Why is an orthogonal basis important?

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- Aug 27th 2011, 05:01 AMEuclidOrthogonal basis
Why is an orthogonal basis important?

- Aug 27th 2011, 06:35 AMAckbeetRe: Orthogonal basis
Because it is particularly easy to write any vector in terms of an orthogonal basis (even easier if it's orthonormal). Here's an application: thinking of the vector space of functions, you can solve differential equations formed by means of, say, an Hermitian operator, by writing down the solution as a sum of eigenvectors of the operator. Physicists do this kind of thing all the time to solve ODE's or even PDE's.

Another application is diagonalization. Suppose you have the first-order system of DE's: $\displaystyle \dot{\mathbf{x}} = A\mathbf{x}$. A solution is $\displaystyle e^{At}\mathbf{x}_{0}$. But how do you compute $\displaystyle e^{At}?$ Well, if $\displaystyle A$ is diagonalizable (you can find an orthogonal matrix $\displaystyle P$ such that $\displaystyle A=PDP^{-1}$ for some diagonal matrix $\displaystyle D$), then $\displaystyle e^{At}=Pe^{Dt}P^{-1},$ which is much easier to compute. And the process of diagonalization, or finding the orthogonal matrix $\displaystyle P$, is a process of finding an orthonormal (which is automatically orthogonal) basis. - Aug 28th 2011, 03:31 PMHallsofIvyRe: Orthogonal basis
If a basis, $\displaystyle \{v_1, v_2, \cdot\cdot\cdot v_n\}$ is

**orthonormal**(all vectors are orthogonal and have length 1), then the vector v can be written as $\displaystyle v= <v,v_1>v_1+ <v,v_2>v_2+\cdot\cdot\cdot+ <v, v_n>v_n$. That is, the coefficients of the $\displaystyle v_i$ vector is just the inner product of v with $\displaystyle v_i$.

If the basis is orthogonal but the vectors are not of unit length, the coefficient is $\displaystyle \frac{<v, v_i>}{<v_i, v_i>}$. - Aug 28th 2011, 10:52 PMFernandoRevillaRe: Orthogonal basis
Also, if $\displaystyle B=\{v_1,\ldots,v_n\}$ is an

**orthogonal**basis of an euclidean space $\displaystyle V$ and $\displaystyle x=x_1v_1+\ldots+x_nv_n$ , $\displaystyle y=y_1v_1+\ldots+y_nv_n$ then, the inner product can be expressed by $\displaystyle <x,y>=d_1x_1y_1+\ldots+d_nx_ny_n$ . If $\displaystyle B$ is an**orthonormal**basis, $\displaystyle <x,y>=x_1y_1+\ldots+x_ny_n$ that is, the inner product works as in the usual case.

By the Gram-Schmidt Theorem always is possible to find an orthonormal basis of $\displaystyle V$ . This means that (technically but nor conceptually) things work as if "there is only one inner product" in $\displaystyle V$ .