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Title: Orthogonal Projection Onto Subspace
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Series: Abstract Linear Algebra
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Chapter: General inner products
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YouTube-Title: Abstract Linear Algebra 15 | Orthogonal Projection Onto Subspace
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Bright video: https://youtu.be/bGfUaaREXhU
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Dark video: https://youtu.be/is5P9BUQwoo
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Quiz: Test your knowledge
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Dark-PDF: Download PDF version of the dark video
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Print-PDF: Download printable PDF version
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Thumbnail (bright): Download PNG
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Thumbnail (dark): Download PNG
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Subtitle on GitHub: ala15_sub_eng.srt missing
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Timestamps (n/a)
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Subtitle in English (n/a)
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Quiz Content
Q1: Let $V$ be a vector space with inner product $\langle \cdot, \cdot \rangle$ and $U \subseteq V$ be a subset. What is never correct?
A1: $U \cap U^\perp = { 0 }$
A2: $U \cap U^\perp \subseteq { 0 }$
A3: $U \cap U^\perp \supseteq { 0 }$
A4: $U^\perp = \emptyset$
Q2: Let $V$ be a vector space with inner product $\langle \cdot, \cdot \rangle$ and $U \subseteq V$ be a subspace. What is always correct?
A1: $U \cap U^\perp = { 0 }$
A2: $U \cap U^\perp = V $
A3: $U \cup U^\perp = V$
A4: $U^\perp = \emptyset$
Q3: Let $V$ be a vector space with inner product $\langle \cdot, \cdot \rangle$ and $U \subseteq V$ be a finite subset given by $U = { b_1, b_2, \ldots, b_k }$. What is not correct in general?
A1: $x \in U^\perp$ implies $x \perp b_j$ for all $j$.
A2: $x \perp b_j$ for all $j$ implies $x \in U^\perp$.
A3: $U^\perp = (\mathrm{Span}(U))^\perp$.
A4: $U \cap U^\perp = {0}$.
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Last update: 2024-10