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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?

JohnD
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Waffleboy
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4 Answers4

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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.

Timbuc
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  • 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? – Waffleboy Nov 30 '14 at 12:20
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    @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. – Timbuc Nov 30 '14 at 14:08
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    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 :) – s0ulr3aper07 Dec 22 '18 at 17:40
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$||A||=\sqrt{\sum_{i,j=1}^n |a_{ij}|^2}$

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 $A=0$

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

user26857
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student
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  • 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? – Alexandre Tourinho Jul 20 '18 at 16:34
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I thought about this proof, but I think it is not totally correct:

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?

Alexandre Tourinho
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