Inner product from ON basis
Task number: 4430
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: \(\langle{(1{,}4)^T}|{(2, 0)^T}\rangle =\frac19(2{\cdot}1\cdot2+1{\cdot}0+4{\cdot}2+5{\cdot}4\cdot0)= \frac19(4+0+8+0)=\frac43 \)
Answer
The value of the inner product \(\langle{(1{,}4)^T}|{(2, 0)^T}\rangle\) is \(\frac43 \).Solution
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 \(c(1,-1)^T\).
Answer
The vectors orthogonal to \((4{,}1)^T\) form the line \(c(1,-1)^T\).
