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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?

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    $\begingroup$ The beginning of your question is curious, because $T$ does not occur in the actual question. Since every basis is a basis of eigenvectors of some linear operator (for instance of the identity operator), you might as well have started with "Let $(e_1,e_2,\ldots, e_n)$ be any basis of a real vector space $V$". The way it is fomulated now might give the impression that the inner product defined is determined by $T$, but that is not true. $\endgroup$ – Marc van Leeuwen Sep 2 '14 at 8:56
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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 $a \neq b$.

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  • $\begingroup$ thx for the help. Another doubt: orthonormal basis $(e_1,...e_n)$ from the question can be obtained from any basis of $V$, for example, $(v_1,...v_n)$, by applying gram-schmidt, using the inner product as given above. This also means that $(e_1,...e_n)$ with respect to this inner product ? $\endgroup$ – ElleryL Sep 1 '14 at 16:56
  • $\begingroup$ Any basis becomes orthonormal this way. $\endgroup$ – Marc Bogaerts Sep 1 '14 at 17:00
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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 $.

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  • $\begingroup$ If we define an inner product, and given any basis of a vector space $V$,if we apply Gram-Schmidt to get the orthonormal basis. We say that it's an orthonormal basis of $V$ with respect to the inner product we defined. Is this correct understanding? $\endgroup$ – ElleryL Sep 2 '14 at 0:56
  • $\begingroup$ Orthogonality and orthonormality are concepts that depend on the inner product that you're working with. The book "Geometry of Minkowski Space-Time", from G.L. Naber has a part of linear algebra in which he does stuff quite generally, you might like it. $\endgroup$ – Ivo Terek Sep 2 '14 at 1:01

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