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Suppose that A is an $n\times m$ matrix with $ n\neq m$.

Here's my reasoning.

Every nonpivot column corresponds to a free variable in the system Ax = 0. Each free variable becomes a parameter, and each parameter is multiplied times a basis vector of null(A). Therefore the number of nonpivot columns equals nullity(A). Since rank(A) + nullity(A) = m, the nullity(A) must be greater than zero.

I'm not sure if I'm justified in stating the last sentence. Any suggestions or can you provide a different proof?

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    $\begingroup$ it is not true that for $n \times m$ matrix with $n \neq m,$ nullity = dimension of kernel of $A$ is positive. for example, take $A$ to be the $2 \times 1$ column vector $[1 \ 0]^T.$ rank of $A$ is $1$ and nullity is zero. $\endgroup$ – abel Nov 9 '14 at 2:23
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Suppose n < m.

Then your vectors that'll span your space will have linearly dependent vectors. You can use these to create a zero vector.

Suppose m < n.

Then your vectors will not completely span your space so nullity(A) >= n - m.

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