Inner product in a Vector Space

Question

Let $X$ is be a vector space over a field $\mathbf F$, and $\{e_1, \dots ,e_k\}$ a basis for $X$. For $x,y \in X $ such that $$x=\sum_{n=1}^k \lambda_n e_n \textrm{ and } y=\sum_{n=1}^k \mu_n e_n$$ define $(x,y) := \displaystyle \sum\limits_{n=1}^k \lambda_n \overline{\mu_n}$. Show that the function $(\cdot,\cdot): X \times X \to \mathbf F$ is an inner product on $X$.

I have recently started studying this topic. Can someone please help me. Thank you.

Answer 1

Let $X$ be a vector space over a field $F$ and let $\beta = \{ v_1, ..., v_k \} $ be a basis for $X$. Let $x,y \in X$ such that exist $\lambda_1,...,\lambda_k,\gamma_1,...,\gamma_k \in F$ that satisfy: $$x = \sum_{i=1}^k \lambda_i v_i \quad y = \sum_{i=1}^k \gamma_i v_i $$

The definition of an inner product is according to Linear Algebra by Stephen H. Friedberg, Arnold J. Insel and Lawrence E. Spence: A function $(,):X \times X \to F$ such that for all $x,y,z \in X$ and $c \in F$:

• $(x+z,y) = (x,y) + (z,y)$
• $(cx,y) = c(x,y)$
• $\overline{(x,y)} = (y,x)$
• $(x,x) > 0$ if $x \neq 0$

To prove the statement of defining for all $x,y \in X$ $(x,y) = \sum_{i=1}^k \lambda_i \overline{\gamma_i}$ you must write step by step each vector and then apply the definition. First two points are directly proven with writing the explicit form of $x+z$ and $cx$ and then applying the definition of (,). Point 3 and 4 are the somewhat tricky ones.

3. $\overline{(x,y)} = \overline {\sum_{i=1}^k \lambda_i \overline { \gamma_i }} = \sum_{i=1}^k \overline{\lambda_i} \overline{\overline { \gamma_i }} = \sum_{i=1}^k \overline { \lambda_i }\gamma_i = (y , x)$

4. To prove this statement, we use the contraposition version: If $(x,x)\leq 0 $ then $x=0_X$ (the zero vector of $X$).

$(x,x) = \sum_{i=1}^k \lambda_i \overline { \lambda_i } = \sum_{i=1}^k |\lambda_i|^2 $. Because for all $\lambda \in F$ $|\lambda|^2 \geq 0$ then $(x,x)$ can't be negative. Therefore $(x,x) = 0$. So $\sum_{i=1}^k |\lambda_i|^2 = 0$ implies for all $i \in \{1,...,n\}$ $|\lambda_i|^2 = 0$ and that occurs only when $\lambda_i = 0_F$ the zero element of $F$. Then , $x = \sum_{i=1}^k 0_F v_i = 0_X$.

Extra: To add just for better understanding of future problems regarding inner products (and linear algebra), I personally recommend 3Blue1Brown's youtube channel, there's been uploaded a list of videos named The essence of linear algebra and explains geometrically and abstractly main chapters of Linear Algebra like inner product in vector spaces. It could help :) dot product video: https://www.youtube.com/watch?v=LyGKycYT2v0

Answer 2

Based on the comments, I noticed that OP is stuck proving that $(x, y) = \overline{(y, x)},$ so I will prove that here.

Proof. By definition of the inner product, we have that $$\overline{(y, x)} = \overline{\sum_{n = 1}^k \mu_n \bar \lambda_n} = \sum_{k = 1}^n \overline{\mu_n \bar \lambda_n} = \sum_{k = 1}^n \bar \mu_n \bar{\bar \lambda}_n = \sum_{k = 1}^n \lambda_n \bar \mu_n = (x, y).$$