-
Title: Steinitz Exchange Lemma
-
Series: Linear Algebra
-
Chapter: Matrices and linear systems
-
YouTube-Title: Linear Algebra 26 | Steinitz Exchange Lemma
-
Bright video: https://youtu.be/d69Bw-n5fvo
-
Dark video: https://youtu.be/OjWEEFASnVo
-
Quiz: Test your knowledge
-
Dark-PDF: Download PDF version of the dark video
-
Print-PDF: Download printable PDF version
-
Thumbnail (bright): Download PNG
-
Thumbnail (dark): Download PNG
-
Subtitle on GitHub: la26_sub_eng.srt missing
-
Timestamps (n/a)
-
Subtitle in English (n/a)
-
Quiz Content
Q1: Let $\mathcal{B} = (\mathbf{u}, \mathbf{v})$ be a basis of a subspace $U\subseteq \mathbb{R}^n$. Which family is also a basis for $U$?
A1: $(3\mathbf{u}, 4\mathbf{v})$
A2: $(\mathbf{u}, \mathbf{u})$
A3: $(\mathbf{u}, -\mathbf{u})$
A4: $(2\mathbf{v}, -\mathbf{v})$
A5: $(2\mathbf{v}, 4\mathbf{v})$
Q2: What is the formulation of Steinitz Exchange Lemma as we have learnt in the video.
A1: Let $\mathcal{B}$ be a basis of a subspace $U$ with $k$ vectors and $\mathcal{A}$ be a family of $\ell$ linearly independent vectors in $U$. Then one can add $k - \ell$ vectors from $B$ to $\mathcal{A}$ to get a new basis of $U$.
A2: Let $\mathcal{B}$ be a basis of a subspace $U$ with $k$ vectors and $\mathcal{A}$ be a family of $\ell$ linearly independent vectors in $U$. Then one can remove $k$ vectors from $B$ to $\mathcal{A}$ to get a new basis of $U$.
A3: Let $\mathcal{B}$ be a basis of a subspace $U$ with $k$ vectors and $\mathcal{A}$ be a family of $\ell$ linearly independent vectors in $U$. Then $\mathcal{A} \cup \mathcal{B}$ is new basis of $U$.
-
Last update: 2024-10