Question

In statistical theory, a common requirement is that a matrix be of full rank. That is, the rank should be as large as possible. Explain why an m n matrix with more rows than columns has full rank if and only if its columns are linearly independent.

Short answer

No, the new non-homogeneous system will not have any solution.

Step-by-step solution

Describe the given statement

It is given that a homogeneous system has full rank. It implies that if the system is $m\times n$, then the rank of the system is $n$ as the rank is equal to the number of pivot positions.

Use the rank theorem

Consider the homogeneous system $Ax=0$, where $A$ is an $m\times n$ matrix. As the system hasfull rank, $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A=n$ and there are $n$ unknowns. By the rank theorem, $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A+\mathrm{d}\mathrm{i}\mathrm{m}\mathrm{N}\mathrm{u}\mathrm{l}A=n$.

Put the values as shown:

$\begin{array}{rl}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A+\mathrm{d}\mathrm{i}\mathrm{m}\mathrm{N}\mathrm{u}\mathrm{l}A& =n\\ \mathrm{d}\mathrm{i}\mathrm{m}\mathrm{N}\mathrm{u}\mathrm{l}A& =n-\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}A\\ \mathrm{d}\mathrm{i}\mathrm{m}\mathrm{N}\mathrm{u}\mathrm{l}A& =n-n\\ \mathrm{d}\mathrm{i}\mathrm{m}\mathrm{N}\mathrm{u}\mathrm{l}A& =0\end{array}$

Draw a conclusion

As $\mathrm{d}\mathrm{i}\mathrm{m}\mathrm{N}\mathrm{u}\mathrm{l}A$ is 0 and the system is homogeneous, there must exist only a trivial solution. This is possible when matrix $A$ has linearly independent columns.