Problem about existence of a inner product in $\mathbb{R}^3$

Question

Let $B=\{e_1=(1,0,0), e_2=(0,1,0),e_3=(0,0,1)\}$ be a base of $\mathbb{R}^3$.

Does there exist an inner product $\langle\cdot,\cdot\rangle:\mathbb{R}^3\to\mathbb{R}$ such that $\langle e_1,e_1\rangle=2$, $\langle e_2,e_2\rangle=3$, $\langle e_3,e_3\rangle=4$, $\langle e_1,e_2\rangle=0$ and $\langle e_2,e_3\rangle=\langle e_1,e_3\rangle=1$?

This problem was taken of an Analisys book (first chapter). I think I need to know some property of inner product to solve it, but the author just defines inner product in the text. So, I would like know if there is a way to ensure the existence of an inner product in this kind of problem.

Answer 1

Hint: $$\begin{pmatrix} a_1 & a_2 & a_3\end{pmatrix} \begin{pmatrix} 2 & 0 & 1\\ 0 & 3 &1\\ 1 & 1 & 4\\ \end{pmatrix}\begin{pmatrix} b_1 \\ b_2 \\ b_3\end{pmatrix}$$

Answer 2

Hint: Try to find an invertible matrix A whose columns $v_1,v_2,v_3$ satisfy the given conditions, and then define $<u, v>=(Au)\cdot (Av)$. For example, you could take $v_1=\begin{pmatrix}1\\1\\0\end{pmatrix}$ and $v_2=\begin{pmatrix}1\\-1\\-1\end{pmatrix}$, and then solve for $v_3$.