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I know $AA^t=0$, so, picking any $X$, $X^tAA^tX=0$. I read that this implies $A^tX=0$, resulting in $A=0$ (what I inicially wanted to prove), but I can't understand that implication I read, so is there an easy way to comprehend it?

Alexandre Tourinho
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  • It's because (if we are dealing with real matrices) $Y^tY=0\implies Y=0$. – Angina Seng Aug 03 '18 at 14:53
  • Note that $(X^tA)=(A^tX)^t$, so that you have $(A^tX)^tA^tX=0$, which is the same as the square of the euclidian norm of $A^tX$. – Suzet Aug 03 '18 at 14:55
  • @LordSharktheUnknown This is exactly what the OP wants to prove if I am not mistaken. – Suzet Aug 03 '18 at 14:56
  • I want to prove that $YY^t=0$ implies $Y=0$, and writing $X^tAA^tX=0$ looks like a good idea. If this is the same as $A^tX=0$ I am done with the original proof, but I can't comprehend they are the same, so I want to understand this implication – Alexandre Tourinho Aug 03 '18 at 15:05

3 Answers3

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Let $A^tX=(b_1,b_2,\cdots,b_n)^t$, then $X^tA=(b_1,b_2,\cdots,b_n)$. So $$X^tAA^tX=\sum_{k=1}^n b_k^2.$$ That is what you need!

Riemann
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If $0=(Y^tY)_{ij}=\sum_k Y_{ki}Y_{kj}$ for all $i,\,j$, in particular $0=(Y^tY)_{ii}=\sum_k Y_{ki}^2$. This sum of squares being zero requires reach entry $Y_{ki}$ to be zero too.

J.G.
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To prove

$Y^TY = 0 \Longrightarrow Y = 0 \tag 1$

without resorting to explicit matrix multiplication or the sums of squares of components, etc, one may pick an arbitrary vector $x$ and write

$\Vert Yx \Vert^2 = \langle Yx, Yx \rangle = \langle x, Y^TY x \rangle = 0, \tag 2$

since we assume that

$Y^TY = 0; \tag 3$

therefore we have, for any vector $x$,

$\Vert Yx \Vert^2 = 0 \Longrightarrow \Vert Yx \Vert = 0 \Longrightarrow Yx = 0 \Longrightarrow Y = 0; \tag 4$

now if we set

$Y = A^TX, \tag 5$,

we have

$Y^T = X^T A, \tag 6$

whence

$Y^TY = (X^T A)(A^T X), \tag 7$

so applying our previous result to $Y = A^TX$ leads to

$A^TX = 0. \tag 8$

Robert Lewis
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