Let $M$ be an $m \times n$ matrix and $N$ an $n \times m$ matrix. Prove that if $NM=I_n$ then the columns of $M$ are linearly independent.
Let $v_1, v_2, v_3, \ldots, v_n$ be $n$-vectors that form the columns of matrix $M$. That is,
$$M=\left(\begin{array}{c|c}v_1 & v_2 & v_3 & \cdots & v_n\end{array}\right).$$
Does anyone have any tips on how to solve that problem?
Best Answer
If $NM=I$, then $N$ is a left inverse of $M$. Thus the equation $Mx=0$ admits only the trivial solution. To prove this, consider the equation $Mx=0$. Now left multiply by $N$ on both sides and you get that $x=0$.
Because $$\begin{bmatrix} A_{1,1} &\cdots & A_{1,n}\ \\ \vdots & \ddots & \vdots \ \ \\ A_{m,1}&\cdots & A_{m,n}\end{bmatrix} \begin{bmatrix}a_1 \\ \vdots \\a_n \end{bmatrix}=a_1\begin{bmatrix}A_{1,1} \\ \vdots \\ A_{m,1}\end{bmatrix}+\dots +a_n\begin{bmatrix}A_{1,n}\\ \vdots \\ A_{m,n} \end{bmatrix}$$
The above result implies that any linear combination of the columns of $M$ equals $0$ if and only if $a_1=\dots=a_n=0$. Thus, the columns are linearly independent.