Definition 8. Span of vectors.
\(\Span(S)\) is the set of linear combinations of vectors in \(S\) i.e.
\begin{equation*}
\left\{\sum_{i=1}^{n} a_{i} \cdot s_{i} \mid a_{i} \in \F, s_i \in S\right\}
\end{equation*}
Section Week 2
This is an outline of the topics we covered in the second week of class.
Subsection Tuesday 1/20
Coding Theorist of the Day.
Marcel J.E. Golay, 1902-1989
-
Swiss mathematician and physicist
-
Worked at Bell Labs (before Hamming & Shannon) briefly; mostly at Perkin-Elmer
-
Worked on problems in radar vessel detection, chromatography, spectroscopy, data processing
-
Invented the famous binary Golay code which generalizes the Hamming code (used in Voyager probes)
Class Materials.
In-person meeting cancelled today. Instead, three videos posted below along with the day’s activity.
Subsection Thursday 1/22
Coding Theorist of the Day.
F. Jessie MacWilliams, 1917-1990
-
English coding theorist
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One of the first women to publish in coding theory
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Programmer at Bell Labs then math PhD
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Important identity relating codes and their duals
-
Wrote “The Theory of Error-Correcting Codes” with Sloane
Reminders.
-
Assign pages to questions in Gradescope
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Problem Set M1 due next Th & Weekly Practice 2 due next T
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LA office hour set for M 2-3 in Clough 280
Notes.
Started with problems 1-3 from Linear Algebra with Finite Fields (sorry about 3).
Definition 9. Linear independence.
A set \(S=\{s_1,\ldots,s_n\}\) is linearly independent if any one its vectors is not in the span of the others i.e.
\begin{equation*}
\sum_{i=1}^{n} a_{i} \cdot s_{i}=0
\end{equation*}
has only the solution \(a_{1}=a_{2}=\ldots=0\)
Definition 10. Basis.
A basis for a vector space \(V\) is equivalently
-
a maximal lin. ind. set from \(V\)
-
a lin. ind. spanning set for \(V\)
-
a minimal spanning set for \(V\)
Definition 11. Dimension.
Definition 12. Rowspace.
Note 13.
In any field \(x y=0 \Rightarrow x=0 \text { or } y=0 \)
Note 14.
Over finite fields, solution sets to systems of linear equations are always finite
Note 15.
For us, vectors in \(\F^{n}\) are rows and we get expressions like
\begin{equation*}
\mathbf{x}M=\mathbf{y}
\end{equation*}
For problem 3, Computed
\begin{align*}
4\cdot(1,2,3,4,5) \amp \\
+5\cdot(1,0,1,2,1) \amp \\
+(0,2,1,0,4) \amp \\
= (2,3,4,5,1)
\end{align*}
A basis of \(\Span(S)\) is \(\{(1,2,3,4,5),(1,0,1,2,1),(0,2,1,0,4)\}\) and the dimension is 3. Other bases are formed by replacing one of these vectors with a linear combination of basis vectors where the coefficient of the replaced vector is nonzero. (Basis Replacement Theorem)
Definition 16. Generator Matrix.
For example, a generator matrix for \(\Span(S)\) is
\begin{equation*}
G=\begin{bmatrix}
1 \amp 2 \amp 3 \amp 4\amp 5\\
1 \amp 0 \amp 1 \amp 2\amp 1\\
0 \amp 2 \amp 1\amp0\amp 4\\
\end{bmatrix}
\end{equation*}
Definition 17. Parity-Check Matrix.
A parity-check matrix for a subspace \(V\leq \F^n\) is a matrix \(H\) such that \(\forall x\in V\)
\begin{equation*}
H\transpose{x}=0\text{.}
\end{equation*}
Equivalently \(G\transpose{H}=0\) where \(G\) is a generator matrix for \(V\)
\(H\) represents the nullspace of \(G\text{:}\) in general \(H\) is any generator matrix for the nullspace of \(V\) (the right nullspace of \(G\)).
Computation of \(H\) for \(\Span(S)\text{:}\)
\begin{align*}
\begin{bmatrix}
1 \amp 2 \amp 3\amp 4 \amp 5 \\
1\amp 0 \amp 1\amp 2\amp 1 \\
0\amp 2 \amp 1\amp 0\amp4
\end{bmatrix} \amp \overset{R_2-R_1}{\longrightarrow} \amp\begin{bmatrix}
1 \amp 2 \amp 3\amp 4 \amp 5 \\
0 \amp 5 \amp 5 \amp 5\amp 3 \\
0\amp 2 \amp 1\amp 0\amp 4
\end{bmatrix}\\
\amp \overset{3R_2}{\longrightarrow} \amp\begin{bmatrix}
1 \amp 2 \amp 3\amp 4 \amp 5 \\
0 \amp 1 \amp 1 \amp 1\amp 2 \\
0\amp 2 \amp 1\amp 0\amp 4
\end{bmatrix}\\
\amp \overset{R3+5R_2}{\longrightarrow} \amp\begin{bmatrix}
1 \amp 2 \amp 3\amp 4 \amp 5 \\
0 \amp 1 \amp 1 \amp 1\amp 2 \\
0\amp 0 \amp 6\amp 5\amp 0
\end{bmatrix}\\
\amp \overset{-R3}{\longrightarrow} \amp\begin{bmatrix}
1 \amp 2 \amp 3\amp 4 \amp 5 \\
0 \amp 1 \amp 1 \amp 1\amp 2 \\
0\amp 0 \amp 1\amp 2\amp 0
\end{bmatrix}\\
\amp \overset{R2+6R_3}{\underset{R1+4R_3}{\longrightarrow}} \amp\begin{bmatrix}
1 \amp 2 \amp 0 \amp 5 \amp 5 \\
0 \amp 1 \amp 0 \amp 6\amp 2 \\
0\amp 0 \amp 1\amp 2\amp 0
\end{bmatrix}\\
\amp \overset{R1+5R_2}{\longrightarrow} \amp\begin{bmatrix}
1 \amp 0 \amp 0 \amp 0 \amp 1 \\
0 \amp 1 \amp 0 \amp 6\amp 2 \\
0\amp 0 \amp 1\amp 2\amp 0
\end{bmatrix}
\end{align*}
We want
\begin{align*}
\begin{bmatrix}
1 \amp 0 \amp 0 \amp 0 \amp 1 \\
0 \amp 1 \amp 0 \amp 6\amp 2 \\
0\amp 0 \amp 1\amp 2\amp 0
\end{bmatrix} \begin{bmatrix} x_1 \\ x_2\\x_3\\x_4\\x_4\end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}
\end{align*}
So
\begin{align*}
\amp x_1+x_5=0 \\
\amp x_2+6x_4+2x_5=0 \\
\amp x_3+2x_4=0
\end{align*}
and this is spanned by
\begin{gather*}
\begin{bmatrix}
0 \\ 1 \\ 5 \\ 1 \\ 0
\end{bmatrix},
\begin{bmatrix}
6 \\ 5 \\ 0 \\ 0 \\ 1
\end{bmatrix}
\end{gather*}
All of this shows that
\begin{align*}
H = \begin{bmatrix}
0 \amp 1 \amp 5\amp 1 \amp 0\\
6 \amp 5 \amp 0 \amp 0 \amp 1
\end{bmatrix} \amp
\end{align*}
