- Vector spaces and fields are like groups
- With 2 operations
- Vector
- direction and magnitude
- numbers with an order
- list = ordered set
- \(N\)x\(1\) matrix
- A vector is not just an \(N\)x\(1\) matrix. A vector exists in a
vector space
- might be full of physics vectors, matrices, or polynomials
- Field
- Addition and multiplication might be different
- might be related to normal addition/multiplication
- might by any binary operation
- Addition is "primary" operation, multiplication is "secondary"
- addition is really good (more group like)
- multiplication needs to exclude the additive identity (because it can't have an inverse)
- questions
- multiplication is repeated addition?
- not necessarily
- binary expressions?
- associative?
- both yes
- multiplication is repeated addition?
- 1.3 demonstrates that the complex numbers are a field
- commutativity
- associativity
- identities
- additive inverse
- multiplicative inverse except additive identity
- distributive
- usually means Reals or Complex
- integers mod 3 are a field
- #bonushw show integers mod 3 are a field
- integers mod 3 are a field
- higher dimensions
- \(R^2\) is a Cartesian plane, \(R^4\) is a space
- operations
- addition is really nice (element wise)
- scalar multiplication is easy enough
- vector vector multiplication is hard to find
- Addition and multiplication might be different
- two square roots of \(i\)
- fundamental theorem of algebra
- (important)
- So, \(i\) should have two square roots
- Powers of \(i\) go in a circle (90 degrees rotation)
- Complex number rotation gives a preferred direction
- So that's why the quadrants are numbered in that direction
- One can be found geometrically 20math530srcSquareRootI.png
- We could also try it algebraically
- \((a+bi)^2=i=a^2-b^2+2abi\)
- so \(a^2-b^2 = 0\) and \(2ab = 1\)
- fundamental theorem of algebra
- dot product
- How much of \(\vec{A}\) is in the direction of \(\vec{B}\) multiplied by the magnitude of \(\vec{B}\)
- \(\vec{A} \cdot \vec{B} = |A||B| cos \theta\)
- #bonushw prove that ^^
- Identity: \(\frac{\vec{A}\cdot\vec{B}}{|A||B|} = cos \theta\)
- cross product
- only works on 3x1 matrices
- steps
- determinant
- \(i\), \(j\), \(k\) are the unit vectors
- \[\begin{bmatrix}2\\1\\0\end{bmatrix}\begin{bmatrix}1\\2\\-1\end{bmatrix} = \left|\begin{bmatrix}i &j &k\\2 &1 &0\\1 &2 &-1\end{bmatrix}\right| = i\left|\begin{bmatrix}1&0\\2&-1\end{bmatrix}\right|-j\left|\begin{bmatrix}2&0\\1&-1\end{bmatrix}\right| + k\left|\begin{bmatrix}2&1\\1&2\end{bmatrix}\right| = \begin{bmatrix}-1\\2\\3\end{bmatrix}\]
- dropping zero: \(0a = (0+0)a = 0a+0a \Rightarrow 0a = 0\), so the additive identity can't have a multiplicative inverse (everything multiplied it will just be the additive identity)
- determinant
- measures the "size" of a matrix, denoted absolute value (relevant to inverse of a matrix multiplication)
- #todo #exr0n #future prove identities are unique
