Could someone please help me untangle the notation with the following?

Let $V$ be a real vector space with inner product $\langle \cdot \,,\cdot \rangle : V \times V \rightarrow \mathbb{R}$. Let $W = V \times V$ with vector addition defined by

$(v_1,w_1) + (v_2,w_2 ) = (v_1 + v_2, \, w_1 + w_2)$

and scalar multiplication defined by $(a + bi)(v,w) = (av - bw, aw + bv)$ and inner product $\langle.\,,.\rangle' : W \times W \rightarrow \mathbb{C}$ defined by

$\langle(v_1 ,w_1 ),(v_2 ,w_2 )\rangle' = (\langle v_1 ,v_2 \rangle + \langle w_1 ,w_2\rangle) + i(-\langle v_1 ,w_2 \rangle + \langle w_1 ,v_2 \rangle )$

If I am to prove positive definiteness then I would start by:

Assume that the inner product on $V$ satisfies the three axioms of inner products. We show that $W$ is positive definite:

$\langle(v_1 ,w_1 ),(v_2 ,w_2 )\rangle' = (\langle v_1 ,v_2 \rangle + \langle w_1 ,w_2\rangle) + i(-\langle v_1 ,w_2 \rangle + \langle w_1 ,v_2 \rangle )$

But, what does $\langle v_1,v_2 \rangle$ equal to? This notation is very difficult for me to follow. Am I to assume that $\langle v_1,v_2 \rangle = (|v_1|+|v_2|)^2$?

Any and help would be very much appreciated.

  • $\begingroup$ Why the inner product should be positive? Do you mean $\langle {u, u}\rangle$ not negative? $\endgroup$
    – dmtri
    Commented Mar 18, 2019 at 10:53
  • $\begingroup$ @dmtri inner products are always positive definite. It is one of the three properties of inner products. $\langle(v_1 ,v_2 ),(v_1 ,v_2 )\rangle'$ should always be greater than or equal to zero. $\endgroup$
    – Idle Fool
    Commented Mar 18, 2019 at 10:59
  • $\begingroup$ @астонвіллаолофмэллбэрг So, please help me. I understand what you wrote to a point. If you expand the inner product of $\langle u,u \rangle$ for some $u \in V$ it would go a long ways towards understanding. Is it $\langle u,u \rangle = (|u|+|u|)^2$, which I believe is the standard inner product? $\endgroup$
    – Idle Fool
    Commented Mar 18, 2019 at 11:06
  • $\begingroup$ @IdleMathGuy I will write a detailed answer, and we'll attach a discussion to that if required. $\endgroup$ Commented Mar 18, 2019 at 11:09
  • $\begingroup$ @астонвіллаолофмэллбэрг Thank you so much! I just want to understand. $\endgroup$
    – Idle Fool
    Commented Mar 18, 2019 at 11:37

1 Answer 1


Untangling the notation

Let $W = V \times V$ with addition defined by $(v_1,w_1) + (v_2,w_2) = (v_1+v_2,w_1+w_2)$.

  • Each element of $W$ is a pair of elements from $V$. So an element of $W$ looks like $w = (v_,v_2)$ where $v_1,v_2 \in V$. Above, all of $v_1,v_2,w_1,w_2$ are in $V$. The addition is componentwise.

and scalar multiplication be defined by $(a+bi)(v,w) = (av-bw,aw+bv)$.

Although not mentioned, $W$ is going to be designed as a complex vector space. So, we have to explain how we multiply a complex number, which is $(a+bi)$, with some element of $W$, say $(v,w)$ where $v,w \in V$. One of course must check the usual properties of multiplication for this , but I won't do it.

and inner product $\langle \cdot,\cdot \rangle' : W \times W \to \mathbb C$ defined by : $$ \langle (v_1,w_1),(v_2,w_2)\rangle' = (\langle v_1,v_2\rangle + \langle w_1,w_2\rangle) + i(-\langle v_1,w_2\rangle + \langle w_1,v_2 \rangle) $$

Over here, you are defining a complex inner product on $W$, taking values in $\mathbb C$, so you are specifying explicitly the real and imaginary parts of the result. Again, $v_1,v_2,w_1,w_2$ are vectors in $V$, and all terms on the right hand side involve the inner product on $V$.

Now, here's the point : we have not shown yet that the above function is an inner product, so the author calling it an inner product is "incorrect writing"(but not an incorrect statement), especially when he wants us to prove that it is positive definite first.

Solving the problem

So we have the real vector space $V$, and it has an inner product $\langle \cdot , \cdot \rangle$. This inner product has nothing to do with the usual $\mathbb R^n$-inner product, or the absolute value $|\cdot|$ : an inner product is one that satisfies (conjugate) symmetry, linearity in the first argument and positive definiteness, and while operating you must assume only these three properties and no more.

For positive definiteness, we need to prove exactly this : for every $w \in W$, we have $\langle w,w\rangle' \geq 0$(NOTE : This involves the fact that $\langle w,w\rangle'$ must in fact be a real number because no order is defined on the complex numbers , but we will see that this is the case here), with equality exactly when $w = 0$.

For this, first since $w \in W$, we will write it as $w=(w_1,w_2)$ where $w_1,w_2 \in V$. Then we use the formula to write down $\langle w,w\rangle'$ : $$ \langle w,w\rangle' = \langle (w_1,w_2),(w_1,w_2)\rangle \\ = \langle w_1,w_1\rangle + \langle w_2,w_2\rangle - i\langle w_1,w_2\rangle + i\langle w_1,w_2\rangle \\ = \langle w_1,w_1 \rangle + \langle w_2,w_2\rangle $$

is therefore a real number, and can be compared with $0$.

However, now things are very simple : because the inner product on $V$ is positive definite, certainly the last line is a sum of two non-negative terms and is hence non-negative.

Furthermore, if $\langle w,w\rangle' = 0$, then the last line is zero, which by non-negativity forces each of $\langle w_1,w_1\rangle$ and $\langle w_2,w_2\rangle$ to be zero, which forces both $w_1=w_2=0$ because the inner product on $V$ is positive definite. Of course, then $w =(0,0)$ is the zero element of $W$.

Thus, we show that positive definiteness holds.

Since I have answered the question, you should now strengthen your grip by proving that the function $\langle \cdot,\cdot\rangle'$ satisfies all the other properties of an inner product on $W$ as well.

  • $\begingroup$ Thank you! I will definitely prove that the rest of the properties. If only math instructors explained things so well... $\endgroup$
    – Idle Fool
    Commented Mar 18, 2019 at 14:38
  • $\begingroup$ You are welcome! Also, when you are done,you can post an answer and close the question. $\endgroup$ Commented Mar 18, 2019 at 14:44

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