Apostol question on alternative definition of dot product

The problem says:

Suppose we define the dot product by $A\cdot B = \sum_{k=1}^n |a_kb_k|$. Which of the following properties hold with this new defition? Does the Cauchy-Schwarz inequality still hold?

(a) Commutativity: $A\cdot B = B\cdot A$

(b) Distributivity: $A\cdot (B+C) = A\cdot B + A\cdot C$

(c) Homogeneity: $c(A\cdot B) = (cA)\cdot B = A\cdot(cB)$

(d) Positivity: $A\cdot A > 0$ if $A\neq O$

(e) $A\cdot A = O$ if $A = O$.

Note that $A,B,C \in V_n$ and $c$ is a scalar.

The properties (a), (d), and (e) still hold.

Distributivity does not hold due to the triangle inequality. Observe that \begin{align*} A\cdot (B+C) &= \sum_{k=1}^n |a_k(b_k+c_k)| \\ &= \sum_{k=1}^n |a_kb_k+a_kc_k| \\ &\leq \sum_{k=1}^n \left(|a_kb_k| + |a_kc_k|\right) \\ &= \sum_{k=1}^n |a_kb_k| + \sum_{k=1}^n |a_kc_k| \\ &= A\cdot B + A\cdot C \ . \end{align*} We can see that equality will not hold for all $A,B,$ and $C$.

The homogeneity property does not hold due to the absolute value signs. Assume $c<0$. Then \begin{align*} (cA)\cdot B &= \sum_{k=1}^n |(ca_k)b_k| \\ &= |c|\sum_{k=1}^n |a_kb_k| \\ &\neq c\sum_{k=1}^n |a_kb_k| \\ &= c(A\cdot B) \ . \end{align*} The back of the book states that only (b) does not hold. I fail to see how (c) could hold. Is this a typo or am I incorrect? I can prove the Cauchy-Schwarz inequality if (c) does hold, but I don't think Cauchy-Schwarz holds in the case that (c) does not.

EDIT: Actually, since homogeneity holds for $c\geq 0$, I am inclined to believe that Cauchy-Schwarz does hold (also since I can't find a counter-example). But I am having difficulty proving it since distributivity does not hold. Any hints?

EDIT2: I found that Cauchy-Schwarz does hold. I put my proof as an answer. I appreciate any critiquing of it.

• Easiest to see distributivity fail by having $B=-C$. Aug 6, 2013 at 18:22
• Assuming by "scalar" the book allows negative numbers, then yes, the book appears to be wrong. Aug 6, 2013 at 18:23
• If you only consider non-negative $c$, then homogeneity holds.
– robjohn
Aug 6, 2013 at 18:45
• In this chapter of Apostol's book, he seems to be considering $c\in\mathbb{R}$. Later, he considers scalars in $\mathbb{C}$.
– user59083
Aug 6, 2013 at 18:47
• Can anyone critique the below proof for Cauchy-Schwarz holding?
– user59083
Aug 7, 2013 at 16:57

The Cauchy-Schwarz Inequality does hold.

Define $A^+ = ( |a_1| , |a_2| , \dots |a_n| )$.

Then our dot product $A \cdot_n B$ is the same as $A^+ \cdot_o B^+$ with the old dot product. Also, the norm for the new and old dot products are equal, since $\sum |a_i a_i| = \sum a_i a_i$.

Since the (original) Cauchy Schwarz inequality holds for $\cdot_o$:

\begin{align*} A \cdot_n B &= A^+ \cdot_o B^+ \\ &\le \Vert A^+ \Vert \Vert B^+ \Vert \\ &= \Vert A \Vert \Vert B \Vert \end{align*}

Equality holds iff $B^+ = k A^+$, which is a weaker condition than $B = kA$. In $\mathbb R^2$, equality holds for $A = ( 1,1 )$ and $B = ( 1,-1)$.

• Can you explain why it is important that $B^+ = kA^+$ is a weaker condition than $B=kA$?
– user59083
Aug 9, 2013 at 13:24
• @5space $B = kA \implies B^+ = kA^+$. However, the converse does not hold, which is evident when we take $A = ( 1,1 )$ and $B = ( 1,-1 )$. Now, $B^+ = A^+$ but $B \ne kA$ no matter how we choose $k$. This is important because this means that equality can hold without $B = kA$, unlike the usual Cauchy-Schwarz inequality.
– A.S
Aug 9, 2013 at 13:26

This proves that the Cauchy-Schwarz inequality holds for the new dot product definition:

The result is trivial if either $A$ or $B$ is $O$. Assume $A,B\neq O$.

Let $A',B'\in V_n$ such that $a'_i = |a_i|$ and $b'_i = |b_i|$ for each $i=1,2,...,n$. Let $\theta'$ be the angle between $A'$ and $B'$. Since $A'\cdot B' = \sum_{k=1}^n |a'_kb'_k| = \sum_{k=1}^n a'_kb'_k$, we have $$A'\cdot B' = ||A'||\,||B'||\,\cos\theta'.$$ But notice that $A'\cdot B' = A\cdot B$, $||A'|| = ||A||$, and $||B'|| = ||B||$. Therefore $$A\cdot B = ||A||\,||B||\,\cos\theta',$$ which implies that $$A\cdot B = |A\cdot B| \leq ||A||\,||B||$$ since $\theta \in \left[ 0, \frac{\pi}{2} \right]$. Thus the Cauchy-Schwarz inequality holds. Equality holds if and only if $\cos\theta'=1$. That is, if and only if $A'||B'$.

Equality does not hold iff $A||B$, since $A'||B'$ does not imply $A||B$.

• $|A \cdot B| \ne \sum_i |a_i b_i|$ in general.
– A.S
Aug 9, 2013 at 6:48
• @AndrewSalmon: But isn't it the case with the new definition? Since $A\cdot B = \sum_i |a_ib_i|$ by definition?
– user59083
Aug 9, 2013 at 13:19
• Then $A \cdot B = \Vert A \Vert \Vert B \Vert \cos \theta$ ceases to hold. However, you define $\cdot$ here, it's not going to work as you have here, since this implies that equality holds iff $B = kA$, but note the counterexample: $A = ( 1,1 )$ and $B = ( 1,-1 )$.
– A.S
Aug 9, 2013 at 13:24
• @AndrewSalmon: Doesn't $A\cdot B = ||A||\,||B||\,\cos\theta$ hold for $\theta$ between $A'$ and $B'$ since we are just substituting $A'\cdot B' = A\cdot B$ and $||A'|| = ||A||$, $||B'||=||B||$? Because $A'\cdot B' = ||A'||\,||B'||\,\cos\theta$ does hold. So this does prove the Cauchy-Schwarz inequality, but it does not imply that equality holds iff $B=kA$. The problem with my proof (of equality holding) is the ($\implies$) direction. The problem is that $B\neq kA$ does not imply $B'\neq|k|A'$. What do you think?
– user59083
Aug 9, 2013 at 13:53
• That works. It's important that by $\theta$, we really mean the angle between $A^+$ and $B^+$ (the absolute value sign around $\cos \theta$ is unnecessary, because $0 \le \theta \le \frac{\pi}{4}$). $\cos \theta = 1$ iff $A^+$ and $B^+$ are parallel, so we get the same conditions here as we did in my answer.
– A.S
Aug 9, 2013 at 13:59