Linear Final

Linear Combination

Given vectors v1, v2, v3…vp in Rn, the vector y will be defined as a LC by y=c1v1+c2v2+…+cpvp where c1…cp are in R and are scalar weights

Span

Collection of all vectors that can be written in the form c1v1+c2v2+…+cpvp where v1…vp is in Rn and c1…cp is in R.

Thm 4

For each B in Rm, the equation Ax=b has a solution

Each b in Rm is a LC of the columns of A

The columns of A span Rm

A has a pivot position in every row

Is it a linear transformation?

Let u, v, be in R3, u=[ ] and v = [ ], c is in R

1. T(0) = 0

2. T(u+v) = T(u) + T(v)

3. T(cu) = cT(u)

If A is an invertible nxn matrix, prove that for each b in Rn the equation Ax=b has the unique solution x=A-1b

1. Existence - if x=A-1b, Ax=AA-1b, Ax=b, b=b

2. Uniqueness - let u be another solution Au=b. If Au=b, A-1Au=A-1b, u=A-1b, u=x

Eigenvector

A nonzero vector x of nxn matrix A such that Ax=lambdax, a vector transformed by matrix A into a scalar multiple of itself

eigenvalue

lambda the scalar multiple for Ax=lambdax so that x vector stretches or shrinks but stays in the same direction

Diagonizable

A square matrix is diagonalizable if it is similar to a diagonal matri

Null space

set of all solutions of Ax=0 in Rn

column space

The set of all Lcs of the columns of A in Rm

Basis

Let H be a subspace of a vector space V. An indexed set of vectors B=[b1…bp} in V is a basis for H if: B is LI and subspace spanned by B coincides with H (H=span{b1…bp})

An orthogonal matrix is a square matrix such that…

G-1=GT

Prove the formula for linear combination weights for an orthogonal basis.

y=c1u1+c2u2+…+cpup

y . u1 = (c1u1+…+cpup) . u1

y . u1 = c1u1 . u1 + c2u2 . u1 + …. + cpup . u1

y . u1 = c1u1 . u1

c1 = y . u1 / u1 . u1