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Suppose $x = 0$ is the only solution to the matrix equation $Ax = 0$ where $A$ is $m \times n$, $x$ is $n \times 1$, and $0$ is $m \times 1$. Then (i) The rank of $A$ is $n$, and (ii) $m \times n$

Why is this the case, can somebody give me an intuition on the two implications.

Ri-Li
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Meera Unni
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    The linear transformation associated with A is one-to-one with domain $\mathbb R^{m}$ and range$\mathbb R^{n}$. It is an isomorphism from $\mathbb R^{m}$ onto its range. Hence the range has dimension n. Rank of A is nothing but the dimension of the range of A so the rank is n. Since the range is a subspace of $\mathbb R^{m}$ it follows that $m \geq n$. – Kavi Rama Murthy Feb 26 '18 at 07:17

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