# 09-240/Classnotes for Tuesday September 29

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## Vector subspaces

Definition. ${\displaystyle \mathbf {W} \subset \mathbf {V} }$ is a "subspace" if it is a vector space under the operations it inherits from V.

Theorem. ${\displaystyle \mathbf {W} \subset \mathbf {V} }$ is a subspace iff it is "closed under addition and vector multiplication by scalars", i.e. ${\displaystyle x,y\in \mathbf {W} \Rightarrow x+y\in \mathbf {W} }$ and ${\displaystyle a\in F,x\in \mathbf {W} \Rightarrow ax\in \mathbf {W} }$.

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 ${\displaystyle \cup }$ 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. ${\displaystyle \{0\}}$
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 ${\displaystyle \mathbf {V} =\mathbb {M} _{n\times n}(F)}$. If ${\displaystyle W=\{A\in \mathbf {V} :A^{\top }=A\}}$, then W is a subspace of V. (W is the set of "symmetric" matrices in V; AT denotes the transpose of A.)
6. ${\displaystyle \mathbf {W} =\{A\in \mathbb {M} _{n\times n}:\operatorname {tr} (A)=0\}}$
where ${\displaystyle \operatorname {tr} (A)=\sum _{i=1}^{n}a_{ii}}$ is the "trace" of A.
Properties of trace:
1. ${\displaystyle \operatorname {tr} (0\cdot A)=0}$
2. ${\displaystyle \operatorname {tr} (cA)=c\cdot \operatorname {tr} (A)}$
3. ${\displaystyle \operatorname {tr} (A+B)=\operatorname {tr} (A)+\operatorname {tr} (B)}$
so W is indeed a subspace.

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

1. ${\displaystyle W_{1}\cap W_{2}=\{x\in \mathbf {V} :x\in \mathbf {W} _{1}{\mbox{ and }}\mathbf {W} _{2}\}}$ is a subspace of V, W1, and W2.
2. But ${\displaystyle W_{1}\cup W_{2}=\{x\in \mathbf {V} :x\in \mathbf {W} _{1}{\mbox{ or }}x\in \mathbf {W} _{2}\}}$ is a subspace of V iff ${\displaystyle \mathbf {W} _{1}\subset \mathbf {W} _{2}}$ or ${\displaystyle \mathbf {W} _{2}\subset \mathbf {W} _{1}}$. (See HW2 pp. 20-21, #19.)

## Linear combinations

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

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

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

Definition: A subset ${\displaystyle S\subset \mathbf {V} }$ "generates" or "spans" V iff the set of linear combinations of elements of S is all of V.

Example: Let ${\displaystyle \mathbf {V} =\mathbb {M} _{n\times n}(\Re )}$
Let ${\displaystyle M_{1}={\begin{pmatrix}1&0\\0&0\end{pmatrix}},M_{2}={\begin{pmatrix}0&1\\0&0\end{pmatrix}},M_{3}={\begin{pmatrix}0&0\\1&0\end{pmatrix}},M_{4}={\begin{pmatrix}0&0\\0&1\end{pmatrix}}}$
Then ${\displaystyle S=\{M_{1},M_{2},M_{3},M_{4}\}}$ generates V.

Proof: Given ${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}\in \mathbb {M} _{2\times 2}(\Re )}$, write
${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}=aM_{1}+bM_{2}+cM_{3}+dM_{4}.}$

Example: Let ${\displaystyle N_{1}={\begin{pmatrix}0&1\\1&1\end{pmatrix}},N_{2}={\begin{pmatrix}1&0\\1&1\end{pmatrix}},N_{3}={\begin{pmatrix}1&1\\0&1\end{pmatrix}},N_{4}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}}$
Does ${\displaystyle \{N_{1},N_{2},N_{3},N_{4}\}}$ generate V?

${\displaystyle M_{1}=-{\frac {2}{3}}N_{1}+{\frac {1}{3}}(N_{2}+N_{3}+N_{4})}$
${\displaystyle M_{2}=-{\frac {2}{3}}N_{2}+{\frac {1}{3}}(N_{1}+N_{3}+N_{4})}$
${\displaystyle M_{3}=-{\frac {2}{3}}N_{3}+{\frac {1}{3}}(N_{1}+N_{2}+N_{4})}$
${\displaystyle M_{4}=-{\frac {2}{3}}N_{4}+{\frac {1}{3}}(N_{1}+N_{2}+N_{3})}$

Then ${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}=aM_{1}+bM_{2}+cM_{3}+dM_{4}=}$

${\displaystyle a\cdot \left(-{\frac {2}{3}}N_{1}+{\frac {1}{3}}(N_{2}+N_{3}+N_{4})\right)+b\cdot \left(-{\frac {2}{3}}N_{2}+{\frac {1}{3}}(N_{1}+N_{3}+N_{4})\right)+\ldots }$

Theorem: If ${\displaystyle S\in \mathbf {V} }$, then ${\displaystyle \operatorname {span} (S)=}$ {all l.c. of elements of S} is a subspace of V.