2
$\begingroup$

I am confused about the hint Spivak adds to problem 1-2 in his Calculus on Manifolds:

When does equality hold in Theorem 1-1(3)? Hint: Re-examine the proof; the answer is not “when $x$ and $y$ are linearly dependent.”

The said (in)equality here would be: $$ \left \lvert x + y \right \rvert \leq \left \lvert x \right \rvert + \left \lvert y \right \rvert $$

My question: Why is $x$ and $y$ being linear not the condition for equality?

If one squares both sides of the equation, one obtains for the LHS: $$ \left \lvert x + y \right \rvert^2 = \left \lvert x \right \rvert^2 + \left \lvert y \right \rvert^2 + 2\sum^n_{i=1} x_i y_i \text{,} $$ and for the RHS $$ \left( \left\lvert x \right\rvert + \left\lvert y \right\rvert \right)^2 = \left \lvert x \right \rvert^2 + \left \lvert y \right \rvert^2 + 2 \left\lvert x \right\rvert \cdot \left\lvert y \right\rvert\text{.} $$

Clearly, the LHS and RHS are equal if: $$ \sum^n_{i=1} x_i y_i = \left\lvert x \right\rvert \cdot \left\lvert y \right\rvert \text{,} $$

which is the case iff $x$ and $y$ are linearly dependent (as per Theorem 1-1(2)). What am I missing? Why is the answer not “when $x$ and $y$ are linearly dependent”? What other condition is there?

$\endgroup$
  • 1
    $\begingroup$ You need to note there's an absolute value in theorem 1-1(2). $\endgroup$ – user121880 May 21 '14 at 21:51
  • $\begingroup$ Yeah, I managed to ignore the absolute value sign after staring at this the entire day. $\endgroup$ – mSSM May 21 '14 at 22:01
3
$\begingroup$

If equality holds, as you say, they are linearly dependent. But the converse doesn't hold. Think of $x=(1,1)$ and $y=(-2,-2)=-2x.$

Suppose $y=\lambda x.$ Then

$$\lambda \sum_{i=1}^n\sum^n_{i=1} x_i^2= \left\lvert x \right\rvert \cdot \left\lvert y \right\rvert= |\lambda||x|^2.$$

What sign has to have $\lambda?$

| cite | improve this answer | |
$\endgroup$

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.