# Prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$

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.

• Obligatory remark: This holds when $A\in\mathbb R^{m\times n}$, but not (for instance) when $A\in\mathbb C^{m\times n}$. – darij grinberg Apr 3 '13 at 4:31
• See this counter-example for the case where $A\in\mathbb C^{m\times n}$. – M.S. Dousti Jan 13 '16 at 22:43
• Or $A^T=(1~~\mathbf i)$. – Marc van Leeuwen Apr 27 '16 at 14:52
• 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 $Ax=0$. – hardmath Jun 8 '17 at 17:44

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}

• How does $(Ax)^T (Ax)=0$ implys $Ax=0$? – user268307 Dec 20 '15 at 21:56
• @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. – Omnomnomnom Dec 20 '15 at 21:58
• The last step $\dim(N(A^TA)) = \dim(N(A)) \implies \text{rank}(A^TA) = \text{rank}(A)$ is due to rank-nullity theorem. – chibicode Apr 2 '18 at 17:57

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 $A$.

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.

• 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. – Marc van Leeuwen Apr 27 '16 at 14:38
• 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. – Xiangru Lian Apr 30 '16 at 4:47
• 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 $\mathrm{rank}(RR^T)=\mathrm{rank}(R)=\mathrm{rank}(A).$ – syeh_106 Jan 14 '17 at 2:07
• @syeh_106 Awesome. Thanks for your contribution. – Xiangru Lian Jan 14 '17 at 10:31