Can someone please explain why this theorem is true?
Theorem: If A is the matrix of coefficients of a system of linear equations, then the system has a solution if and only if the rank of the augmented matrix is equal to the rank of the matrix A?
When does a matrix have a non-trivial solution
linear algebramatrices
Related Solutions
As Andre says this is not true. On the other hand if your question is why is the reason for which a system (of $n$ equation and $n$ unknowns) with complete rank has always a unique solution. In my opinion the better way to see it is by linear maps. Let $n$ be fixed and select a collection of $a_{i,j}\in \mathbb{F}$ for $(i,j) \in \{\ 1, \ldots,n\} \times \{\ 1, \ldots,n\}$. Define the linear map $f$
$$(x_1 \ldots,x_n)\longmapsto \bigg(\sum_{1\le j\le n} a_{1,j}x_j \ldots\sum_{1\le j\le n} a_{n,j}x_j \bigg)$$
When we say that the rank is complete, we say that for all $(c_1 \ldots,c_n)$ always exists a $(x_1 \ldots,x_n)$ such that $f(x_1 \ldots,x_n)=(c_1 \ldots,c_n)$, i.e.,
$$\bigg(\sum_{1\le j\le n} a_{1,j}x_j \ldots\sum_{1\le j\le n} a_{n,j}x_j \bigg)=(c_1 \ldots,c_n)$$
With really means that $\sum_{1\le j\le n} a_{i,j}x_j =c_j$ for our system.
Since the $\text{rank }f=n$. Then by the dimention formula we can conclude that its kernel is trivial, i.e., $\ker f = \{0\}$. Then $f$ is an injective map. So we know that always exists a solution because is surjective (the rank has the same dimension as the target space) and also this solution is unique since is one-to-one.
Using our above notation. What happens when the rank is less than $n$. Then there exists some values in the target space which are "outside" of the image and in these cases there is no solution. Since the rank is less than $n$ so the linear map cannot be surjective and exists vectors in $\mathbb{F}^n \backslash f(\mathbb{F}^n)$.
For the question: when does it have an infinite number of solution? Assuming again that the rank is less than $n$. Let $(c_1, \ldots ,c_n) \in f(\mathbb{F^n})$ (a vector in the image), because is in the image there is a vector in the domain that under $f$ is mapped in $(c_1, \ldots ,c_n)$, let call it $x=(x_1, \ldots, x_n)$. But also we know that the kernel is not trivial, again using the dimension formula we conclude $\dim(\ker f)= n-\text{rank} f>0$. Then there is some vector $(z_1, \ldots, z_n) \not= 0$ such that $f(z_1, \ldots, z_n)= (0, \ldots, 0)$.
Now consider $(x_1,\ldots, x_n)+k(z_1, \ldots, z_n)$, where $k\in \mathbb{F}$ and see what happens under $f$.
\begin{align}f((x_1,\ldots, x_n)+k(z_1, \ldots, z_n))=f(x_1,\ldots, x_n)+kf(z_1, \ldots, z_n)\\ =f(x_1,\ldots, x_n)+0= (c_1 \ldots, c_n) \end{align}
This occurs because $f$ is linear and the vector $(z_1, \ldots, z_n)$ is in the kernel (is zero under $f$). Thus $(x_1,\ldots, x_n)+k(z_1, \ldots, z_n)$ is also a solution. And indeed the set
$$x+\ker f: = \{x+k: x=(x_1,\ldots,x_n)+k, \text{ and } k\in \ker f\}$$
by the same argument as above contain all the solutions. In that case there is a infinite number of solutions.
I like to think of $a$ and $b$ as being real numbers I don't happen to know.
Your task is to write something like "There is a unique solution if and only if [some conditions on $a$ and $b$]" and so on for the other cases.
You can determine the number of solutions from the row echelon form (found via Gaussian elimination, i.e., row operations).
It might not be consistent (no solutions). This occurs if and only if there is a row such as $[0\ 0\ 0\ 1]$ in the row echelon form (which is equivalent to the equation $0=1$).
There might be a unique solution. In this case, there will be three leading entries (one in each row) in row echelon form.
There might be a one-parameter solution. In this case, there will be two leading entries (and thus a row of zeroes) in row echelon form.
And so on.
A "one-parameter solution" is when the solution space is a line. As a more simple example $$ \left( \begin{array}{cc|c} 1 & -1 & 0 \\ 2 & -2 & 0 \\ \end{array} \right) $$ has infinitely many solutions: $(x,y)=(t,t)$ for all $t \in \mathbb{R}$. Here we have one parameter: $t$. The solution space is the line $\{(t,t):t \in \mathbb{R}\}$.
The system of equations $$ \left( \begin{array}{ccc|c} 1 & -1 & 0 & 0 \\ 2 & -2 & 0 & 0 \\ \end{array} \right) $$ would have a two-parameter solution space: $\{(t,t,u):t,u \in \mathbb{R}\}$.
Related Question
- [Math] What does “multiple non-trivial solutions exists mean?”
- Linear Equations Solution – Ax = b Solvability and Rank Condition
- When using Rouché–Capelli theorem should I check rank of augmented matrix if rank of coefficient matrix is max
- Does the system of linear equation $A^TAz=A^Tb$ always have a solution? ( $A^T$ is transpose of the matrix $A$)
Best Answer
Consider the matrix $A$ and the augmented matrix $(\,A\mid b\,)$. Reduce to row-echelon form. The rank (for both of these) is the number of leading (pivot) columns. Now