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a) Let \(A:=\left(\begin{array}{ccc}1 & \frac{1}{2} & 0 \\ \frac{1}{2} & 1 & 0 \\ 0 & 0 & 1\end{array}\right)\). Briefly show that the bilinear map \(\mathbb{R}^{3} \times \mathbb{R}^{3} \rightarrow \mathbb{R}\) defined by \((x, y) \mapsto x^{T} A y\) gives a scalar product.
b) Let \(\alpha: \mathbb{R}^{3} \rightarrow \mathbb{R}\) be the linear functional \(\alpha:\left(x_{1}, x_{2}, x_{3}\right) \mapsto x_{1}+x_{2}\) and let \(v_{1}:=(-1,1,1), v_{2}:=(2,-2,0)\) and \(v_{3}:=(1,0,0)\) be a basis of \(\mathbb{R}^{3}\). Using the scalar product of the previous part, find an orthonormal basis \(\left\{e_{1}, e_{2}, e_{3}\right\}\) of \(\mathbb{R}^{3}\) with \(e_{1} \in \operatorname{span}\left\{v_{1}\right\}\) and \(e_{2} \in \operatorname{ker} \alpha\).
151. Let \(A: \mathbb{R}^{n} \rightarrow \mathbb{R}^{k}\) be a linear map defined by the matrix \(A\). If the matrix \(B\) satisfies the relation \(\langle A X, Y\rangle=\langle X, B Y\rangle\) for all vectors \(X \in \mathbb{R}^{n}, Y \in \mathbb{R}^{k}\), show that \(B\) is the transpose of \(A\), so \(B=A^{T}\). [This basic property of the transpose,
\langle A X, Y\rangle=\left\langle X, A^{T} Y\right\rangle
is the only reason the transpose is important.]
in Mathematics by Platinum (129,882 points) | 181 views

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