# Orthogonal basis

• August 27th 2011, 06:01 AM
Euclid
Orthogonal basis
Why is an orthogonal basis important?
• August 27th 2011, 07:35 AM
Ackbeet
Re: 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: $\dot{\mathbf{x}} = A\mathbf{x}$. A solution is $e^{At}\mathbf{x}_{0}$. But how do you compute $e^{At}?$ Well, if $A$ is diagonalizable (you can find an orthogonal matrix $P$ such that $A=PDP^{-1}$ for some diagonal matrix $D$), then $e^{At}=Pe^{Dt}P^{-1},$ which is much easier to compute. And the process of diagonalization, or finding the orthogonal matrix $P$, is a process of finding an orthonormal (which is automatically orthogonal) basis.
• August 28th 2011, 04:31 PM
HallsofIvy
Re: Orthogonal basis
If a basis, $\{v_1, v_2, \cdot\cdot\cdot v_n\}$ is orthonormal (all vectors are orthogonal and have length 1), then the vector v can be written as $v= v_1+ v_2+\cdot\cdot\cdot+ v_n$. That is, the coefficients of the $v_i$ vector is just the inner product of v with $v_i$.

If the basis is orthogonal but the vectors are not of unit length, the coefficient is $\frac{}{}$.
• August 28th 2011, 11:52 PM
FernandoRevilla
Re: Orthogonal basis
Also, if $B=\{v_1,\ldots,v_n\}$ is an orthogonal basis of an euclidean space $V$ and $x=x_1v_1+\ldots+x_nv_n$ , $y=y_1v_1+\ldots+y_nv_n$ then, the inner product can be expressed by $=d_1x_1y_1+\ldots+d_nx_ny_n$ . If $B$ is an orthonormal basis, $=x_1y_1+\ldots+x_ny_n$ 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 $V$ . This means that (technically but nor conceptually) things work as if "there is only one inner product" in $V$ .