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How can I understand that $A^TA$ is invertible if $A$ has independent columns? I found a similar question, phrased the other way around, so I tried to use the theorem

$$ rank(A^TA) \le min(rank(A^T),rank(A)) $$

Given $rank(A) = rank(A^T) = n$ and $A^TA$ produces an $n\times n$ matrix, I can't seem to prove that $rank(A^TA)$ is actually $n$.

I also tried to look at the question another way with the matrices

$$ A^TA = \begin{bmatrix}a_1^T \\ a_2^T \\ \ldots \\ a_n^T \end{bmatrix} \begin{bmatrix}a_1 a_2 \ldots a_n \end{bmatrix} = \begin{bmatrix}A^Ta_1 A^Ta^2 \ldots A^Ta_n\end{bmatrix} $$

But I still can't seem to show that $A^TA$ is invertible. So, how should I get a better understanding of why $A^TA$ is invertible if $A$ has independent columns?

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6 Answers 6

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Consider the following: $$A^TAx=\mathbf 0$$ Here, $Ax$, an element in the range of $A$, is in the null space of $A^T$. However, the null space of $A^T$ and the range of $A$ are orthogonal complements, so $Ax=\mathbf 0$.

If $A$ has linearly independent columns, then $Ax=\mathbf 0 \implies x=\mathbf 0$, so the null space of $A^TA=\{\mathbf 0\}$. Since $A^TA$ is a square matrix, this means $A^TA$ is invertible.

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  • $\begingroup$ This answer uses vocabulary that is much more familiar than the other answer you linked in the comments. Thanks! $\endgroup$
    – Chewers Jingoist
    Commented Jun 26, 2016 at 23:54
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    $\begingroup$ I liked @littleO's elegant proof but, going against popular votes, I'm accepting this answer because it gave me a better understanding. $\endgroup$
    – Chewers Jingoist
    Commented Jun 27, 2016 at 0:16
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    $\begingroup$ @CharlieParker I am pretty sure your proof is flawed. You do not consider the case where $Ax \neq 0$, but $A^TAx=0$ because $Ax$ is in the null space of $A^T$. In order to show this can't happen, I made an argument using the null space of $A^T$ to show $Ax$ must equal $0$. $\endgroup$
    – Noble Mushtak
    Commented Oct 18, 2017 at 21:54
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    $\begingroup$ @CharlieParker Sorry for the late response. $Ax \in N(A^T)$ because we have $A^T(Ax)=0$. Also, $Ax \in C(A)$, as you said. Therefore, $Ax$ is in both $N(A^T)$ and $C(A)$. However, the only vector that can be in both a subspace and the orthogonal complement of its subspace is the zero vector. Thus, $Ax$ must be the zero vector. $\endgroup$
    – Noble Mushtak
    Commented Nov 5, 2017 at 14:50
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    $\begingroup$ @CharlieParker Here is an article proving that the intersection of a subspace and its orthogonal complement is just the zero vector. $\endgroup$
    – Noble Mushtak
    Commented Nov 5, 2017 at 14:52
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If $A $ is a real $m \times n $ matrix then $A $ and $A^T A $ have the same null space. Proof: $A^TA x =0\implies x^T A^T Ax =0 \implies (Ax)^TAx=0 \implies \|Ax\|^2 = 0 \implies Ax = 0 $.

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    $\begingroup$ I was also wondering why $N(A^TA) = N(A)$ but I didn't think that would be related to my question. Thanks! $\endgroup$
    – Chewers Jingoist
    Commented Jun 27, 2016 at 0:03
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    $\begingroup$ Is it true that anything multiplied on the left leaves the null space unaffected? i.e. $N(CA) = N(A)$ for all $C$? $\endgroup$
    – Charlie Parker
    Commented Oct 18, 2017 at 21:09
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    $\begingroup$ Also, at what point are you using that the matrix $A$ has column rank $r$? I'm curious because your proof seems a lot less intuitive than what the simpler argument than: If we want $A^TAx=0$ only with the zero vector then its obvious thats the case by definition because $A$ has independent columns, so only $Ax=0$ and thats true too for $A^T A $. $\endgroup$
    – Charlie Parker
    Commented Oct 18, 2017 at 21:11
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    $\begingroup$ @CharlieParker Also, in this answer, littleO only showed that $A$ and $A^TA$ had the same null space. However, the full proof is a bit more intricate than this: Since $A$ has column rank $r$ (i.e. independent columns), it has a trivial null space. Thus, by the above, $A^TA$ also has a trivial null space. Therefore, since $A^TA$ is a square matrix and has a trivial null space, it is invertible. $\endgroup$
    – Noble Mushtak
    Commented Oct 18, 2017 at 21:58
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    $\begingroup$ @CharlieParker Yes, an invertible $C$ would not change the null space. However, it is not the only type of matrix that works. As littleO showed, $A^T$ also works, even if $A$ does not have independent columns. $\endgroup$
    – Noble Mushtak
    Commented Oct 18, 2017 at 22:10
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Let $A \in \mathbb R^{m \times n}$. Note that

$$f (x) := x^T A^T A x = \|A x\|_2^2$$

is positive semidefinite. Function $f$ vanishes when $A x = 0_m$. If $A$ has full column rank, i.e., if its $n$ columns are linearly independent, then $A x =0_m$ implies that $x = 0_n$, i.e., $f$ is positive definite and, hence, $A^T A$ is positive definite and, thus, invertible.

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I think it need be mentioned that we deal with the real field, as the mentioned both facts may not be true for arbitrary field $F$.

Example 1. Let $F=\mathbb Z_5$ and $A=\left[\begin{smallmatrix} 1 & 3\\ 2 & 1\\ 0 & 1 \end{smallmatrix}\right]$. Then $A$ has two independent columns, but $A^T A=\left[\begin{smallmatrix} 0 & 0\\ 0 & 1 \end{smallmatrix}\right]$ is not invertible.

Example 2. Let $F$ and $A$ be the same as above. Then $A=\left[\begin{smallmatrix} 1 & 3\\ 2 & 1\\ 0 & 1 \end{smallmatrix}\right]\sim \left[\begin{smallmatrix} 1 & 0\\ 0 & 1\\ 0 & 0 \end{smallmatrix}\right]$, i.e., ${\rm null}(A)=\{0\}$. But ${\rm null}(A^TA)={\rm span}(\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right])$ not equal to ${\rm null}(A)$.

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  • $\begingroup$ P.S. I see that user1551 has already mentioned that the first fact is not true for the complex field. $\endgroup$
    – V. Mikaelian
    Commented Jun 2, 2021 at 6:51
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Suppose A is a $m \times n$ matrix ($m\geq n$). Since $A$ has linearly independent columns, by QR decomposition $A=QR$ where $Q$ is a $m \times n$ matrix with orthonormal columns and $R$ is a $n \times n$ invertible triangular matrix.

Thus $A^TA=(QR)^T(QR)=R^T(Q^TQ)R=R^TR$. Since $R^T$ and $R$ are both invertible,$A^TA$ is invertible.

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It's easy to see that $A^TA$ is symetric and positive definite. Hence orthogonally diagonalizable with all eigenvalues $>0$. Hence the determinant of $A^TA$, which equals the product of the eigenvalues is positive. Thus $A^TA$ is invertible.

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