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I know from $rank(A)+nullity(A)=n$ that is $rank(A)=rank(A^T)=n$ and $nullity(A^T) ≠ 0$.Is there any other useful connection to solve this problem?

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  • $\begingroup$ What is def(A)?? $\endgroup$ Dec 27 '17 at 19:34
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    $\begingroup$ Dimension of $N(A)$.My professor use that term to describe a dimension of nullspace. $\endgroup$
    – VidaXpo
    Dec 27 '17 at 19:37
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    $\begingroup$ @DevendraSinghRana "deficiency", aka nullity $\endgroup$ Dec 27 '17 at 19:38
  • $\begingroup$ Rank A is n then it is not possible to have rank of AA^T to be m $\endgroup$ Dec 27 '17 at 19:40
  • $\begingroup$ @DevendraSinghRana I corrected mistake. $\endgroup$
    – VidaXpo
    Dec 27 '17 at 19:43
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Suppose $A^TAv=0$; then also $v^TA^TAv=(Av)^T(Av)=0$, which implies $Av=0$. Since the null space of $A$ is trivial, you're done: $v=0$, so the nullity of $A^TA$ is zero and the rank is full.

Suppose $x=(x_1,\dots,x_n)\in\mathbb{R}^n$ and $x^Tx=0$; this means $$ x_1^2+x_2^2+\dots+x_n^2=0 $$ so $x_1=0, x_2=0,\dots,x_n=0$ and $x=0$. Apply it to $x=Av$.

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  • $\begingroup$ $(Av)^T(Av)=0$, which implies $Av=0$ ; please explain this I didn't understand, sir! $\endgroup$
    – Learning
    Dec 29 '17 at 10:20
  • $\begingroup$ @Abhishek I added the explanation. $\endgroup$
    – egreg
    Dec 29 '17 at 11:47
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Well ! Use the fact that $$RankA+RankB -k<=Rank (AB)<= min (Rank A , Rank B)$$ Where $A$ is $n×k$ and $B$ is $k×m$ matrix.

It will work surely!

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You can define:

$\mathbb{A}_{m\times n}(\mathbb{R})$ be the matrix of any linear transformation $T:\mathbb{R^n}\to\mathbb{R^m}$ .So by Sylvester's law $rank(A)+nullity(A)=n\implies rank(A)=n$

and we know that $rank(A)=rank(A^TA)$=$n$; [1]: Prove rank $A^TA$ = rank $A$ for any $A_{m \times n}$

here , $A^TA$ is $n\times n$ , has full rank.

Therefore , $A^TA$ is invertible. (Hope it may help you.)

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