## Inner product from ON basis

Assume that $$B =\{(1, 1)^T, (2,-1)^T\}$$ is an orthonormal basis of the space $$\mathbb R^2$$ with respect to some non-standard inner product.

In other words, you do not have a formula for the inner product, but you do know that some vectors are orthogonal to each other.

With respect to this inner product, determine:

• the value of $$\langle(1{,}4)^T|(2, 0)^T\rangle$$,
• the set of vectors orthogonal to $$\mathcal L\{(4, 1)^T\}$$.
• #### Solution 1

It is necessary to determine the coordinates of the given vectors with respect to the basis $$B$$. $\left( \begin{array}{cc|cc} 1 & 2 & 1 & 2 \\ 1 & -1 & 4 & 0 \end{array} \right) \sim \left( \begin{array}{cc|cc} 1 & 2 & 1 & 2 \\ 0 & -3 & 3 & -2 \end{array} \right) \sim \left( \begin{array}{cc|cc} 1 & 2 & 1 & 2 \\ 0 & 1 & -1 & \frac23 \end{array} \right) \sim \left( \begin{array}{cc|cc} 1 & 0 & 3 & \frac23 \\[1mm] 0 & 1 & -1 & \frac23 \end{array} \right)$

We get $$[(1{,}4)^T]_B=(3,-1)^T$$ and $$[(2{,}0)^T]_B=\bigl(\frac23 ,\frac23 \bigr)^T$$

Now $$\langle(1{,}4)^T|(2, 0)^T\rangle=[(2{,}0)^T]_B^T[(1{,}4)^T]_B=\bigl(\frac23 ,\frac23 \bigr)(3,-1)^T=\frac43$$.

• #### Solution 2

Let us denote the matrix assembled from the basis $$B$$ by the symbol$$\mathbf{B}= \begin{pmatrix} 1 & 2 \\ 1 & -1 \end{pmatrix}$$

Observe that for any vector $$\mathbf{u}\in \mathbb R^2$$ it holds that $$[\mathbf{u}]_B= \mathbf{B}^{-1}\mathbf{u}= \begin{pmatrix} 1 & 2 \\ 1 & -1 \end{pmatrix}^{-1} \mathbf{u} = \begin{pmatrix} \frac13 & \frac23 \\[1mm] \frac13 & -\frac13 \end{pmatrix} \mathbf{u}$$.

Then the formula for the inner product can be derived as follows: $$\langle\mathbf{u}|\mathbf{v}\rangle=[\mathbf{v}]_B^T[\mathbf{u}]_B= (\mathbf{B}^{-1}\mathbf{v})^T\mathbf{B}^{-1}\mathbf{u}= \mathbf{v}^T(\mathbf{B}^{-1})^T\mathbf{B}^{-1}\mathbf{u}$$ $$= \mathbf{v}^T \begin{pmatrix} \frac13 & \frac13 \\[1mm] \frac23 & -\frac13 \end{pmatrix} \begin{pmatrix} \frac13 & \frac23 \\[1mm] \frac13 & -\frac13 \end{pmatrix} \mathbf{u} = \mathbf{v}^T \begin{pmatrix} \frac29 & \frac19 \\[1mm] \frac19 & \frac59 \end{pmatrix} \mathbf{u} =\frac19(2u_1v_1+u_1v_2+u_2v_1+5u_2v_2)$$

Indeed: $$\langle{(1{,}4)^T}|{(2, 0)^T}\rangle =\frac19(2{\cdot}1\cdot2+1{\cdot}0+4{\cdot}2+5{\cdot}4\cdot0)= \frac19(4+0+8+0)=\frac43$$

The value of the inner product $$\langle{(1{,}4)^T}|{(2, 0)^T}\rangle$$ is $$\frac43$$.
Using the calculations from the previous section, every vector $$\mathbf{u}$$ orthogonal to $$(4{,}1)^T$$ must satisfy $$(4{,}1) \begin{pmatrix} \frac29 & \frac19 \\[1mm] \frac19 & \frac59 \end{pmatrix} \mathbf{u} =0$$
That is, $$(1{,}1)\mathbf{u}=0$$, which (taken as a system of one equation with two unknowns) yields $$c(1,-1)^T$$.
The vectors orthogonal to $$(4{,}1)^T$$ form the line $$c(1,-1)^T$$.