Strange proof of Schwarz Inequality with Pythagorean Theorem

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $B-\frac{B\cdot A}{A\cdot A}A =0$.