Inner Product proof, axiom 1

Question

I hate to argue that my text book is wrong. That being said, I am going to try and do just that.

The text book says that this IS a valid inner product, I disagree.

The vectors u and v are defined as: $$\mathbf u = (u_1, u_2),\mathbf v = (v_1, v_2)$$

And they define the inner product <u, v> to be: $$<\mathbf u, \mathbf v> = 2u_1v_2 + u_2v_1 + u_1v_2 + 2u_2v_2$$

I claim that this fails Axiom 1, where <u, v> = <v, u>.

Proof: $$<\mathbf v, \mathbf u> = 2v_1u_2 + v_2u_1 + v_1u_2 + 2v_2u_2$$ and these are not equal because of the first part on each side $2u_1v_2 \neq 2v_1u_2$

so even though the rest of the terms are commutative, this would fail axiom 1 because the first ones are not.

$$2u_1v_2 + u_2v_1 + u_1v_2 + 2u_2v_2 \neq 2v_1u_2 + v_2u_1 + v_1u_2 + 2v_2u_2$$ If I am wrong it would be much appreciated if someone could show me why. Please and thank you.

Answer 1

You see, writing the inner product in a more illuminating manner will explain why your textbook either has a typo, as others point out, or is simply wrong, which i salute you for pointing out.

So we know $\langle u,v\rangle = 2u_1v_2 + u_2v_1 + u_1v_2 + 2u_2v_2$. We rewrite this as: $$ \langle u,v\rangle = \begin{pmatrix} u_1 \ u_2 \end{pmatrix} \begin{pmatrix} 0\ 3 \\ 1 \ 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} $$ Now write $\langle v,u\rangle $ as: $$ \langle v,u\rangle = \begin{pmatrix} v_1 \ v_2 \end{pmatrix} \begin{pmatrix} 0\ 3 \\ 1 \ 2 \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} $$

Now, $\langle v,u\rangle = \langle u,v\rangle $ would imply that: $$ \begin{pmatrix} v_1 \ v_2 \end{pmatrix} \begin{pmatrix} 0\ 3 \\ 1 \ 2 \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \begin{pmatrix} u_1 \ u_2 \end{pmatrix} \begin{pmatrix} 0\ 3 \\ 1 \ 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} $$ Both sides are numbers (or $1 \times 1 $ matrices), so taking transpose on one side doesn't affect the equation, whence on taking the transpose of the left hand side, $$ \begin{pmatrix} u_1 \ u_2 \end{pmatrix} \begin{pmatrix} 0\ 1 \\ 3 \ 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} u_1 \ u_2 \end{pmatrix} \begin{pmatrix} 0\ 3 \\ 1 \ 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} $$ But this is not true, because the matrices in the middle are not the same. Hence your textbook is wrong. But the illuminating fact here, is that if the middle matrix were the same as it's transpose, then you would have the property satisfied. These matrices are called symmetric matrices. Next time, all you need to do is write your inner product in this form and check whether the resulting matrix is symmetric or not.

Besides, others have pointed out the typo there could be in your textbook. Please ask if any doubts.

Answer 2

You're right that as stated it is not an inner product for exactly the reason you gave. However, most probably the first term is typed incorrectly and it should have been "$2 u_1 v_1$", which then makes sense as the four terms correspond to the four possible products of $u_1,u_2$ with $v_1,v_2$.

Answer 3

Examine it again: $\langle\mathbf u, \mathbf v\rangle = 2\underline{u_1v_2} + u_2v_1 + \underline{u_1v_2} + 2u_2v_2$

The underlined terms are a clue that there may be a typo in the expression.   Why else are they separated?

The symmetry requirement suggests that the first term was intended to be $u_1v_1$.

$$\langle\mathbf u, \mathbf v\rangle = 2u_1v_1 + u_2v_1 + u_1v_2 + 2u_2v_2$$