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Let $X$ be a square symmetric positive definite matrix.

Show that, $\forall x,y\in\mathbb{R}^n$:

$(a^TXb)^2 \leq (a^TXa)(b^TXb)$ with equality holding iff $a$ and $b$ are linearly dependent.

I'm struggling with this one! Please help.

For equality:

$a$, $b$ linearly dependent $\Leftrightarrow a = kb$.

$\therefore$ on LHS, $(a^TXb)^2 = (a^TXb)(a^TXb)=(kb^TXb)(kb^TXb)=k^2(b^TXb)(b^TXb)$

and on RHS: $(a^TXa)(b^TXb)=(kb^TX(kb))(b^TXb)=k^2(b^TXb)(b^TXb)$ = LHS.

++++++

Question: for a positive definite matrix, X, is it true that, $a^TXa = ||X||.||a||^2$, where ||X|| is the 2-norm?

If so, then I would like to do the following:

$(a^TXb)^2 = |a.Xb|^2 \leq ||a||^2.||Xb||^2 = ||a||^2.||X||^2.||b||^2 = ||a^TXa||.||b^TXb|| = (a^TXa)(b^TXb)$

with equality iff a and Xb are linearly dependent.

my only problem then would be to connect this somehow to the fact that a and b are linearly dependent..? Can someone please tell me if this is completely wrong?

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    Two things to try: i) Inner product and ii) Cauchy-Schwartz inequality. Are you sure the condition given in the question is correct?2012-10-08
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    Ah, I mistyped. Sorry. Updated question now. It's dependent, not independent.2012-10-08
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    The inequality doesn't have any inner products/dot products in it though?2012-10-08
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    I think your statement should be "with equality holding if and only if $a$ and $b$ are linearly dependent“. Do you know what an inner product abstractly is?2012-10-08
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    Sigh. Yep, I see, my mistake again. Hmm.. abstractly, I think it's the angle between two vectors weighted according to the norms of those vectors?2012-10-08
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    @Tapu do positive definite matrices really preserve linear dependence/independence? (I saw the solution you suggested that you subsuquently deleted). I'd never heard of this property.2012-10-08
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    @jellyfish Positive definite matrices are invertible (=regular, automorphisms), so yes they do.2012-10-08
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    Sorry, maybe I worded that question wrong. What I meant to ask was: if X is p.d. and if a, b are linearly independent, then does that mean that a, Xb are linearly independent? (rather than Xa, Xb)2012-10-08
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    No; given $b$, you could take $a$ to be $Xb$, then $a$ and $Xb$ are equal, so for sure they are not linearly independent. Also, you've edited the question, but the title doesn't reflect the editing. And dot product is just one example of an inner product --- if $X$ is p.d., then $a^tXb$ has all the usual properties of the dot product and qualifies as an inner product. Check it out!2012-10-08
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    Thanks @GerryMyerson I changed the title, and checked the properties of an inner product, as suggested. :)2012-10-08
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    Good. Now, did you find $a^tXb$ to be an inner product? And did you find out anything about Cauchy-Schwarz for inner products?2012-10-08
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    @GerryMyerson yep, found $a^TXb$ to be an inner product. I've now tried to combine it with Cauchy-Schwarz in the working presented below the question text on this page. seems I get stuck trying to establish a connection between the linear dependence between a and Xb with that of a linear dependence between a and b.2012-10-08
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    jellyfish: Considering the succession of comments above, I wonder: did you read the answers posted?2012-10-08
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    @did I did. But I didn't really understand those solutions. I was hoping there would be a simpler approach..2012-10-08
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    So... when you *do not really understand* something which is written explicitely for your intention, you just... keep silent? This is my turn to *not really understand*.2012-10-08

2 Answers 2

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The map $(x,y)\mapsto x^TXy$ is an inner product: bilinearity is obvious, and positive definiteness follows from the assumption on $X$. We can apply Cauchy-Schwarz inequality: if $B(\cdot,\cdot)$ is a positive definite linear form, we have $$|B(x,y)|^2\leq B(x,x)B(y,y),$$ with equality if and only if $x$ and $y$ are linearly dependent. Indeed, we write $$0\leq B(B(x,x)y-B(y,y)x,B(x,x)y-B(y,y)x)$$ and we expand. This gives $$RHS=B(x,x)^2B(y,y)-B(x,x)B(y,y)B(y,x)-B(y,y)B(x,x)B(x,y)+B(y,y)^2B(x,x),$$ hence $$RHS=B(x,x)B(y,y)\left(B(x,x)-2B(x,y)+B(y,y)\right).$$ If $x$ and $y$ are nonzero, this gives $2B(x,y)\leq B(x,x)+B(y,y)$. Now apply the latest inequality to $\frac 1{\sqrt{B(x,x)}}x$ and $\frac 1{\sqrt{B(y,y)}}y$ instead of $x$ and $y$ respectively.

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    I have no idea what this is: $0≤B(B(x,x)y−B(y,y)x,B(x,x)y−B(y,y)x)$2012-10-08
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    It the bilinear form $B$ applied to specific vectors.2012-10-08
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    thanks. I just expanded it and tried simplifying, and have no idea what the point of this is...2012-10-08
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    There should be some simplifications (don't forget that $B$ is bilinear and symmetric). The idea is: if $x$ and $y$ are nonzero and colinear, what could be the constant link them?2012-10-08
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    I'm not getting anywhere with the expansion and simplification.2012-10-08
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    I get to 0< B(x-y,x-y) but that seems useless.2012-10-08
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    wow! @DavideGiraudo Can you explain how you came up with such a proof solution? Or even thought take those square roots? I did initially reduce to the last inequality that you utilised but would have never have imagined to take the reciprocal square root terms..2012-10-08
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One could copy a classical method of proof of Cauchy-Schwarz inequality: consider the function $P:\mathbb R\to\mathbb R$ defined by $P(t)=(a+tb)^TX(a+tb)$. Then $P(t)\geqslant0$ for every $t$ (why?), $P$ is a second degree polynomial (why?), hence its discriminant is nonpositive, that is... (to be completed). Furthermore $P(t)=0$ for some $t$ iff $a$ and $b$ are linearly dependent (why?) hence... (to be completed).

If some steps need more explanations, just yell.

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    non-positive discriminant? Not sure what this relates to.2012-10-08
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    Ah. OK. You know, this thing with the solutions of ax^2+bx+c=0 being, like, minus b plus or minus square root of blablabla.2012-10-08