91影视

To find another orthonormal basis, we can use a rotation approach. Let's consider a rotation by an angle 胃. For an angle 胃 = 45掳 or 蟺/4, we can apply the rotation matrix: \[R(\theta) = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\] and apply it to the given orthonormal basis vectors (1,0) and (0,1). 1. Find the coordinates of the new basis vectors v鈧� and v鈧�: \[v_1 = R(\theta) \cdot \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix}\] With 胃 = 蟺/4, we have: \[v_1 = \begin{bmatrix} \sqrt{2}/2 \\ \sqrt{2}/2 \end{bmatrix}\] 2. Calculate v鈧� in a similar manner: \[v_2 = R(\theta) \cdot \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -\sin(\theta) \\ \cos(\theta) \end{bmatrix}\] With 胃 = 蟺/4, we have: \[v_2 = \begin{bmatrix} -\sqrt{2}/2 \\ \sqrt{2}/2 \end{bmatrix}\] So, another orthonormal basis for 鈩澛� is {(\( \sqrt{2}/2 \), \( \sqrt{2}/2 \)), (-\( \sqrt{2}/2 \), \( \sqrt{2}/2 \))}.

We will now use the Gram-Schmidt process to orthonormalize the given orthogonal vectors. 1. Normalize u鈧�: \[v_1 = \frac{u_1}{||u_1||} = \frac{\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}}{\sqrt{1^2+1^2+1^2+1^2}} = \frac{\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}}{2}\] So, \(v_1 = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})\). 2. Define w鈧� = u鈧� - (u鈧傗媴v鈧�)v鈧� and normalize it to obtain v鈧�: \[w_2 = \begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \end{bmatrix} - \left(\begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \end{bmatrix} \cdot \begin{bmatrix} 1/2 \\ 1/2 \\ 1/2 \\ 1/2 \end{bmatrix}\right)\begin{bmatrix} 1/2 \\ 1/2 \\ 1/2 \\ 1/2 \end{bmatrix}\] \[w_2 = \begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \end{bmatrix} - 0\begin{bmatrix} 1/2 \\ 1/2 \\ 1/2 \\ 1/2 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \end{bmatrix}\] Now, normalize w鈧� to obtain v鈧�: \[v_2 = \frac{w_2}{||w_2||} = \frac{\begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \end{bmatrix}}{\sqrt{1^2+(-1)^2+1^2+(-1)^2}} = \frac{\begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \end{bmatrix}}{2}\] So, \(v_2 = (\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{2})\). 3. Define w鈧� = u鈧� - (u鈧冣媴v鈧�)v鈧� - (u鈧冣媴v鈧�)v鈧� and normalize it to obtain v鈧�: \[w_3 = \begin{bmatrix} 1 \\ 2 \\ -1 \\ -2 \end{bmatrix} - \left(\begin{bmatrix} 1 \\ 2 \\ -1 \\ -2 \end{bmatrix} \cdot \begin{bmatrix} 1/2 \\ 1/2 \\ 1/2 \\ 1/2 \end{bmatrix}\right)\begin{bmatrix} 1/2 \\ 1/2 \\ 1/2 \\ 1/2 \end{bmatrix} - \left(\begin{bmatrix} 1 \\ 2 \\ -1 \\ -2 \end{bmatrix} \cdot \begin{bmatrix} 1/2 \\ -1/2 \\ 1/2 \\ -1/2 \end{bmatrix}\right)\begin{bmatrix} 1/2 \\ -1/2 \\ 1/2 \\ -1/2 \end{bmatrix}\] \[w_3 = \begin{bmatrix} 1 \\ 2 \\ -1 \\ -2 \end{bmatrix} - 0\begin{bmatrix} 1/2 \\ 1/2 \\ 1/2 \\ 1/2 \end{bmatrix} - 2\begin{bmatrix} 1/2 \\ -1/2 \\ 1/2 \\ -1/2 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \\ -2 \end{bmatrix} -\begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ -2 \\ -1 \end{bmatrix}\] Now normalize w鈧� to obtain v鈧�: \[v_3 = \frac{w_3}{||w_3||} = \frac{\begin{bmatrix} 0 \\ 3 \\ -2 \\ -1 \end{bmatrix}}{\sqrt{0^2+3^2+(-2)^2+(-1)^2}} = \frac{\begin{bmatrix} 0 \\ 3 \\ -2 \\ -1 \end{bmatrix}}{\sqrt{14}}\] So, \(v_3 = (0, \frac{3}{\sqrt{14}}, -\frac{2}{\sqrt{14}}, -\frac{1}{\sqrt{14}})\). The orthonormal basis obtained from the given orthogonal vectors is: {(1/2, 1/2, 1/2, 1/2), (1/2, -1/2, 1/2, -1/2), (0, 3/鈭�14, -2/鈭�14, -1/鈭�14)}.