then ${\mathbf u}- {\mathbf v}$ is a solution of the system $A {\mathbf x}= {\mathbf 0}.$ That is, ${\mathbf u}- {\mathbf v}= {\mathbf x}_h$ for some solution ${\mathbf x}_h$ of $A {\mathbf x}= {\mathbf 0}.$ That is, any two solutions of $A {\mathbf x}= {\mathbf b}$ differ by a solution of the associated homogeneous system $A {\mathbf x}= {\mathbf 0}.$

   In conclusion, for ${\mathbf b}\neq {\mathbf 0},$ the set of solutions of the system $A {\mathbf x}= {\mathbf b}$ is of the form, $\{{\mathbf x}_0 + {\mathbf x}_h\};$ where ${\mathbf x}_0$ is a particular solution of $A {\mathbf x}= {\mathbf b}$ and ${\mathbf x}_h$ is a solution $A {\mathbf x}= {\mathbf 0}.$

EXERCISE 2.5.6  

1. For what values of $c$ and $k$ -the following systems have $i)$ no solution, $\;\; ii)$ a unique solution and $\;\; iii)$ infinite number of solutions.
   1. $x + y + z = 3, \; \; x + 2 y + c z = 4, \; \; 2 x + 3 y + 2 c z = k.$
   2. $x + y + z = 3, \; x + y + 2 c z = 7, \; x + 2y + 3 c z = k.$
   3. $x + y + 2z = 3, \; x + 2y + c z = 5, \; x + 2y + 4 z = k.$
   4. $k x + y + z = 1, \; x + k y + z = 1, \; x + y + k z = 1.$
   5. $x + 2 y - z = 1, \; 2 x + 3 y + k z = 3, \; x + k y + 3 z = 2.$
   6. $x - 2 y = 1, \; x - y + k z = 1, \; k y + 4 z = 6.$
2. Find the condition on $a, b, c$ so that the linear system
   $$x + 2y - 3 z = a, \; 2 x + 6 y - 11 z = b, \; x - 2 y + 7 z = c$$
   is consistent.
3. Let $A$ be an $n \times n$ matrix. If the system $A^2 {\mathbf x}= {\mathbf 0}$ has a non trivial solution then show that $A {\mathbf x}= {\mathbf 0}$ also has a non trivial solution.