# If $A\in M_{m\times n}(\mathbb{R})$ and $m>n$ and $Ker(A)=0$ then $A^TA$ is invertible.

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?

• What is def(A)?? Dec 27 '17 at 19:34
• Dimension of $N(A)$.My professor use that term to describe a dimension of nullspace. Dec 27 '17 at 19:37
• @DevendraSinghRana "deficiency", aka nullity Dec 27 '17 at 19:38
• Rank A is n then it is not possible to have rank of AA^T to be m Dec 27 '17 at 19:40
• @DevendraSinghRana I corrected mistake. Dec 27 '17 at 19:43

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 $x=Av$.

• $(Av)^T(Av)=0$, which implies $Av=0$ ; please explain this I didn't understand, sir! Dec 29 '17 at 10:20
• @Abhishek I added the explanation. Dec 29 '17 at 11:47

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!

• Can you give me a source of $RankA+RankB−k<=Rank(AB)$ ? Dec 27 '17 at 19:57
• quora.com/How-do-I-show-rank-AB-≥rank-A-+rank-B-−n Dec 27 '17 at 20:01

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.)