Suppose that $A$ is an $m \times n$ matrix. I want to show that $A^TA$ is non-singular if and only if $A$ has full rank.

Now `full rank' can mean two things. I know that saying that $A$ has full rank means that $\text{rank}(A)=\min\{m,n\}$. If $m \geq n$ then $\text{rank}(A)=n$, i.e., $A$ has full column rank, so $A$ is injective. I have proven this case with no issue.

What I am struggling to prove is that $A^TA$ is non-singular if and only if $A$ has full row rank, i.e., in the case where $m <n$, $\text{rank}(A)=m$, meaning that $A$ is surjective.

Thank you, in advance.

  • $\begingroup$ Row rank is equal to column rank.... $\endgroup$ – MathematicsStudent1122 Dec 14 '17 at 18:20
  • $\begingroup$ Should it not be $AA^T$? $\endgroup$ – Roberto Rastapopoulos Dec 14 '17 at 18:25
  • $\begingroup$ @RobertoRastapopoulos No, the general problem was given to me as stated in the first sentence. $\endgroup$ – Sarah Dec 14 '17 at 18:34
  • $\begingroup$ Why is there a mention of "row rank" in the last paragraph but only of "rank" in the first one? $\endgroup$ – Roberto Rastapopoulos Dec 14 '17 at 18:45
  • 2
    $\begingroup$ You should really specify the field you are working over. The claim simply is not true in finite characteristic (or over $\mathbb{C}$ for that matter) $\endgroup$ – Sebastian Schoennenbeck Dec 14 '17 at 18:46

Note that:

if $x \neq0 \in Null(A)$


thus $A^TA$ is singular

if $x \neq0$ and $A^TA$ is singular

$$A^TAx=0 \implies x^TA^TAx=0 \implies Ax=0$$

thus $x\in Null(A)$


$A^TA$ is singular $\iff$ A is not full column rank

that's equivalent to

$A^TA$ is not-singular $\iff$ A is full column rank


if A is full row rank and $n\neq m \implies Null(A)\neq {0}$ thus $A^TA$ is singular

  • $\begingroup$ This is exactly the method I used for one direction when I knew that $A$ has full column rank. How does this help me show that $A$ has full row rank, i.e. that $A$ is surjective? $\endgroup$ – Sarah Dec 14 '17 at 18:36
  • $\begingroup$ This is exactly my proof for when $A$ has full column rank, i.e., $A$ is injective. How does this work for when $A$ is surjective? $\endgroup$ – Sarah Dec 14 '17 at 18:38
  • $\begingroup$ Thanks for the correction, obviously is true when A is full column rank! $\endgroup$ – user Dec 14 '17 at 18:42
  • $\begingroup$ I think that when A is full row rank the statement is true only if n=m. Otherwise Null(A) is not 0 and you can use the same proof. $\endgroup$ – user Dec 14 '17 at 18:44
  • $\begingroup$ Okay, I see what you mean. If $\text{rank}(A)=m$, and $m <n$ then by the Dimension Theorem, $$n=\text{nullity}(A)+\text{rank}(A),$$ which implies that $0<n-m=\text{nullity}(A)$. $\endgroup$ – Sarah Dec 14 '17 at 18:47

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