This is possibly a stupid question. I've read the question and answer in this link If $A$ is full column rank, then $A^TA$ is always invertible already.

My question is from $x=0$ how can we conclude $A^TA$ is invertible (or nonsingular)? I hope to get a simple (uncomplicated) explanation. Many thanks!

  • 1
    $\begingroup$ For a linear endomorphism on a finite-dimensional vector space to be invertible, it suffices that it is injective. You can see this via the rank--nullity theorem! $\endgroup$ Feb 2, 2021 at 11:41
  • $\begingroup$ I am not sure that I get the meaning of ``linear endomorphism" and your point . Can you provide a simpler explanation? $\endgroup$
    – MinHaw
    Feb 2, 2021 at 11:48
  • $\begingroup$ If $V$ is a finite-dimensional vector space, and you have a linear map $V \to V$, then it is an isomorphism once it is injective. In terms of matrices: If you have an $n \times n$ matrix defining a map $k^n \to k^n$ then this map is invertible as soon as it is injective. $\endgroup$ Feb 2, 2021 at 11:51
  • $\begingroup$ $A^TA$ is a square matrix. For square matrices $X$, being invertible is equivalent to $\ker X = \{ 0 \}$. $\endgroup$
    – twosigma
    Feb 2, 2021 at 12:17
  • $\begingroup$ @JeroenvanderMeer I got it. Thanks! $\endgroup$
    – MinHaw
    Feb 2, 2021 at 13:26

1 Answer 1


If the columns $c_1, \ldots, c_n$ of an $n \times n$ matrix $B$ do not generate the whole space, then they must be linearly dependent. Thus, there exist $x_1, \ldots,x_n$ not all zero for which

$$ x_1c_1 + \ldots + x_n c_n = 0. $$

In other words, if $x := (x_1, \ldots, x_n)$ then $x \neq 0$ and

$$ Bx = 0. $$

This way, we see that if $B$ does not have full rank, then there exits $x \neq 0$ for which $Bx = 0$.

The answer you cite shows that if $A^T Ax = 0$, then $x = 0$, hence $A^T A$ must have full rank by our previous remark.

Now, if a matrix $B$ has full rank, it is invertible: for each $e_i = (0,\ldots, \overbrace{1}^i,\ldots 0)$ there exists $(a_{ij})_i$ for which

$$ d_{1i}c_1 + \ldots + d_{ni}c_n = e_i, $$

since $c_1, \ldots, c_n$ generate the whole space. If $C$ is the matrix defined by $D_{ij} = d_{ij}$, we see that

$$ BD = I. $$

This in turn shows that $B$ is invertible.

  • $\begingroup$ What is $B$ here? We need to explain $A^TA$ is invertible, not $B$ is invertible. Do you mean $B = A^TA$? The last your two sentences are not trivial to me. $\endgroup$
    – MinHaw
    Feb 2, 2021 at 13:30
  • $\begingroup$ $B$ is an arbitrary matrix, and we can apply this to $B = A^T A$. I can change the name of the inverse to avoid confusion $\endgroup$
    – qualcuno
    Feb 2, 2021 at 13:45
  • $\begingroup$ What we have shown is: $(i)$ for a non-full rank matrix $B$ there exists $x \neq 0$ for which $Bx = 0$ and $(ii)$ if $B$ is square has full rank, it is invertible $\endgroup$
    – qualcuno
    Feb 2, 2021 at 13:47
  • $\begingroup$ $A^T A$ does not satisfy $(i)$, so it must be of full rank. By $(ii)$, it is invertible. Please do ask if this is still unclear $\endgroup$
    – qualcuno
    Feb 2, 2021 at 13:48
  • $\begingroup$ It makes more sense to me now. Thanks you. $\endgroup$
    – MinHaw
    Feb 2, 2021 at 14:37

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