+68 Let u and v be vectors such that ||u||=3 and ||v||=2 such that the angle between u and v when placed tail to tail is 60 degrees.
Let A be a matrix such that row_1(A)=u and row_2(A)=v.
Then what are Au, Av in that order? (Your answers should be numerical.)
+9673 Let \(\langle \cdot, \cdot \rangle\) denote the inner product. Then
\(\dfrac{\langle \mathbf u, \mathbf v \rangle}{\|\mathbf u\|\|\mathbf v\|} = \cos 60^\circ = \dfrac12\\ \langle \mathbf u ,\mathbf v \rangle = \dfrac{3\cdot 2}2 = 3\)
\(A = \begin{pmatrix}\mathbf u^T\\\mathbf v^T\end{pmatrix}\\ A \mathbf u = \begin{pmatrix}\mathbf u^T\\\mathbf v^T\end{pmatrix} \mathbf u = \begin{pmatrix}\langle \mathbf u,\mathbf u \rangle\\\langle \mathbf v, \mathbf u\rangle\end{pmatrix} = \begin{pmatrix}\|\mathbf u\|^2\\\langle \mathbf u, \mathbf v\rangle\end{pmatrix} = \begin{pmatrix}9\\3\end{pmatrix}\\\)
Similarly we have
\(A \mathbf v = \begin{pmatrix}\langle \mathbf u, \mathbf v\rangle \\\|\mathbf v\|^2\end{pmatrix} = \begin{pmatrix}3\\4\end{pmatrix}\)
If this is not the notation you are used to, \(\langle \mathbf u,\mathbf v \rangle \) is just \(\mathbf u \cdot \mathbf v\).
+9673 Let \(\langle \cdot, \cdot \rangle\) denote the inner product. Then
\(\dfrac{\langle \mathbf u, \mathbf v \rangle}{\|\mathbf u\|\|\mathbf v\|} = \cos 60^\circ = \dfrac12\\ \langle \mathbf u ,\mathbf v \rangle = \dfrac{3\cdot 2}2 = 3\)
\(A = \begin{pmatrix}\mathbf u^T\\\mathbf v^T\end{pmatrix}\\ A \mathbf u = \begin{pmatrix}\mathbf u^T\\\mathbf v^T\end{pmatrix} \mathbf u = \begin{pmatrix}\langle \mathbf u,\mathbf u \rangle\\\langle \mathbf v, \mathbf u\rangle\end{pmatrix} = \begin{pmatrix}\|\mathbf u\|^2\\\langle \mathbf u, \mathbf v\rangle\end{pmatrix} = \begin{pmatrix}9\\3\end{pmatrix}\\\)
Similarly we have
\(A \mathbf v = \begin{pmatrix}\langle \mathbf u, \mathbf v\rangle \\\|\mathbf v\|^2\end{pmatrix} = \begin{pmatrix}3\\4\end{pmatrix}\)
If this is not the notation you are used to, \(\langle \mathbf u,\mathbf v \rangle \) is just \(\mathbf u \cdot \mathbf v\).
