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If $A$ is a $7 \times 5$-matrix, what is the lowest possible dimension of Nul $A$?

We have that every column can be a pivot column, so we are left with no free variables. For the similar task

If $A$ is a $3 \times 7$-matrix, what is the lowest possible dimension of Nul $A$?

it was easy to conclude with the answer "4". However, in this case, we have that every column can be a pivot column, no free variables. Would this imply that the lowest possible dimension is 0?

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Yes, that is true. That was the basis of my argument. – Andrew Thompson Mar 6 '14 at 12:33
Yes you are right – Francisco Mar 6 '14 at 12:33
I deleted the comment when i realized that your last sentence was referring to the 7 x 5 matrix. :) – neofoxmulder Mar 6 '14 at 12:49
up vote 4 down vote accepted

By the rank-nullity theorem for an $m\ x \ n$ matrix $$Rank \ A + Nullity \ A = n$$

Since the Rank of a $7x5$ can be up to and including 5 , the Nullity can be 0.

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in $A$ $3\times 7$-matrix ,even if every column be pivot column , rank cannot exceed $3$, just as rank of $A$ in first case cannot exceed $5$.
moreover $Nul(A)=0$ implies rank if full, which is possible only in square matrix and since $3\times 7$ is not square so $Nul(A)$ cannot not be $0$ Therefore, $2\le dim(Nul(A)) \le 3$

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Ok, I see your argument, but this is not what I asked. – Andrew Thompson Mar 6 '14 at 12:33
@AndrewThompson: You just need to apply the rank-nullity theorem to complete the answer. – Najib Idrissi Mar 6 '14 at 12:48
my answer was for 3x7 case – ketan Mar 6 '14 at 13:00
@AndrewThompson If this is not what you asked then why did you remarked it in blue in your question? That is very confusing... – DonAntonio Mar 6 '14 at 13:08
Well @AndrewThompson, at least two here (ketan and I) thought that was integral part of the question. As you'd probably know, many of us read very quickly only the main question. This is why that was confusing, in particular that that scond blue part ends with a question mark... – DonAntonio Mar 6 '14 at 17:19

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