1

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!

MinHaw
  • 23
  • 1
    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! – Jeroen van der Meer Feb 02 '21 at 11:41
  • I am not sure that I get the meaning of ``linear endomorphism" and your point . Can you provide a simpler explanation? – MinHaw Feb 02 '21 at 11:48
  • 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. – Jeroen van der Meer Feb 02 '21 at 11:51
  • $A^TA$ is a square matrix. For square matrices $X$, being invertible is equivalent to $\ker X = { 0 }$. – twosigma Feb 02 '21 at 12:17
  • @JeroenvanderMeer I got it. Thanks! – MinHaw Feb 02 '21 at 13:26
  • @twosigma I got your point. Thanks! – MinHaw Feb 02 '21 at 13:26
  • This is duplicate. Additionally, this seems like a homework problem due to the number of posts with this question. If that is the case, you should work on it on your own, otherwise you are robbing yourself from the experience of learning the material –  Feb 02 '21 at 14:13
  • @crystal_math As I wrote in my question "Revisit", so I intend to explore more about an existing question. And yes, it is a homework problem, the link I cited contains an answer. But I don't seek the existing answer, I seek for understanding. Of course, I've worked on it by my own before posting the question here. – MinHaw Feb 02 '21 at 14:24
  • @MinHaw apologies for misunderstanding –  Feb 02 '21 at 16:58

1 Answers1

2

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.

qualcuno
  • 17,121
  • 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. – MinHaw Feb 02 '21 at 13:30
  • $B$ 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 – qualcuno Feb 02 '21 at 13:45
  • 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 – qualcuno Feb 02 '21 at 13:47
  • $A^T A$ does not satisfy $(i)$, so it must be of full rank. By $(ii)$, it is invertible. Please do ask if this is still unclear – qualcuno Feb 02 '21 at 13:48
  • It makes more sense to me now. Thanks you. – MinHaw Feb 02 '21 at 14:37