## 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 $$[\mathbf x]_B=\mathbf{B}^T\mathbf x$$.

The coordinates sought are: $$[(0{,}0,1)^T]_B=\Bigl(\frac1{\sqrt3},0,-\frac2{\sqrt6}\Bigr)^T$$
$$[(2{,}1,0)^T]_B=\Bigl(\sqrt3,\frac1{\sqrt2},\sqrt{\frac32}\Bigr)^T$$
$$[(1{,}2,3)^T]_B=\Bigl(2\sqrt3,-\frac1{\sqrt2},-\sqrt{\frac32}\Bigr)^T$$
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.
The orthogonal projections of the given vectors are $$\Bigl(\frac13,\frac13,\frac13\Bigr)^T$$ and $$\Bigl(\frac32,\frac12{,}1\Bigr)^T$$.