Andrew Thangaraj

Aug-Nov 2020

- Vector space \(V\) over a scalar field \(F\)
- \(F\): real field \(\mathbb{R}\) or complex field \(\mathbb{C}\) in this course

- \(m\times n\) matrix A represents a linear map \(T:F^n\to F^m\)
- dim null \(T\) + dim range \(T\) = dim \(V\)

- Linear equation: \(Ax=b\)
- Solution (if it exists): \(u+\) null\((A)\)

- Four fundamental subspaces of a matrix
- Column space, row space, null space, left null space

- Determinant of a square matrix
- Function with many interesting properties

- Change of basis for linear map
- Can result in simpler matrix representation

*constant*: \(a\), \(a\in F\)

*Roots*: all scalars if \(a=0\); no roots for nonzero\(a=0\):

*zero polynomial*

*linear*: \(ax+b\), \(a,b\in F\), \(a\ne 0\)

*Roots*: one root \(=-b/a\)

*quadratic*: \(ax^2+bx+c\), \(a,b,c\in F\), \(a\ne 0\)

*Roots*: \((-b\pm\sqrt{b^2-4ac})/2a\)Two roots (repeated or multiplicity 2 if \(b^2=4ac\))

\(F=\mathbb{R}\): two complex roots or two real roots

*cubic*: \(ax^3+bx^2+cx+d\), \(a\ne0\)

*Roots*: there is a formula involving cube roots etcThree roots (multiplicities: 1,1,1 or 1,2 or 3)

\(F=\mathbb{R}\): at least one real root

*quartic*: \(ax^4+bx^3+cx^2+dx+e\), \(a\ne0\)

*Roots*: there is a formula involving fourth roots etc- Four roots (multiplicities: 1,1,1,1 or 1,1,2 or 2,2 or 1,3 or 4)

*quintic and higher*: \(p_nx^n+\cdots+p_1x+p_0\)

*Roots*: there is**no formula involving radicals**(Abel-Ruffini theorem)

*modern numerical methods*: compute roots for polynomials

- Try MATLAB, Mathematica, Python numpy/scipy

*Polynomials in one variable with coefficients from a field*

\(p(x)=p_0+p_1x+p_2x^2+\cdots+p_nx^n,\quad p_i\in F\)

*Degree of \(p(x)\)*: deg \(p(x)\) \(=\) largest \(n\) such that \(p_i\ne 0\)

degree(zero polynomial) \(=-\infty\)

*Leading term of \(p(x)\)*: LT \(p(x)=p_nx^n\), where \(n=\) deg \(p(x)\)

Addition, Multiplication (well-known)

- deg \(a(x)b(x)\ge\) deg \(a(x)\), if \(b(x)\ne 0\)

*Division algorithm: \(p(x)\) by \(a(x)\)*

- \(i=0\), \(q(x)=0\), \(p_0(x)=p(x)\)
- while deg \(p_i(x)\ge\) deg \(a(x)\)
- \(t_i(x)=\text{LT}(p_i(x))/\text{LT}(a(x))\)
- \(q(x)=q(x)+t_i(x)\)
- \(p_{i+1}(x)=p_i(x)-t_i(x)a(x)\)
- \(i=i+1\)

- return \(q(x)\), \(r(x)=p_i(x)\)

Given two polynomials \(p(x)\) and \(a(x)\), there exist unique polynomials \(q(x)\) and \(r(x)\) such that \[p(x)=q(x)a(x)+r(x),\] with \(r(x)=0\) or deg \(r(x)<\) deg \(a(x)\).

\(q(x)\): quotient, \(r(x)\): remainder

*Proof*: Use division algorithm with induction

*Roots and factors*

\(\lambda\in F\) is a

*root*of \(p(x)\) if \(p(\lambda)=0\)If \(p(x)\) divided by \(a(x)\) results in remainder zero,

\(p(x)=q(x)a(x)\)

\(a(x)\)

*divides*\(p(x)\), denoted \(a(x)\,|\,p(x)\)\(a(x)\) is a

*factor*of \(p(x)\)

*Fundamental theorem of algebra*

Every non-constant polynomial with complex coefficients has a complex root.

*Proof*: Involves complex analysis.

\(\lambda\) is a root of \(p(x)\) iff \(x-\lambda\) is a factor of \(p(x)\).

degree-1 factor: called linear factor

*Proof*: Divide \(p(x)\) by \(x-\lambda\).

\(p(x)=p_nx^n+\cdots+p_1x+p_0\), \(p_i\in\mathbb{C}\), \(p_n\ne 0\)

There exists \(n\) complex roots \(\lambda_1,\ldots,\lambda_n\) and \[p(x)=p_n(x-\lambda_1)\cdots(x-\lambda_n)\]

*Proof*: Use fundamental theorem and degree-1 factor result repeatedly

\(p(x)=p_nx^n+\cdots+p_1x+p_0\), \(p_i\in\mathbb{R}\), \(p_n\ne 0\)

Need not have real roots (eg: \(x^2+1\))

If \(\lambda\) is a root, \(\bar{\lambda}\) is a root

*Factors with real coefficients*

If \(\lambda\) is a real root, \(x-\lambda\) is a factor

If \(\lambda\) is a complex root, \(x^2-(\lambda+\bar{\lambda})x+\lambda\bar{\lambda}\) is a factor

*Factorization*

\(p(x)=\) (quadratic factors) (linear factors)