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A is an $m \times n$ matrix. If $\operatorname{rank}(A)=n$ and $AX=0$ where $X$ is $n \times k$, then $X = 0$. Below is how I conclude $X = 0$.

Since $\mathsf{rank}(A)=n$ and $A$ is $m \times n$, so we know $A^TA$ is an $n \times n$ invertible matrix. Then $A^TAX$ = $A^T0$, which means $A^TAX = 0$. Then $X = 0$ because $A^TA$ is invertible.

Am I right? If $X$ in $\mathbb{R}^n$, then there is no doubt this statement holds, but $X$ is a matrix, I am not 100% sure.

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Notice that in the equation $AX=0$ for $X$ the columns of $X$ are completely unrelated. – Mariano Suárez-Alvarez Feb 11 '12 at 23:45
What do you mean the columns of X are completely unrelated? Could you specify? – Shannon Feb 12 '12 at 0:45
I think what Mariano means is that you can consider the columns of $X$ individually. Let $x$ be a column of $X$. Then your equation $AX = 0 \in \mathbb{R}^{m \times k}$ implies $Ax = 0 \in \mathbb{R}^{m \times 1}$. – William DeMeo Feb 12 '12 at 1:19
Right. Thank you so much. – Shannon Feb 12 '12 at 2:31
@williamdemeo I still dont get it why AX=0 implies Ax = 0. Since x_i is column of X, so we getAx_1 + Ax_2 +... +Ax_k=0, but how can you conclude Ax_i = 0 from this? – Shannon Feb 12 '12 at 20:54

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