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Title: Rank gives Equivalence
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Series: Abstract Linear Algebra
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YouTube-Title: Abstract Linear Algebra 29 | Rank gives Equivalence
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Bright video: https://youtu.be/ah9o9O0VS5w
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Dark video: https://youtu.be/qyibwU7OSVg
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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: ala29_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 $A, T \in \mathbb{F}^{n \times n}$ and $T$ be invertible. What is always correct?
A1: $\mathrm{Ker}(T A) = \mathrm{Ker}(A)$
A2: $\mathrm{Ker}( A T) = \mathrm{Ker}(A)$
A3: $\mathrm{Ran}(T A) = \mathrm{Ran}(A)$
A4: $\mathrm{Ker}(T A) = \mathrm{Ran}(A T)$
Q2: Let $A, T, S \in \mathbb{F}^{n \times n}$ and $S, T$ be invertible. What is always correct?
A1: $\mathrm{rank}(SAT) = \mathrm{rank}(A)$
A2: $\mathrm{rank}(SAT) = n$
A3: $\mathrm{rank}(SAT) = \mathrm{rank}(T)$
A4: $\mathrm{rank}(A) + \mathrm{dim} \mathrm{Ker}(T) = n$
Q3: Are the matrices $A = \begin{pmatrix} 1 & 0 & 1 \ 0 & 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 & 0\ 1 & 1 & 0 \end{pmatrix}$ equivalent?
A1: No!
A2: Yes!
A4: One needs more information.
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Last update: 2024-10