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Title: Matrices induce linear maps
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Series: Linear Algebra
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Chapter: Matrices and linear systems
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YouTube-Title: Linear Algebra 19 | Matrices induce linear maps
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Bright video: Watch on YouTube
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Dark video: Watch on YouTube
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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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Exercise Download PDF sheets
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Thumbnail (bright): Download PNG
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Thumbnail (dark): Download PNG
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Subtitle on GitHub: la19_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: Consider the map $f_A: \mathbb{R}^n \rightarrow \mathbb{R}^m$ given by $f_A(\mathbf{x}) = A \mathbf{x}$ with a matrix $ A \in \mathbb{R}^{m \times n}$. Is the map linear?
A1: Yes, it is always a linear map.
A2: It’s only a linear map for $n = m$.
A3: It’s only well-defined for $n = m$.
A4: No, it’s never a linear map.
A5: One needs more information.
Q2: Consider the maps $f_A: \mathbb{R}^n \rightarrow \mathbb{R}^n$ and $f_B: \mathbb{R}^n \rightarrow \mathbb{R}^n$ given by $f_A(\mathbf{x}) = A \mathbf{x}$ and $f_B(\mathbf{x}) = B \mathbf{x}$ with matrices $ A, B \in \mathbb{R}^{n \times n}$. What can one say about the composition $f_A \circ f_B$?
A1: $f_A \circ f_B = f_{AB}$
A2: It’s not well-defined.
A3: $f_A \circ f_B = f_{A+B}$
A4: $f_A \circ f_B = f_{BA}$
A5: $f_A \circ f_B = f_{A}$
Q3: Consider the matrix $A = \begin{pmatrix} 2 & 1 \ 1 & 1\end{pmatrix}$. Which of the following claims is correct?
A1: $ f_A(\mathbf{x}) = \begin{pmatrix} 2x_1 + x_2 \ x_1 + x_2 \end{pmatrix} $
A2: $ f_A(\mathbf{x}) = \begin{pmatrix} 2x_1 - x_2 \ x_1 \cdot x_2 \end{pmatrix} $
A3: $f_A: \mathbb{R}^2 \rightarrow \mathbb{R}$
A4: $f_A: \mathbb{R} \rightarrow \mathbb{R}^2$
A5: $f(\mathbf{x}) = 0$ for all $\mathbf{x} \in \mathbb{R}^2$
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Last update: 2024-10
