It looks like there is a completely miswritten formula in Apostol's Calculus, Volume II, regarding inner products. The formula states that: "If $x=(x_1,x_2)$ and $y=(y_1,y_2)$ are any two vectors in $V_2$, define $(x,y)$ by the formula $(x,y)=2x_1y_1+x_1y_2+x_2y_1+x_2y_2$. Through online research and consulting with a friend, it looks like $(x,y)=x_1y_1+x_2y_2$ is in fact the correct formula for inner products, but I wanted to make sure the book's version was not also correct, because this is more than a minute typo. Is there any validity to the book's definition?


No, no typos. One can define different inner product in $\mathbb{R}^2$ than the standard one. See for instance this question.

If you read carefully what the book by Apostol (page 14 and page 15) says, you would find that he gives you both the standard and non-standard examples:

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Here is the definition in his book: enter image description here

  • 1
    $\begingroup$ Okay, I reread the statement that the inner product is defined axiomatically rather than by a specific formula, and this makes sense. Thank you! $\endgroup$
    – user376580
    Dec 20 '16 at 15:50

The canonical inner product in $\mathbb R^2$ is given by your formula, i.e.

$$\langle x,y\rangle = x_1y_1+x_2y_2$$

However, there are many other possible inner products in $\mathbb R^2$. Apostol is giving an example of a second choice of inner product in $\mathbb R^2$.


One should check that the proposed formula actually defines a (real) inner product.


Does $(x,y) = (y,x)$ ? Yes, it is easily checked that exchanging roles of $x,y$ does not change the result defined by $(x,y)=2x_1y_1+x_1y_2+x_2y_1+x_2y_2$.

Linearity in each argument

Given symmetry, it suffices to check linearity in just one argument:

$$ (au+bv,y) = a(u,y) + b(v,y) $$

Because the formula is homogeneous of degree $1$ in the components of each argument, this check is straightforward.


Usually the most interesting property to check, as is the case here. We want to show:

$$ (x,x) \ge 0 \text{ with equality only when } x=0 $$

Note that by the formula $(x,x) = 2x_1^2 + 2x_1x_2 + x_2^2$. It suffices to show that this expression is nonnegative and only zero when $x_1 = x_2 = 0$.

To see this easily, rewrite the expression:

$$ 2x_1^2 + 2x_1x_2 + x_2^2 = x_1^2 + (x_1 + x_2)^2 $$

From this clearly we get a nonnegative result, and we only get zero if both $x_1=0$ and $x_1+x_2 = 0$, implying that $x_1=x_2=0$.


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