Let $V$ be a vector space and let $v_1, v_2, \dots , v_n$ be a basis in $V$. For $x = \sum_{k=1}^{n}{\alpha_kv_k}, y = \sum_{k=1}^{n}{\beta_kv_k}$, define $\langle x,y \rangle := \sum_{k=1}^{n}{\alpha_k\bar{\beta_k}}$. Prove that $\langle x, y \rangle$ defines an inner product in V.

When I tried doing this question, I thought the best thing to do is to see if this definition satisfies the rules of an inner product. This wasn't entirely an issue when proving conjugate symmetry:

\begin{align*} \overline{\langle y,x \rangle} &= \overline{\sum_{k=1}^{n}{\beta_k\overline{\alpha_k}}} \\ &= \overline{\beta_1 \overline{\alpha_1}} + \dots + \overline{\beta_n \overline{\alpha_n}} \\ &= \alpha_1 \overline{\beta_1} + \dots + \alpha_n \overline{\beta_n} \\ &= \sum_{k=1}^{n}{\alpha_k \overline{\beta_k}} \\ &= \langle x,y \rangle \end{align*}

What I seem to be stuggling with is trying to prove linearity, non-negativity and non-degeneracy for this because for linearity the expression seems to become too big and complicated and for the other two properties I would need to have the inner product of a vector $x$ by itself.

Any suggestion of how I can go forward with this?

  • $\begingroup$ Linearity , non negativity are also as simple as you have done the conjugate symmetry part. Why do you think that those one will be lengthy? $\endgroup$
    – A learner
    Commented Jul 20, 2021 at 16:20

1 Answer 1



Define $z = \sum_k \gamma_k v_k$, then $y + z = \sum_k (\beta_k + \gamma_k) v_k$ and

\begin{eqnarray} \langle x, y + z\rangle &=& \sum_k a_k\overline{(\beta_k + \gamma_k)} \\ &=& \sum_k (a_k \overline{\beta_k} + a_k \overline{\gamma}) \\ &=& \sum_k a_k \overline{\beta_k} + \sum_k a_k \overline{\gamma_k} \\ &=& \langle x, y\rangle + \langle x, z\rangle \end{eqnarray}


\begin{eqnarray} \langle x, x\rangle &=& \sum_k a_k\overline{a_k} \\ &=& \sum_k |a_k^2| \ge 0 \end{eqnarray}

which is only zero iff $a_k = 0$


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