# Prove that $AA^T=0\implies A = 0$ [duplicate]

Let $A$ be an $n \times n$ matrix with real entries, where $n\geq2$. Let $AA^T = [b_{ij}]$, where $A^T$ is the transpose of $A$. If $b_{11} + b_{22 }+\cdots+ b_{nn} = 0$, show that $A = 0$.

From what I've gleaned so far, $AA^T$ is a symmetric matrix, and the diagonals are zero. I can't figure out how to solve this question. Is there some property that exists that I'm missing for handling this question?

• Hint: If $x,y,...,z$ are real numbers, then the solution of the equation $x^2 +y^2 +...+z^2=0$ is $x=0,y=0,...,z=0$
– user195546
Commented Nov 30, 2014 at 4:11
• For some basic information about writing math at this site see e.g. here, here, here and here.
– user195546
Commented Nov 30, 2014 at 4:18
• Learn latex quickly Commented Nov 30, 2014 at 4:29
• Your problems on title and the content are different. Commented Nov 30, 2014 at 6:37
• Sorry for the improper formatting! Thanks for the link Kamal, They were really helpful :) and thanks JohnD for the edit! Commented Nov 30, 2014 at 12:18

Let

$$A=(a_{ij})\implies A^t=(a_{ji})\implies AA^t=(b_{ij})=\left(\sum_{k=1}^ma_{ik}a_{jk}\right)$$

so that

$$0=\sum_{i=1}^nb_{ii}=\sum_{i=1}^n\sum_{k=1}^na_{ik}a_{ik}$$

Complete the proof now.

• Forgive me, my linear algebra is pretty bad, thats the sum of the multiplication of diagonal entries = 0 yes? is it sufficient then to say that as the sum is zero, either aik or ajk must equal to zero. Thus, since A = Atranspose, Both A and A transpose are zero? Commented Nov 30, 2014 at 12:20
• @Waffleboy, this is not a matter of linear algebra but only algebra: we have $\;a_{ik}a_{ik}=a_{ik}^2\;$ , and since we're in the reals, the sum of squares equals zero iff all the summands are zero. Commented Nov 30, 2014 at 14:08
• I'm also new to Linear Algebra and I'm wondering whether you intended to say $(b_{ij})=\left(\sum_{k=1}^ma_{ik}a_{kj}\right)$ instead? Also, this isn't explicitly stated, but you've assumed A to be a matrix with n rows and m columns right? And of course, thanks for the answer :) Commented Dec 22, 2018 at 17:40

$||A||=\sqrt{\sum_{i,j=1}^n |a_{ij}|^2}$

i.e. $||A||=\sqrt {trace(A^t A)}$ .Now by the given condition $b_{11}+b_{22}+...+b_{nn}=0$

i.e. trace of $A^tA=0$

i.e. $||A||=0$ if and only if $A=0$

$$\forall X \in \mathbb{R}^n$$, $$X^tAA^tX=0$$, i.e. $$\left=0$$, hence $$A^tX=0, \forall X \in \mathbb {R}^n$$, so $$A^t=0$$, thus $$A=0$$.

• I dont't really see why if $X^tAA^tX=0$ $\forall X$, then $A^tX=0$. For me, if you call $A^tX=B$, we are just resaying that $BB^t=0$ implies $B=0$. Can you clarify? Commented Jul 20, 2018 at 16:34

Supose $A≠0$.

Then, for any $X$, $A+X≠X$.

Multiplying by $A^T$ at the right, we get:

$(A+X)A^T≠XA^T$

$AA^T+XA^T≠XA^T$

$XA^T≠XA^T$

Arriving in a contradiction, we get $A=0$. I realise it may not be totally correct to multiply by $A^T$ without making some proper assumptions, but I know I can pick any $X$. Is there a way to formalise this proof?