ON basis and Fourier coefficients
Task number: 4431
Let \(\mathbf{b}_1 = \Bigl(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\Bigr)^T\), \(\mathbf{b}_2 = \Bigl(\frac1{\sqrt2},-\frac1{\sqrt2},0\Bigr)^T\), \(\mathbf{b}_3 = \Bigl(\frac1{\sqrt6},\frac1{\sqrt6},-\frac2{\sqrt6}\Bigr)^T\), be vectors in the real arithmetic vector space \(\mathbb R^3\) with the standard inner product.
- Verify that \(B = \{\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3\}\) is an orthonormal basis of \(\mathbb R^3\),
- calculate the coordinates of the vectors \((0, 0, 1)^T\), \((2, 1, 0)^T\) and \((1, 2, 3)^T\) relative to the orthonormal basis \(B\),
- determine the orthogonal projections of the vectors \((0, 0, 1)^T\) and \((2, 1, 0)^T\) into the subspace \(\mathcal L(\{\mathbf{b}_1,\mathbf{b}_2\})\)
Solution
Denote the matrix \(\mathbf{B}= (\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3)= \begin{pmatrix} \frac1{\sqrt3} & \frac1{\sqrt2} & \frac1{\sqrt6}\\ \frac1{\sqrt3} & -\frac1{\sqrt2} & \frac1{\sqrt6}\\ \frac1{\sqrt3} & 0 & -\frac2{\sqrt6} \end{pmatrix} \)
It is sufficient to verify by using the matrix product that \(\mathbf{B}^T\mathbf{B}=\mathbf I\).
Solution
The coordinates, here the Fourier coefficients, we get be successively the inner product with the vectors of the basis \(B\)
\([(0{,}0,1)^T]_B= (\langle(0{,}0,1)^T|\mathbf{b}_1\rangle, \langle(0{,}0,1)^T|\mathbf{b}_2\rangle, \langle(0{,}0,1)^T|\mathbf{b}_3\rangle)^T =\Bigl(\frac1{\sqrt3},0,-\frac2{\sqrt6}\Bigr)^T\) and so on
\([(2{,}1,0)^T]_B=\Bigl(\sqrt3,\frac1{\sqrt2},\sqrt{\frac32}\Bigr)^T\),
\([(1{,}2,3)^T]_B=\Bigl(2\sqrt3,-\frac1{\sqrt2},-\sqrt{\frac32}\Bigr)^T\).They can also be calculated at once with the matrix product \([\mathbf x]_B=\mathbf{B}^T\mathbf x\).
Answer
The coordinates sought are: \([(0{,}0,1)^T]_B=\Bigl(\frac1{\sqrt3},0,-\frac2{\sqrt6}\Bigr)^T\)
\([(2{,}1,0)^T]_B=\Bigl(\sqrt3,\frac1{\sqrt2},\sqrt{\frac32}\Bigr)^T\)
\([(1{,}2,3)^T]_B=\Bigl(2\sqrt3,-\frac1{\sqrt2},-\sqrt{\frac32}\Bigr)^T\)Solution
In the previous consideration we found that \((0{,}0,1)^T = \frac1{\sqrt3}\mathbf{b}_1 + 0\mathbf{b}_2 - \frac2{\sqrt6}\mathbf{b}_3\) , hence the orthogonal projection of the vector \((0{,}0,1)^T\) yields the vector \(\frac1{\sqrt3}\mathbf{b}_1 = \Bigl(\frac13,\frac13,\frac13\Bigr)^T\)
Similarly because \((2{,}1,0)^T= \sqrt3\mathbf{b}_1 + \frac1{\sqrt2} \mathbf{b}_2 + \sqrt{\frac32} \mathbf{b }_3\), the vector \(\sqrt3\mathbf{b}_1 + \frac1{\sqrt2} \mathbf{b}_2 = \Bigl(\frac32,\frac12{,}1\Bigr)^T\) is the desired orthogonal projection.
Answer
The orthogonal projections of the given vectors are \(\Bigl(\frac13,\frac13,\frac13\Bigr)^T\) and \(\Bigl(\frac32,\frac12{,}1\Bigr)^T\).
