• Title: Rank gives Equivalence

  • Series: Abstract Linear Algebra

  • YouTube-Title: Abstract Linear Algebra 29 | Rank gives Equivalence

  • Bright video: https://youtu.be/ah9o9O0VS5w

  • Dark video: https://youtu.be/qyibwU7OSVg

  • Quiz: Test your knowledge

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  • Subtitle on GitHub: ala29_sub_eng.srt missing

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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.

  • Last update: 2024-10

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