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Week of...

Notes and Links

1

Sep 7

Tue, About, Thu

2

Sep 14

Tue, HW1, HW1 Solution, Thu

3

Sep 21

Tue, HW2, HW2 Solution, Thu, Photo

4

Sep 28

Tue, HW3, HW3 Solution, Thu

5

Oct 5

Tue, HW4, HW4 Solution, Thu,

6

Oct 12

Tue, Thu

7

Oct 19

Tue, HW5, HW5 Solution, Term Test on Thu

8

Oct 26

Tue, Why LinAlg?, HW6, HW6 Solution, Thu

9

Nov 2

Tue, MIT LinAlg, Thu

10

Nov 9

Tue, HW7, HW7 Solution Thu

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Nov 16

Tue, HW8, HW8 Solution, Thu

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Nov 23

Tue, HW9, HW9 Solution, Thu

13

Nov 30

Tue, On the final, Thu

S

Dec 7

Office Hours

F

Dec 14

Final on Dec 16

To Do List

The Algebra Song!

Register of Good Deeds

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WARNING: The notes below, written for students and by students, are provided "as is", with absolutely no warranty. They can not be assumed to be complete, correct, reliable or relevant. If you don't like them, don't read them. It is a bad idea to stop taking your own notes thinking that these notes can be a total replacement  there's nothing like one's own handwriting!
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Vector subspaces
Definition. $\mathbf {W} \subset \mathbf {V}$ is a "subspace" if it is a vector space under the operations it inherits from V.
Theorem. $\mathbf {W} \subset \mathbf {V}$ is a subspace iff it is "closed under addition and vector multiplication by scalars", i.e. $x,y\in \mathbf {W} \Rightarrow x+y\in \mathbf {W}$ and $a\in F,x\in \mathbf {W} \Rightarrow ax\in \mathbf {W}$.
Goal: Every VS has a "basis", so while we don't have to use coordinates, we always can.
Examples of what is not a subspace (without diagrams):
 A unit circle is not closed under addition of scalar multiplication.
 The xaxis $\cup$ yaxis is closed under scalar multiplication, but not under addition.
 A single quadrant of the Cartesian plane is closed under addition, but not under scalar multiplication.
Examples of subspaces:
 $\{0\}$
 Any VS (which is a subspace of itself)
 A line passing through the origin (if it does not pass through the origin, then it is not closed under scalar multiplication)
 A plane
 Let $\mathbf {V} =\mathbb {M} _{n\times n}(F)$. If $W=\{A\in \mathbf {V} :A^{\top }=A\}$, then W is a subspace of V. (W is the set of "symmetric" matrices in V; A^{T} denotes the transpose of A.)
 $\mathbf {W} =\{A\in \mathbb {M} _{n\times n}:\operatorname {tr} (A)=0\}$
 where $\operatorname {tr} (A)=\sum _{i=1}^{n}a_{ii}$ is the "trace" of A.
 Properties of trace:
 $\operatorname {tr} (0\cdot A)=0$
 $\operatorname {tr} (cA)=c\cdot \operatorname {tr} (A)$
 $\operatorname {tr} (A+B)=\operatorname {tr} (A)+\operatorname {tr} (B)$
 so W is indeed a subspace.
Claim: If W_{1} and W_{2} are subspaces of V, then
 $W_{1}\cap W_{2}=\{x\in \mathbf {V} :x\in \mathbf {W} _{1}{\mbox{ and }}\mathbf {W} _{2}\}$ is a subspace of V, W_{1}, and W_{2}.
 But $W_{1}\cup W_{2}=\{x\in \mathbf {V} :x\in \mathbf {W} _{1}{\mbox{ or }}x\in \mathbf {W} _{2}\}$ is a subspace of V iff $\mathbf {W} _{1}\subset \mathbf {W} _{2}$ or $\mathbf {W} _{2}\subset \mathbf {W} _{1}$. (See HW2 pp. 2021, #19.)
Linear combinations
Definition: A vector u is a "linear combination" (l.c.) of vectors u_{1}, ..., u_{n} if there exists scalars a_{1}, ..., a_{n} such that
 $u=a_{1}u_{1}+a_{2}u_{2}+\ldots +a_{n}u_{n}$
Example: $\mathbb {P} _{n}(F)=\{{\mbox{Polynomials of degree at most }}n{\mbox{ with coefficients in }}F\}$
$=\left\{\sum _{i=0}^{n}a_{i}x^{i}:a_{i}\in F\right\}$
Definition: A subset $S\subset \mathbf {V}$ "generates" or "spans" V iff the set of linear combinations of elements of S is all of V.
Example: Let $\mathbf {V} =\mathbb {M} _{n\times n}(\Re )$
Let $M_{1}={\begin{pmatrix}1&0\\0&0\end{pmatrix}},M_{2}={\begin{pmatrix}0&1\\0&0\end{pmatrix}},M_{3}={\begin{pmatrix}0&0\\1&0\end{pmatrix}},M_{4}={\begin{pmatrix}0&0\\0&1\end{pmatrix}}$
Then $S=\{M_{1},M_{2},M_{3},M_{4}\}$ generates V.
Proof: Given ${\begin{pmatrix}a&b\\c&d\end{pmatrix}}\in \mathbb {M} _{2\times 2}(\Re )$, write
${\begin{pmatrix}a&b\\c&d\end{pmatrix}}=aM_{1}+bM_{2}+cM_{3}+dM_{4}.$
Example: Let $N_{1}={\begin{pmatrix}0&1\\1&1\end{pmatrix}},N_{2}={\begin{pmatrix}1&0\\1&1\end{pmatrix}},N_{3}={\begin{pmatrix}1&1\\0&1\end{pmatrix}},N_{4}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}$
Does $\{N_{1},N_{2},N_{3},N_{4}\}$ generate V?
 $M_{1}={\frac {2}{3}}N_{1}+{\frac {1}{3}}(N_{2}+N_{3}+N_{4})$
 $M_{2}={\frac {2}{3}}N_{2}+{\frac {1}{3}}(N_{1}+N_{3}+N_{4})$
 $M_{3}={\frac {2}{3}}N_{3}+{\frac {1}{3}}(N_{1}+N_{2}+N_{4})$
 $M_{4}={\frac {2}{3}}N_{4}+{\frac {1}{3}}(N_{1}+N_{2}+N_{3})$
Then ${\begin{pmatrix}a&b\\c&d\end{pmatrix}}=aM_{1}+bM_{2}+cM_{3}+dM_{4}=$
 $a\cdot \left({\frac {2}{3}}N_{1}+{\frac {1}{3}}(N_{2}+N_{3}+N_{4})\right)+b\cdot \left({\frac {2}{3}}N_{2}+{\frac {1}{3}}(N_{1}+N_{3}+N_{4})\right)+\ldots$
Theorem: If $S\in \mathbf {V}$, then $\operatorname {span} (S)=$ {all l.c. of elements of S} is a subspace of V.