1
$\begingroup$

Does anyone know what is going on in this proof of the Schwarz inequality? Most importantly: how can one assume that $c^2\leqq \|A\|^2$, or later on, that $c^2\|B\| \leqq \|A\|^2$? This would imply that $\|A-cE\|^2$, or in the latter case $\|A-cB\|^2$, could also be equal to zero (seeing as it's a smaller-than-or-equal-to sign). But since we're using the Pythagorean theorem to arrive at these relations, $\|A-cE\|^2$ or $\|A-cB\|^2$ could not possibly be zero, as then one of the sides of our 'triangle' would be zero, and we wouldn't be able to use the Pythagorean theorem to justify this conclusion. So why isn't it just a '<'-sign? As I said, I don't see how vector $A-cE$ or $A-cB$ can have length zero if we're using the Pythagorean theorem, that is, if we're presupposing the existence of a triangle. If either $A-cE$ or $A-cB$ equals zero, there wouldn't be a triangle anymore, and we wouldn't be able to apply the Pythagorean theorem.

Edit: fixed link.

Edit2: I must add that the author maintains an unusual definition for the scalar component; instead of the conventional $\frac{\mathbf{A}\cdot \mathbf{B}}{\|B\|}$, i.e., $\mathbf{A}\cdot \mathbf{\hat{B}}$, the author defines $\frac{\mathbf{A}\cdot \mathbf{B}}{\mathbf{B} \cdot \mathbf{B}}$ to be the scalar component, i.e., $\frac{\mathbf{A}\cdot \mathbf{\hat{B}}}{\|B\|}$. In other words, the author divides the 'conventional' scalar component by the the norm of $\mathbf{B}$, thus making 'his' scalar component a number which is in fact the proportion of the 'regular' scalar component to the entire length of the vector on which this component is projected.

Edit3: Since people don't seem to understand my confusion, let me phrase my question as explicitely as possible. My question is: why is the $\geq$ sign used? Why not just ">"? In what situation could $\|A\|^2$ equal $t^2\|E\|^2$ (the $\geq$ sign means greater than OR EQUAL TO, so in what situation could the left hand side ever equal the right hand side?) The problem is: since we're using the Pythagorean theorem for our proof, we're presupposing the existence of a triangle (or otherwise we wouldn't be able to use the Pythagorean theorem), and as such, $\|A\|^2$ can never equal $t^2\|E\|^2$, because this would mean that $\|A-tE\|^2$ is equal to zero, and that one of the sides of our triangle has length zero. Then we wouldn't have a triangle, and we wouldn't be justified in using the Pythagorean theorem in our particular situation. So let me ask it again: why is the $\geq$ sign used, instead of the ">"-sign?

$\endgroup$
  • 1
    $\begingroup$ "STILL UNANSWERED"? It was just 57 minutes since you posted it. $\endgroup$ – Asaf Karagila Jul 23 '14 at 23:34
  • $\begingroup$ My apologies, I was under the impression that that was a relatively long time for stackexchange. I have changed it now. $\endgroup$ – Ius Klesar Jul 23 '14 at 23:36
  • $\begingroup$ @Luke very rarely questions get answers after 2 years of posting here on SE (true story). $\endgroup$ – user132181 Jul 23 '14 at 23:38
  • $\begingroup$ @user132181 do you care to elaborate on that? $\endgroup$ – John Smith Jul 24 '14 at 8:08
  • $\begingroup$ @JohnSmith I've seen a question or two posted on the site in 2012 but answered only recently. $\endgroup$ – user132181 Jul 24 '14 at 10:26
1
$\begingroup$

I'll consider only your third edit. So, why is $\leq$ necessary in the C-S inequality? The theorem you are studying does not say this, but the case of equality in the C-S inequality is standard: equality holds if and only if $A$ and $B$ are linearly dependent vectors.

Of course the proof cannot be based, in this particular case, on the construction of a triangle that is degenerate.

But the proof of the book is correct even in this case, since the quantity $\|A-cB\|$ vanishes for some $c$ if $A$ and $B$ are collinear, and you must use the weak inequality sign.

$\endgroup$
  • $\begingroup$ Your answer was most helpful, I think I'm starting to see what's going on. But how is the proof in the book still correct even in the case that $\|A-cB\|=0$? If it does vanish, wouldn't that mean that we don't have a triangle anymore, and that using the Pythagorean theorem to construct any such (in)equality would be an invalid thing to do? Just like you can't use $2\pi r$ to find the circumference of a square, you'd be wrong for using $a^2+b^2=c^2$ on something other than a square triangle, and since we don't have a triangle if $\|A-cB\|=0$, wouldn't using the Pythagorean theorem be incorrect? $\endgroup$ – Ius Klesar Jul 25 '14 at 7:33
  • $\begingroup$ The name "Pythagorean theorem" usually refers to an abstract result: if $x$ and $y$ are elements of any inner product space and $x \per y$, then $\|x+y\|^2 = \|x\|^2+\|y\|^2$. This is trivially true when either $x=0$ or $y=0$. You shouldn't really think of a geometric triangle, in the degenerate case. $\endgroup$ – Siminore Jul 25 '14 at 8:19
  • $\begingroup$ I see, thanks a lot, you really managed to get down to the core of what I was having trouble with! Just a quick follow-up question: if it's an abstract result, what justifies it's validity as being an identity? $\endgroup$ – Ius Klesar Jul 25 '14 at 8:50
0
$\begingroup$

These inequalities hold because 1) $\lVert A \rVert^2$ equals $c^2$ plus something positive (top of page 27) and 2) $c^2\lVert B \rVert^2$ equals $\lVert A \rVert^2$ plus something positive (bottom of page 27).

They are not assumptions, but follow from the subadditivity, positivity and homogeneity of the norm.

$\endgroup$
  • $\begingroup$ Please read my question again, I don't think you really see my problem. What you just wrote down is very obvious and straight-forward: norm of A squared equals $c^2$ plus something positive. My question is: why is the $\leqq$ sign used? Why not just '<'? As you said: norm of A squared equals $c^2$ plus something positive, and seeing as that 'something' can't be zero (then the Pythagorean theorem wouldn't hold), why is the $\leqq$ used? $\endgroup$ – Ius Klesar Jul 23 '14 at 23:18
0
$\begingroup$

They've made this just a little too verbose, and they've ignored special cases along the way, and that's why you're getting focused on that. Just hit the high points:

If both $A$ and $B$ are $0$, then $|A\cdot B| \le \|A\|\|B\|$ is trivial to show because both sides are $0$. So, without loss of generality, assume $A \ne 0$ (rename if necessary.) Then $$ B = \left(B-\frac{B\cdot A}{A\cdot A}A\right)+\frac{B\cdot A}{A\cdot A}A. $$ The two vectors on the right are orthogonal. So, by the Pythagorean Theorem, $$ \|B\|^{2} = \left\|B-\frac{B\cdot A}{A\cdot A}A\right\|^{2}+\left\|\frac{B\cdot A}{A\cdot A}A\right\|^{2} \ge \left\|\frac{B\cdot A}{A\cdot A}A\right\|^{2}=\frac{|B\cdot A|^{2}}{\|A\|^{2}}, $$ which is equivalent to $|A\cdot B| \le \|A\|\|B\|$. And and you have equality iff $B-\frac{B\cdot A}{A\cdot A}A =0$.

$\endgroup$
  • $\begingroup$ Could you please try to stick to the given method? I appreciate your effort, but it's important that I understand this particular approach. Furthermore, why does the author write down so explicitely that $A=A-cE+cE$? And why does he then proceed to conclude that A-cE and cE are orthogonal? Is this somehow evident from $A=A-cE+cE$? $\endgroup$ – Ius Klesar Jul 23 '14 at 23:26
  • $\begingroup$ Let's go back to what you learned in 3D Calculus. You have a line through the origin in the direction of $E$. Every point on that line has the form $tE$ for some real $t$. You want to find the orthogonal projection of $A$ onto that line. Then you need to find $t$ so that $(A-tE) \perp E$, which gives $ (A-tE)\cdot E=0$ or $t=A\cdot E/E\cdot E$. The orthogonal projection of $A$ onto that line is the point $tE$ where $t=A\cdot E/E\cdot E$. By design, $A-tE \perp tE$ and $A=A-tE+tE$. $\endgroup$ – DisintegratingByParts Jul 23 '14 at 23:50
  • $\begingroup$ I see, thank you for confirming this. Unfortunately, I still dont have an answer to my main question (about the $\leqq$ sign) $\endgroup$ – Ius Klesar Jul 24 '14 at 0:00
  • $\begingroup$ This is the Pythagorean theorem now: $\|A\|^{2}=\|A-tE\|^{2}+t^{2}\|E\|^{2} \ge t^{2}\|E\|^{2}$. If $E$ is a unit vector in the direction of the line, then $\|A\|^{2} \ge t^{2}$. And you know what $t$ is. All of this is the same as the diagram from 3d projection of a point onto a line. $\endgroup$ – DisintegratingByParts Jul 24 '14 at 0:03
  • $\begingroup$ T.A.E., you don't understand: I fully get everything he does. Read my question. My question is: why are you using a $\geq$ sign? Why not just ">"? $\endgroup$ – Ius Klesar Jul 24 '14 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.