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Title: Conservation of Dimension
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Series: Linear Algebra
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Chapter: Matrices and linear systems
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YouTube-Title: Linear Algebra 28 | Conservation of Dimension
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Bright video: https://youtu.be/XtiteY3n0Oc
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Dark video: https://youtu.be/Ba4ycLfI5Rc
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Ad-free video: Watch Vimeo video
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Forum: Ask a question in Mattermost
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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: la28_sub_eng.srt missing
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Download bright video: Link on Vimeo
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Download dark video: Link on Vimeo
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Timestamps (n/a)
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Subtitle in English (n/a)
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Quiz Content
Q1: What is not a property of a bijective linear map $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$?
A1: $f(\lambda x) = \lambda + f(x)$
A2: $f(x+y) = f(x) + f(y)$
A3: $f^{-1}$ exists.
A4: $f^{-1}$ is linear.
A5: $f^{-1}$ is bijective.
Q2: If there is a bijective linear map $f: U \rightarrow V$ between two subspaces $U, V \subseteq \mathbb{R}^n$, then $\mathrm{dim}(U) = \mathrm{dim}(V)$. Is this correct?
A1: Yes!
A2: No, there are counterexamples.
A3: One needs more information.
Q3: Let $U = { \mathbf{x} \in \mathbb{R}^2 \mid \langle \mathbf{x} , \binom{2}{1} \rangle = 0}$. Now, take another subspace $V \subseteq U$ with $\mathrm{dim}(V) = 1$. Which claim is correct?
A1: $V = U$.
A2: $\mathrm{dim}(U) = 2$.
A3: $\mathrm{dim}(U) = 3$.
A4: $V$ is not a subspace.
A5: $V = { 0 }$.
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Last update: 2024-10
