[Math] Abstract Inner Product Proof

linear algebra

Let $B$ be a basis for the vector space $V$ and define $\langle,\rangle$ by $\langle u,v \rangle = [u]_B \cdot [v]_B$ for $u,v \in V$. Show that $\langle,\rangle $ is an inner product on $V$.

What I know is that it must adhere the following axioms in order to be considered an inner product:

  • $\langle u,v \rangle = \langle v,u \rangle$ [Symmetry axiom]
  • $\langle u + v,w \rangle = \langle u, w \rangle + \langle v, w \rangle$ [Additivity axiom]
  • $\langle ku +,v\rangle = k \langle v,u \rangle$ [Homogeneity axiom]
  • $\langle v,v\rangle \geq 0$ [Positivity axiom]

The abstractness of this proof is throwing me off, any help would be greatly appreciated.

Best Answer

@AidanSims's comment is awesome! However, I would like to share my step-by-step answer.

Let $B=\{v_1,v_2,\ldots,v_n\}$. Given $u,v,w\in V$, write $u=\sum_{i=1}^na_iv_i$, $v=\sum_{i=1}^nb_iv_i$, and $w=\sum_{i=1}^nc_iv_i$ for some scalars $a_i,b_i,c_i\in\mathbb{R}$.

  • We have \begin{align} \left\langle u,v\right\rangle &=[u]_B\cdot[v]_B =\begin{pmatrix}a_1\\a_2\\\vdots\\a_n \end{pmatrix}\cdot \begin{pmatrix}b_1\\b_2\\\vdots\\b_n \end{pmatrix} =\sum_{i=1}^na_ib_i\\ &=\sum_{i=1}^nb_ia_i =\begin{pmatrix}b_1\\b_2\\\vdots\\b_n \end{pmatrix}\cdot\begin{pmatrix}a_1\\a_2\\\vdots\\a_n \end{pmatrix} =[v]_B\cdot[u]_B =\left\langle v,u\right\rangle. \end{align}
  • We have \begin{align} \left\langle u+v,w\right\rangle &=[u+v]_B\cdot[w]_B =\begin{pmatrix}a_1+b_1\\a_2+b_2\\\vdots\\a_n+b_n \end{pmatrix}\cdot \begin{pmatrix}c_1\\c_2\\\vdots\\c_n \end{pmatrix} =\sum_{i=1}^n(a_i+b_i)c_i\\ &=\sum_{i=1}^na_ic_i+\sum_{i=1}^nb_ic_i =\begin{pmatrix}a_1\\a_2\\\vdots\\a_n \end{pmatrix}\cdot\begin{pmatrix}c_1\\c_2\\\vdots\\c_n \end{pmatrix}+ \begin{pmatrix}b_1\\b_2\\\vdots\\b_n \end{pmatrix}\cdot\begin{pmatrix}c_1\\c_2\\\vdots\\c_n \end{pmatrix}\\ &=[u]_B\cdot[w]_B+[v]_B\cdot[w]_B =\left\langle u,w\right\rangle+\left\langle v,w\right\rangle. \end{align}
  • Given $k\in\mathbb{R}$, then we have \begin{align} \left\langle ku,v\right\rangle &=[ku]_B\cdot[v]_B =\begin{pmatrix}ka_1\\ka_2\\\vdots\\ka_n \end{pmatrix}\cdot \begin{pmatrix}b_1\\b_2\\\vdots\\b_n \end{pmatrix} =\sum_{i=1}^n(ka_i)b_i\\ &=k\sum_{i=1}^na_ib_i =k\begin{pmatrix}a_1\\a_2\\\vdots\\a_n \end{pmatrix}\cdot\begin{pmatrix}b_1\\b_2\\\vdots\\b_n \end{pmatrix} =k\left([u]_B\cdot[v]_B\right) =k\left\langle u,v\right\rangle. \end{align}
  • We have \begin{align} \left\langle v,v\right\rangle &=[v]_B\cdot[v]_B =\begin{pmatrix}b_1\\b_2\\\vdots\\b_n \end{pmatrix}\cdot \begin{pmatrix}b_1\\b_2\\\vdots\\b_n \end{pmatrix} =\sum_{i=1}^nb_i^2\geq 0. \end{align}
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