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True or False:

Let $A$ be an $m\times n$ matrix. If $m then $\dim(\ker(A)) > 0$ (i.e the dimension of the null space of is positive).


This is true? Because the $n$-$m$ extra columns are not linearly independent and can be constructed by some combination of the $m$ columns if those are independent and the null space will have dimension $n-m$?

Trying to get an intuitive handle on this stuff...

2 Answers 2

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Another way, $\dim(\ker A) =0$ means $\ker A=\{0\}$, that is, $Av=0$ implies $v=0$.

The best if $A$ is viewed as the linear mapping $\mathbb R^n\to\mathbb R^m$ given by $v\mapsto A\cdot v$. In these terms, $\ker A=\{0\}$ is equivalent that any linearly independent set of vectors in $\mathbb R^n$ are mapped to linearly independent set. Since $m, $\mathbb R^m$ doesn't have such set of size $n$, whereas $\mathbb R^n$ has.

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The dimension theorem says that $\dim(\ker A) + \dim(\mathrm{im\,} A) = n$. Since the image space ($\mathrm{im\,} A$) is contained in $\mathbb R^m$, $\dim(\mathrm{im\,} A) \le m$.