Americas
Europe
Problem 311
Show geometrically that for each angle \(\theta\), the transformation \(\mathrm{T}_{\theta}: \mathrm{R}^{2} \rightarrow \mathrm{R}^{2}\), defined by $(\mathrm{x}, \mathrm{y}) \mathrm{T}_{\theta}=(\mathrm{x} \cos \theta-\mathrm{y} \sin \theta, \mathrm{x} \sin \theta+\mathrm{y} \cos \theta)$, is an orthogonal transformation.
What do you think about this solution?
We value your feedback to improve our textbook solutions.
Let $\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots \mathrm{x}_{\mathrm{n}}\right),\left(\mathrm{y}_{1}, \mathrm{y}_{2}, \ldots, \mathrm{y}_{\mathrm{n}}\right),\left(\mathrm{z}_{1}, \mathrm{z}_{2}, \ldots \ldots \mathrm{z}_{\mathrm{n}}\right)$ be three vectors in \(\mathrm{R}^{\mathrm{n}}\). Verify the following properties of the dot product using these vectors : a) \(\mathrm{X} \cdot \mathrm{X} \geq 0 ; \mathrm{X} \cdot \mathrm{X}=0\) if and only if \(\mathrm{X}=0\) b) \(X \cdot Y=Y \cdot X\) c) $(\mathrm{X}+\mathrm{Y}) \cdot \mathrm{Z}=\mathrm{X} \cdot \mathrm{Y}+\mathrm{X} \cdot \mathrm{Z}$ d) $(\mathrm{CX}) \cdot \mathrm{Y}=\mathrm{X} \cdot(\mathrm{cY})=\mathrm{c}(\mathrm{X} \cdot \mathrm{Y})$
Consider the vector space \(\mathrm{C}[0,1]\) of all continuous functions defined on \([0,1]\). If \(\mathrm{f} \in \mathrm{C}[0,1]\), show that $\left({ }^{1} \int_{0} \mathrm{f}^{2}(\mathrm{x}) \mathrm{d} \mathrm{x}\right)^{1 / 2}$ defines a norm on all elements of this vector space.
Show that the functions $1, \cos \pi \mathrm{x}, \cos 2 \pi \mathrm{x}, \ldots, \cos \mathrm{n} \pi \mathrm{x}$, form an orthogonal set over \([0,1]\). Then normalize them to obtain an orthogonal set.
Let \(\mathrm{S}=\operatorname{Sp}[(1,0,1),(0,2,1)]\). Then \(\mathrm{S}\) is a subspace of \(\mathrm{R}^{3}\). Find the orthogonal complement of \(\mathrm{S}\).
Let \(\mathrm{u}=(2,-1,3)\) and \(\mathrm{v}=(4,-1,2)\) be vectors in \(\mathrm{R}^{3}\). Find the orthogonal projection of \(\mathrm{u}\) on \(\mathrm{v}\) and the component of \(\mathrm{u}\) orthogonal to v.
The first learning app that truly has everything you need to ace your exams in one place.