Problem of the Week #37 - December 10th, 2012

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Chris L T521

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Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Apply the Gram-Schmidt process to the basis $\{1,t,t^2\}$ for the Euclidean space $P_2$ (the space of degree 2 or less polynomials with real coefficients) and obtain an orthonormal basis for $P_2$.

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In this problem, the inner product on $P_2$ is given by

$\langle f,g\rangle = \int_0^1 f(t)g(t)\,dt.$

Remark:
Recall that if we have a basis $\{\mathbf{u}_1,\ldots,\mathbf{u}_n\}$, then the orthogonal basis obtained by applying Gram-Schmidt consists of vectors $\{\mathbf{v}_1,\ldots, \mathbf{v}_n\}$ where

$\mathbf{v}_1=\mathbf{u}_1\qquad\text{and}\qquad \mathbf{v}_k = \mathbf{u}_k-\sum_{i=1}^{k-1}\frac{\langle \mathbf{u}_k,\mathbf{v}_i\rangle}{\langle\mathbf{v}_i,\mathbf{v}_i\rangle}\mathbf{v}_i;\qquad k=2,\ldots, n$

The orthonormal basis would then be $\{\mathbf{w}_1,\ldots,\mathbf{w}_n\}$ with $\mathbf{w}_j= \dfrac{\mathbf{v}_j}{\langle \mathbf{v}_j,\mathbf{v}_j\rangle}$.

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Chris L T521

Well-known member
Staff member
This week's question was correctly answered by Ackbach. You can find his solution below:

I will perform the orthonormalized version, not just the orthogonalized version. Note that $\int_{0}^{1}1^{2}\,dt=1$, so we're fine to start with $e_{1}=1$. Next we take$$u_{2}=t-\langle t,1\rangle\,1=t-\frac{1}{2}.$$
That is, we take the starting vector $t$ and subtract off how much of it is in the $1$ direction. Next, we normalize:
$$\|u_{2}\|=\sqrt{\int_{0}^{1}(t-1/2)^{2}\,dt}=\sqrt{\int_{-1/2}^{1/2}v^{2}\,dv}=\sqrt{2\int_{0}^{1/2}v^{2}\,dv}=\sqrt{2(1/2)^{3}/3}=\frac{1}{\sqrt{12}}.$$
Hence, we have that
$$e_{2}=2\sqrt{3}\left(t-\frac{1}{2}\right).$$
Finally, we have
$$u_{3}=t^{2}-\langle t^{2},1\rangle \cdot 1-\left\langle t^{2},2\sqrt{3}\left(t-\frac{1}{2}\right)\right\rangle \cdot 2\sqrt{3}\left(t-\frac{1}{2}\right)$$
$$=t^{2}-\frac{1}{3}-(t-1/2)=t^{2}-t+\frac{1}{6}.$$
The norm is
$$\|u_{3}\|=\sqrt{\int_{0}^{1}\left(t^{2}-t+\frac{1}{6}\right)^{2}\,dt}=\frac{1}{\sqrt{180}}=\frac{1}{6\sqrt{5}}.$$
Hence, we have that
$$e_{3}=6\sqrt{5}\left(t^{2}-t+\frac{1}{6}\right).$$
Check: $\|e_{3}\|=1$, as required.
Also:
\begin{align*}
\langle e_{1},e_{2}\rangle&=0,\\
\langle e_{1},e_{3}\rangle&=0, \;\text{and}\\
\langle e_{2},e_{3}\rangle&=0,
\end{align*}
as needed. Hence, we have achieved the orthonormalization of the given linearly independent set.

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