Home / Mathematics /Problem 3 Let \( \langle\cdot, \cdot\rangle \) be a scalar product in \( \mathbb{R}^{n} \) (not necessarily the standard one). Let \( V \) be a real vector space (i.e. a vector space over \( \mathbb{R} \) ), and let \( f: V \rightarrow \mathbb{R}^{n} \) be a linear transformation. We define \( \langle\cdot, \cdot\rangle_{f}: V \times V \rightarrow \mathbb{R} \) by setting \[ \langle\mathbf{u}, \mathbf{v}\rangle_{f}=\langle f(\mathbf{u}), f(\mathbf{v})\rangle \] for all \( \mathbf{u}, \mathbf{v} \in V \). Prove that \( \langle\cdot, \cdot\rangle_{f} \) is a scalar product on \( V \) if and only if \( f \) is one-to-one. Problem \( 4 \quad \) Let \( Q_{1}, Q_{2} \in \mathbb{R}^{n \times m} \) be matrices with orthonormal columns, \( { }^{1} \) and let \( Q \in \mathbb{R}^{m \times m} \) be such that \( Q_{1}=Q_{2} Q \). Prove that \( Q \) is an orthogonal matrix. Hint: Consider the product \( Q_{1}^{T} Q_{1}=\left(Q_{2} Q\right)^{T}\left(Q_{2} Q\right) \).

Question Problem 3 Let \( \langle\cdot, \cdot\rangle \) be a scalar product in \( \mathbb{R}^{n} \) (not necessarily the standard one). Let \( V \) be a real vector space (i.e. a vector space over \( \mathbb{R} \) ), and let \( f: V \rightarrow \mathbb{R}^{n} \) be a linear transformation. We define \( \langle\cdot, \cdot\rangle_{f}: V \times V \rightarrow \mathbb{R} \) by setting \[ \langle\mathbf{u}, \mathbf{v}\rangle_{f}=\langle f(\mathbf{u}), f(\mathbf{v})\rangle \] for all \( \mathbf{u}, \mathbf{v} \in V \). Prove that \( \langle\cdot, \cdot\rangle_{f} \) is a scalar product on \( V \) if and only if \( f \) is one-to-one. Problem \( 4 \quad \) Let \( Q_{1}, Q_{2} \in \mathbb{R}^{n \times m} \) be matrices with orthonormal columns, \( { }^{1} \) and let \( Q \in \mathbb{R}^{m \times m} \) be such that \( Q_{1}=Q_{2} Q \). Prove that \( Q \) is an orthogonal matrix. Hint: Consider the product \( Q_{1}^{T} Q_{1}=\left(Q_{2} Q\right)^{T}\left(Q_{2} Q\right) \).

RL8FDJ The Asker · Advanced Mathematics

Transcribed Image Text: Problem 3 Let \( \langle\cdot, \cdot\rangle \) be a scalar product in \( \mathbb{R}^{n} \) (not necessarily the standard one). Let \( V \) be a real vector space (i.e. a vector space over \( \mathbb{R} \) ), and let \( f: V \rightarrow \mathbb{R}^{n} \) be a linear transformation. We define \( \langle\cdot, \cdot\rangle_{f}: V \times V \rightarrow \mathbb{R} \) by setting \[ \langle\mathbf{u}, \mathbf{v}\rangle_{f}=\langle f(\mathbf{u}), f(\mathbf{v})\rangle \] for all \( \mathbf{u}, \mathbf{v} \in V \). Prove that \( \langle\cdot, \cdot\rangle_{f} \) is a scalar product on \( V \) if and only if \( f \) is one-to-one. Problem \( 4 \quad \) Let \( Q_{1}, Q_{2} \in \mathbb{R}^{n \times m} \) be matrices with orthonormal columns, \( { }^{1} \) and let \( Q \in \mathbb{R}^{m \times m} \) be such that \( Q_{1}=Q_{2} Q \). Prove that \( Q \) is an orthogonal matrix. Hint: Consider the product \( Q_{1}^{T} Q_{1}=\left(Q_{2} Q\right)^{T}\left(Q_{2} Q\right) \).
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Transcribed Image Text: Problem 3 Let \( \langle\cdot, \cdot\rangle \) be a scalar product in \( \mathbb{R}^{n} \) (not necessarily the standard one). Let \( V \) be a real vector space (i.e. a vector space over \( \mathbb{R} \) ), and let \( f: V \rightarrow \mathbb{R}^{n} \) be a linear transformation. We define \( \langle\cdot, \cdot\rangle_{f}: V \times V \rightarrow \mathbb{R} \) by setting \[ \langle\mathbf{u}, \mathbf{v}\rangle_{f}=\langle f(\mathbf{u}), f(\mathbf{v})\rangle \] for all \( \mathbf{u}, \mathbf{v} \in V \). Prove that \( \langle\cdot, \cdot\rangle_{f} \) is a scalar product on \( V \) if and only if \( f \) is one-to-one. Problem \( 4 \quad \) Let \( Q_{1}, Q_{2} \in \mathbb{R}^{n \times m} \) be matrices with orthonormal columns, \( { }^{1} \) and let \( Q \in \mathbb{R}^{m \times m} \) be such that \( Q_{1}=Q_{2} Q \). Prove that \( Q \) is an orthogonal matrix. Hint: Consider the product \( Q_{1}^{T} Q_{1}=\left(Q_{2} Q\right)^{T}\left(Q_{2} Q\right) \).
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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2First, we prove that if \( \mathrm{{\left(,\right)}} \) is a scalar product on \( \mathrm{{V}} \), then \( \mathrm{{f}} \) is one-to-one.Suppose that \( \mathrm{{\left(,\right)}} \) is a scalar product on \( \mathrm{{V},} \) is not one-to-one. Then there exist vectors \( \mathrm{{u}_{{1}}{u}_{{2}}} \) in V such that . Then we have\( \mathrm{{\left({u}_{{1}}-{u}_{{2}},{u}_{{1}}-{u}_{{2}}\right)}={\left({u}_{{1}},{u}_{{1}}\right)}-{2}{\left({u}_{{1}},{u}_{{2}},\right)}+{\left({u}_{{2}},{u}_{{2}}\right)}>{0}} \)Since \( \mathrm{{\left(,\right)}} \) is a positive definite scalar product. On the other hand, we have\( \mathrm{{\left({f{{\left({u}_{{1}}\right)}}}-{f{{\left({u}_{{2}}\right)}}},{f{{\left({u}_{{1}}\right)}}}-{f{{\left({u}_{{2}}\right)}}}\right)}={\left({f{{\left({u}_{{1}}\right)}}},{f{{\left({u}_{{2}}\right)}}}\right)}-{2}{f{{\left({\left({u}_{{1}}\right)},{f{{\left({u}_{{2}}\right)}}}\right)}}}+{\left({f{{\left({u}_{{2}}\right)}}},{f{{\left({u}_{{2}}\right)}}}\right)}} \)\( \mathrm{={\left({u}_{{1}},{u}_{{1}}\right)}-{2}{\left({u}_{{1}},{u}_{{2}}\right)}+{\left({u}_{{2}},{u}_{{2}}\right)}} \)since \( \mathrm{{\left(,\right)}} \) is induced by f. ExplanationTherefore, we obtain a contradiction, which implies that f must be one-to-one.Next, we prove that if f is one-to-one, then \( \mathrm{{\left(,\right) ... See the full answer

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