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I know that if a matrix is not of a full rank than it is singular, but is it always true that singular matrices are of non-full ranks?

asaf92
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1 Answers1

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Using rank theorem, $$n=\dim \ker A+\dim Rg(A).$$ Therefore, $$\text{Singular} \implies \ker A\neq \{0\}\implies \dim\ker A\geq 1$$ $$\implies n-\dim\ker(A)\leq n-1\implies \dim Rg(A)<n.$$

Surb
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  • I know that $\operatorname{rank}(A)<n$ means that $\ker \not= {0}$, but how do you get from singular (which as far as I understand means that there is no $A^{-1}$ such that $AA^{-1}=I$ to $\ker\not = 0$? – asaf92 Jul 04 '16 at 11:56
  • I've just made it (i.e. singular $\implies $ not full rank). How can this very easy proof can be not clear ? (it's difficult to make it easier since it's already very natural...) @PanthersFan92 – Surb Jul 04 '16 at 12:04