09-240/Classnotes for Tuesday September 29

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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):

1. A unit circle is not closed under addition of scalar multiplication.
2. The x-axis $\cup$ y-axis is closed under scalar multiplication, but not under addition.
3. A single quadrant of the Cartesian plane is closed under addition, but not under scalar multiplication.

Examples of subspaces:

1. $\{0\}$
2. Any VS (which is a subspace of itself)
3. A line passing through the origin (if it does not pass through the origin, then it is not closed under scalar multiplication)
4. A plane
5. 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.)
6. $\mathbf {W} =\{A\in \mathbb {M} _{n\times n}:\operatorname {tr} (A)=0\}$

   where $\operatorname {tr} (A)=\sum _{i=1}^{n}a_{ii}$ is the "trace" of A.

   Properties of trace:

   1. $\operatorname {tr} (0\cdot A)=0$
   2. $\operatorname {tr} (cA)=c\cdot \operatorname {tr} (A)$
   3. $\operatorname {tr} (A+B)=\operatorname {tr} (A)+\operatorname {tr} (B)$

   so W is indeed a subspace.

Claim: If W₁ and W₂ are subspaces of V, then

1. $W_{1}\cap W_{2}=\{x\in \mathbf {V} :x\in \mathbf {W} _{1}{\mbox{ and }}\mathbf {W} _{2}\}$ is a subspace of V, W₁, and W₂.
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. 20-21, #19.)

Linear combinations

Definition: A vector u is a "linear combination" (l.c.) of vectors u₁, ..., u_n if there exists scalars a₁, ..., a_n such that

$u=a_{1}u_{1}+a_{2}u_{2}+\ldots +a_{n}u_{n}$

Example: $\mathbb {P} _{n}(F)=\{{\mbox{Polynomials of degree at most }}n{\mbox{ with coefficients in }}''F''\}$
$=\{\sum _{i=0}^{n}a_{i}x^{i}:a_{i}\in F\}$

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.