inner product
motivation of define inner product
for w and z in complex space, we expect that the inner product of w with z equals the complex conjugate of the inner product of z with w
property of inner product(by define
only in the first slot:
$
adjoint
adjoint of T:= T*
property:
eigenvalue of T = complex conjugate of eigenvalue of T*
self adjoint: T=T*
in complex inner product space,T=T* =0 <=> v=0
T: SELF adjoint operator, T=T*
in real inner product space(ie, v=v_bar),T=T* =0 <=,(!=>) v=0
property:
T is self adjoint => T has an eigenvalue (prove in p135)
symmetry matrix = self adjoint operator T
property:
$$
normal operator
An operator on an inner-product space is called normal if it commutes with its adjoint:= $A^*A = AA^*$
A commute with B <=> ab=ba
Any two diagonal matrices commute
A is normal operator <=> $|Av|=|A*v|$
property:
for any T: eigenvalue of T = complex conjugate of eigenvalue of T
if engenvalue is real => equals
T is normal => eigenvalue of T = eigenvalue of T, eigenvector eaquls too
spectral theorem:
motivation: want an operator has a diagonal matrix with respect to an orthonormal basis (precisely, the basis consists the eigenvectors of T) in spectral theorem, $T=XDX^T$
F=C + requirement <=> T is normal operator – complex spectral theorem
F=R + requirement <=> T is self-adjoint operator – real spectral theorem
every symmetry matrix has the factorization $A=Q\LambdaQ^T$, Q: orthonormal eigenvectors matrix
$A = A^T$ <=> A have n eigenvalues and n real orthogonal eigenvectors
positive matrix: T is self-adjoint and $\ >= 0$ for all v
every orthogonal projection is positive
theorem
below are equivalent:
1. T is positive
2. $v^T T v>0$ for all v!=0
3. self-adjoint(ie. symmetry for real space) + all eigenvalues are positive
self-adjoint => exist orthonormal basis X(eigenvectors) and diagonal matrix(eigenvalues), st, $TX=X\Lambda$
4. can be writen as $R^T R$, R has independent cloumns
5. determinants of submatrix are positive
determine whether pd?
cal all eigenvalue - should be all positive – aviod
see pivots - should be all positive
####
application: a twice continuously defferentiable multivariable function is convex iff it hessian is PD.
:= property of convex function
if A PD => A can be written as $A = A^{1/2}A^{1/2}$ A^{1/2}>0;
orthonormal and orthogonal
vectors {v1,v2,v3,…vn} are orthonormal <=> = vi^T vj=0, j!=i
matrix wirh orithonomal cols => orthogonal matrix <=> $QQ^T=I$
Gram-schmidt process - create orthonormal vectors
determinant:
singular, non-invertible <=> detA=0
non-singular, invertible <=> detA!=0
property:
|A|=|A^T|
THREE way to cal determinant
1.elimination: $A=P\bar{A}$, $|A|=|P||\bar{A}|$, $|\bar{A}|$ is upper triangle => product of diagonal entries
determinant of corner submatrix $A_k$ is d_1d_2…d_k
pivors from determinant: kth pivot $d_k=|Ak|/|A{K-1}|$
2. big formular
n! trem operation
3. determinant by cofactors formular
eigenvalue
calculation: $det(A-\lambdaI)=0$
elimination does not preserve eigenvalues => row operation change eigenvalues
=> eigenvalue of A do not equals $\bar{A}$
the product of the n eigenvalues equals the determinant of A (no row change)
<= $|A|=|\lambda||I|$
=> eigenvalues of $\bar{A}$ = (-1 if row change exist)product of diagonal entries = determinant
=> no eigenvalues = 0 <=> product of eigenvalues != 0 <=> A is not invertiable
=> pivots and eigenvalues have the same sigh
sum of n eigenvalues equals the sum of the n diagonal entries of A
p279 Gilbert Strang
diagonalizable
A have n independent eigenvectors <=> A is diagonalizable
=> about eigenvectors enough or not
=> symmetry matrix is always diagonalizable
eigenvectors can be chosen orthonormal
$AX=X\Lambda$, X is invertiable => $A=X\LambdaX^{-1}$ => A is diagonalizeble
remark:
diagonalizability - eigenvectors
| no connection
invertiablity - eigenvalues