ALG1_L6

$A$ is an $r\times c$ matrix.
If $b_1,b_2\in \mathrm{Im}(A)$, $b_1=A{x}_1,b_2=A{x}_2,x_1,x_2\in {\mathbb{R}}^c$
$b_1+b_2=A{x}_1+A{x}_2=A(x_1+x_2)\in \mathrm{Im}(A)$.
similar argument for multiplication by scalars.

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Linear combinations, span

Defined linear combinations.

Definition 1(Span).

Span of a set of vectors $S=\{v_1,v_2,\dots ,v_n\},v_i\in V$ is the set of all linear combinations of the vectors in the set.

Observe:

• span(S) is a subspace of $V$.
  • $0\in \text{span}(S)$
  • $\text{span}(S)$ is closed under addition and multiplication by a scalar.
• Claim: $\text{span}(S)$ is the smallest subspace of $V$ that contains $S$, i.e, if a subspace $W\subset V$, contains $S$, then $W$ contains $S$.
  Realize: $\text{span}(S)$ could have been defined as the smallest subspace of $V$ that contains $S$.
• For a matrix $A$, $\text{range}(f_A)=\text{Im}(A)=\text{span}(\text{column vectors of A})$.
• $f_A$ is surjective $⟺$ Span of column vectors of $A$ = ${\mathbb{R}}^r$
• $f_A$ is injective $⟺$ only the trivial linear combination of the column vectors of $A$ results in $0$.

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Linear independence

Defined linear independence.

Claim: $v_1,v_2,\dots v_n\in V$ are linearly dependent $⟺$ one of them can be expressed as a linear combination of the others.
Proof:
2 $⟹$ 1
let $v_k$ be expressible as a linear combination of the remaining vectors. Then, by pushing it to the other side of the equation, a non-trivial linear combination which equals $0$ can be produced.
1 $⟹$ 2
A nontrivial linear combination exists which equals $0$. Pick any $v_k$ with non zero coefficient $c_1$. Divide the equation by $c_1$ and it is easy to see that $v_1$ can be expressed as a linear combination of the other vectors.

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Examples of vector spaces: ${\mathbb{R}}^n,{\mathbb{P}}_n$, homogeneous polynomials of degree $d$ in $k$ variables.