LA

LinAlg Chicken Scratch

On Linear Transformations

Let dim U = n, dim W = m, T element of L(U, W).

T can be defined by transformations of basis vectors {u_1...u_n} of U into vectors in W. As such,
(Tu_1 = a_1*w_1 + ... + a_m*w_m) for all basis vectors u defines a linear transformation.

Null space = all vectors u s.t. Tu = 0
Range = all vectors w s.t. there exists some u where Tu = w

T maps subspaces to subspaces, and the inverse is true as well. Hence Range T is
a subspace of W and null T is a subspace of U

Fundamental theorem (rank-nullity): dim U = dim Null T + dim Range T
Implications:
1) if n > m, Tx = 0 must have a solution (dim null T > 0)
2) if n < m, Tx = c may not have a solution for some c (dim range T < dim w)

T is injective or one-to-one if Tu = Tv implies u = v. Injective implies null space is {0}.
T is surjective or onto if Range T = W.

T is invertible if it is injective and surjective (bijective). An isomorphism between U and W 
exists if there is an invertible linear map between the two.
S is the inverse of T if ST is identity on U and TS is identity on W.

If T' element of L(V, V), then injective/surjective/bijective/invertible each imply the others,
because of rank-nullity theorem.

Cool: 
If T is injective, it has a left inverse (ST is identity on U). Hence T has a left inverse when null T = 0,
or in other words dim U = dim range T <= dim W. (dim U <= dim W; corresponding matrix *cannot* be fat)

If T is surjective, it has a right inverse (TS is identity on W). Hence T has a right inverse when
dim U = dim null T + dim range T >= dim range T = dim W (dim U >= dim W; corresponding matrix
*cannot* be skinny)

These two show again that T is invertible iff it is injective and bijective, which
means that dim U = dim W (the corresponding matrix, see below, must be square).

On Matrices

M(T, u_1...u_n, w_1...w_m) is a matrix for T for the given basis vectors.

M is defined as follows: 

Let Tu_i = sum c_ji * w_ji, as mentioned in the previous section. This defines
each basis vector of U in terms of a vector in W. The corresponding matrix for T
is [c_*1 ... c_*n], where c_*i denotes the column vector [c_1i ... c_mi]'

Similarly, a vector u in U can be represented by the nx1 column vector consisting
of the n coefficients of the linear combination of the n basis vectors of u.

FROM the above definitions can be derived the definition of matrix addition, scalar multiplication, and matrix multiplication! (Matrix multiplication corresponds to linear transformation
composition. Note: If L and S are both linear, LS is linear)

When multiplying Mv, the result is a linear combination of the columns of M. The possible
outputs, then, or 'range', is spanned by the columns of M. *Rank* is the dimension of this
columnspace. Column rank = row rank (proof on wikipedia).

Rank corresponds to dim Range T. hence m - rank = dim null T (rank nullity theorem).

If a matrix has rank = n, then it is injective because n - (dim range T = n) = 0 = dim null T. Hence
it has a left inverse.

If a matrix has rank = m, it is surjective, because rank = m = dimW. Thus,
it has a right inverse.

If a matrix has rank = m = n, it has full rank and has the following properties:
1) is square, because m = n
2) surjective (see above)
3) injective (see above)
4) invertible