How can I prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$?

This is an exercise in my textbook associated with orthogonal projections and Gram-Schmidt process, but I am unsure how they are relevant.

  • 55
    $\begingroup$ Obligatory remark: This holds when $A\in\mathbb R^{m\times n}$, but not (for instance) when $A\in\mathbb C^{m\times n}$. $\endgroup$ Apr 3, 2013 at 4:31
  • 8
    $\begingroup$ See this counter-example for the case where $A\in\mathbb C^{m\times n}$. $\endgroup$ Jan 13, 2016 at 22:43
  • 7
    $\begingroup$ Or $A^T=(1~~\mathbf i)$. $\endgroup$ Apr 27, 2016 at 14:52
  • 6
    $\begingroup$ The conclusion also fails for fields of nonzero characteristic (finite fields in particular). See the comment by @Member below on the Accepted Answer, identifying the crucial step that $(Ax)^T(Ax) = 0$ implies $Ax=0$. $\endgroup$
    – hardmath
    Jun 8, 2017 at 17:44

3 Answers 3


Let $\mathbf{x} \in N(A)$ where $N(A)$ is the null space of $A$.

So, $$\begin{align} A\mathbf{x} &=\mathbf{0} \\\implies A^TA\mathbf{x} &=\mathbf{0} \\\implies \mathbf{x} &\in N(A^TA) \end{align}$$ Hence $N(A) \subseteq N(A^TA)$.

Again let $\mathbf{x} \in N(A^TA)$

So, $$\begin{align} A^TA\mathbf{x} &=\mathbf{0} \\\implies \mathbf{x}^TA^TA\mathbf{x} &=\mathbf{0} \\\implies (A\mathbf{x})^T(A\mathbf{x})&=\mathbf{0} \\\implies A\mathbf{x}&=\mathbf{0}\\\implies \mathbf{x} &\in N(A) \end{align}$$ Hence $N(A^TA) \subseteq N(A)$.

Therefore $$\begin{align} N(A^TA) &= N(A)\\ \implies \dim(N(A^TA)) &= \dim(N(A))\\ \implies \text{rank}(A^TA) &= \text{rank}(A)\end{align}$$

  • 18
    $\begingroup$ How does $(Ax)^T (Ax)=0$ implys $Ax=0$? $\endgroup$
    – user268307
    Dec 20, 2015 at 21:56
  • 19
    $\begingroup$ @AnuragJain note that for any vector $y$, $y^Ty = \|y\|^2$. Also, note that this question was asked two years ago, so you're unlikely to catch the attention of the original asker. $\endgroup$ Dec 20, 2015 at 21:58
  • 10
    $\begingroup$ The last step $\dim(N(A^TA)) = \dim(N(A)) \implies \text{rank}(A^TA) = \text{rank}(A)$ is due to rank-nullity theorem. $\endgroup$
    – chibicode
    Apr 2, 2018 at 17:57
  • $\begingroup$ Hi @A.D how it will work for $A^*$ , i mean how to show that $rank(A)=rank(AA^*)=rank(A^*A)$, is it the same? $\endgroup$
    – user652838
    Dec 3, 2020 at 16:36
  • $\begingroup$ @user726608 Same manner $\endgroup$ Sep 28, 2021 at 12:22

Let $r$ be the rank of $A \in \mathbb{R}^{m \times n}$. We then have the SVD of $A$ as $$A_{m \times n} = U_{m \times r} \Sigma_{r \times r} V^T_{r \times n}$$ This gives $A^TA$ as $$A^TA = V_{n \times r} \Sigma_{r \times r}^2 V^T_{r \times n}$$ which is nothing but the SVD of $A^TA$. From this it is clear that $A^TA$ also has rank $r$. In fact the singular values of $A^TA$ are nothing but the square of the singular values of $A$.

  • 4
    $\begingroup$ Note that from Strang's textbook, it actually use the fact that, there're $r$ non-zero eigenvalues of $A^TA$ i.e. $rank(A^TA)=rank(A)$, to decide the size of $\Sigma_{r \times r}$ and prove the SVD. To avoid circular argument here it would require a different SVD proof. $\endgroup$
    – Weishi Z
    Jan 6, 2021 at 12:03

Since elementary operations do not change the rank of a matrix we have $\text{rank}(A^TA) = \text{rank}(E^TA^TAE)$, where $E$ is a multiplication of several elementary operations which make $AE = [A_1, A_2]$, where $A_1$ is a column full rank matrix with $\text{rank}(A_1) = \text{rank}(A)$.

Thus we can find a matrix $P$ such that $A_1P= A_2$ and $AE = [A_1, A_1P] = A_1[I, P]$.

Thus $\text{rank}(E^TA^TAE) = \text{rank}(A_1[I, P])^T(A_1[I, P])$. In this equation, the matrices are all of full rank and the rank equals $\text{rank}(A)$, so on a real space $\text{rank}(A^TA) = \text{rank}(A)$, completing the proof.

  • 2
    $\begingroup$ I cannot decipher what is said here, but it must be wrong since it never uses that the matrices are over $\Bbb R$ (or more generally an ordered field) rather than for instance over $\Bbb C$ where the result is not true. $\endgroup$ Apr 27, 2016 at 14:38
  • 1
    $\begingroup$ The last theorem actually implicitly uses they are over real space. Thank you for pointing that out. I have added that prerequisite into my answer. $\endgroup$ Apr 30, 2016 at 4:47
  • 3
    $\begingroup$ An alternative simple way to see it: Matrix $A^T$ may be reduced to its reduced row-echelon form, $R$, by $PA^T=R$, where $P$ is the product of a sequence of elementary matrices. So, $A^T=P^{-1}R$ and hence $$\mathrm{rank}(A^TA)=\mathrm{rank}(P^{-1}RR^T(P^{-1})^T)=\mathrm{rank}(RR^T).$$ The result then follows easily from this, since clearly $\mathrm{rank}(RR^T)=\mathrm{rank}(R)=\mathrm{rank}(A).$ $\endgroup$
    – syeh_106
    Jan 14, 2017 at 2:07
  • 1
    $\begingroup$ @syeh_106 Why is it that $\operatorname{rank}(RR^T)=\operatorname{rank}(R)$? I'm sorry if this is too basic. $\endgroup$
    – JPYamamoto
    Aug 17, 2020 at 19:08
  • 2
    $\begingroup$ @JPYamamoto If $R^T$ is full column rank, this is clearly true: $RR^Tx=0 \Rightarrow x^TRR^Tx=\Vert R^Tx\Vert^2=0 \Rightarrow R^Tx = 0 \Rightarrow x = 0$, i.e. $RR^T$ is still full column rank. Otherwise, $R^T= [R_1^T\, 0]$, where $R_1^T$ is full column rank, and it's easily verified that $\mathrm{rank}(RR^T)=\mathrm{rank}(R_1R_1^T)=\mathrm{rank}(R_1)=\mathrm{rank}(R).$ $\endgroup$
    – syeh_106
    Aug 19, 2020 at 4:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .