• Title: Orthogonal Projection Onto Subspace

  • Series: Abstract Linear Algebra

  • Chapter: General inner products

  • YouTube-Title: Abstract Linear Algebra 15 | Orthogonal Projection Onto Subspace

  • Bright video: https://youtu.be/bGfUaaREXhU

  • Dark video: https://youtu.be/is5P9BUQwoo

  • Quiz: Test your knowledge

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  • Subtitle on GitHub: ala15_sub_eng.srt missing

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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: $U \cap U^\perp = {0}$.

  • Last update: 2024-10

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