Revisit: If $A$ is full column rank, then $A^TA$ is always invertible

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!

• 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! Feb 2, 2021 at 11:41
• I am not sure that I get the meaning of linear endomorphism" and your point . Can you provide a simpler explanation? Feb 2, 2021 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. Feb 2, 2021 at 11:51
• $A^TA$ is a square matrix. For square matrices $X$, being invertible is equivalent to $\ker X = \{ 0 \}$. Feb 2, 2021 at 12:17
• @JeroenvanderMeer I got it. Thanks! Feb 2, 2021 at 13:26

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.

• 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. Feb 2, 2021 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 Feb 2, 2021 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 Feb 2, 2021 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 Feb 2, 2021 at 13:48
• It makes more sense to me now. Thanks you. Feb 2, 2021 at 14:37