Clarification on Some Definition of Inner Product Space

Question

Suppose $V$ is finite-dimensional Real vector space and $T\in \mathcal{L}(V)$. Suppose that $V$ has a basis $(e_1,e_2,\ldots, e_n)$ of eigenvectors of $T$, every element of $V$ can be written as a linear combination of $(e_1,\ldots, e_n)$. Thus we define an inner product of $V$ by $$\langle a_1e_1+\cdots + a_ne_n,b_1e_1+\cdots + b_ne_n\rangle=a_1b_1+\cdots + a_nb_n$$

Verify that this is an inner product on $V$,and that $(e_1,\ldots, e_n)$ is orthonormal basis of $V$ with respect to this inner product.

---

To verify that this is an inner product, we need to show that four inner product properties hold for it. This part is easy.

My confusion is the meaning of "$(e_1,\ldots, e_n)$ is orthonormal basis of $V$ with respect to this inner product." I don't understand the meaning of "with respect to" here. Does it imply that, with different basis will give different inner product ? How to verify that $(e_1,\ldots, e_n)$ is orthonormal basis of $V$ with respect to this inner product?

Answer 1

Here "with respect to" just means "using" (and emphasizes that whether this will not in general hold for other inner products).

To show that a basis $(e_a)$ is orthonormal, we just apply the definition, that is, we check that

$\langle e_a, e_a \rangle = 1$ for all indices $a$, and

$\langle e_a, e_b \rangle = 0$ for all indices $a, b$ such that $a \neq b$.

Answer 2

Two vectors can be orthogonal according to one inner product, but not to other. Example: in $\Bbb R^2$ consider: $$\langle (x_1,y_1), (x_2, y_2)\rangle_1 = x_1x_2 + y_1y_2 \qquad \langle (x_1, y_1), ( x_2, y_2)\rangle_2 = x_1x_2 + 2 y_1y_2$$ Then $(1,1)$ and $(1,-1)$ are $\langle \cdot , \cdot \rangle_1-$orthogonal, but not $\langle \cdot, \cdot \rangle_2-$orthogonal.

In general, if you have a non-degenerate (and symmetric, for avoiding technical problems) billinear form $g$, you can say that two vectors $u$ and $v$ are $g-$orthogonal if $g(u,v) = 0 $.