# Spivak Calculus on Manifolds, problem 1-2

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?

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

## 1 Answer

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 $\lambda?$