Linear Algebra Flashcards
What does Sylvestors criterion determine?
If a matrix is positive-definite
What 3 things are equivalent in terms of if A is diagonalisable?
A is similar to D
The eigen values of A are distinct
A is real, symmetric
A and B are similar if there exists an invertible matrix M such that…
A=M^-1BM
A and B will have the same eigenvalues
K vectors in Rn. If k>n, k=n, k<n
k>n = not linearly independent
k<n = don’t span
k=n then if basis then matrix non singular
Dim V =
The cardinality of any basis for V
If u1,u2,..,uk are in v, then the span(u1,…,uk) is
A subspace of v
Subspace criteria
Existence of origin
Closure under addition
Closure under scalar multiplication
Properties of groups
Closure under .
Associativity
Identity
Inverse
Two groups are isomorphic if
There exists a bijection between the groups preserving group operation
Lagrange for groups
If H is a subgroup of G then |H| divides |G|
Direct product
= every possible pair of elements, one from each group
Sign of a permutation
+1 if sigma is the product of an even number of transpositions
-1 if an odd number
If T is a linear map then the following are equivalent
ket(T)=0
T is injective
If v1,v2,..,vk are LI in v then T(v1),T(v2),…,T(vk) are LI in w
Symmetric group sn
Elements are all the permutations on n distinct symbols
|Sn|= n!
Transposition
A permutation that exchanges 2 elements and leaves the rest alone
Alternating group
Subset of Sn containing all even permutations
|An|= n!/2
Permutation of a set
A bijection to itself
If L is hermitean then <Lv,v> is real
If L is anti-hermitean then <Lv,v> is purely imaginary
Basis
= LI and span
rank and nullity = k
rk(A) = number of leading 1’s in RREF
nullity(A) = number of columns in RREF without a leading 1
dimU+dimV=
dim(U+V)+dim(U intersect W)
If V is the direct sum of U and W then
If u1,uk basis of U and w1,wp basis for w then u1,uk,w1,wp basis for v
dimV=dimU+dimW
each v can be written uniquely as v=u+w
Area of a triangle with vertices a,b,c
1/2 |det(b-a,c-a)|
Scalar triple
a dot bxc
If u is the orthogonal projection of v onto U then
||u||^2<=||v||^2
Bessel’s inequality
If u1,…,unis an orthonormal basis for u and v= k1u1+k2u2+…knun then
the sum of ki^2 <=||v||^2
If u0 is the orthogonal projection of v0 onto U then for all u
||u-v0||>=||u0-v0||
Projection operator P:V to V
im(P) = U ker(P) = U complement thing
P^2=P
P(u)=u
(I-P)P=P(I-P)=0
P(v)= (v,u1)u1 +..+(v,uk)uk where ui is an orthonormal basis of U
V = direct sum of U and U orthogonal complement
U orthogonal complement =
{v in V: (u,v)=0 for all u in U}
Pythagoras
||u+v||^2 = ||u||^2+||v||^2
Real cauchy-schwartz inequality
(u,v)^2<=(u,u)(v,v)
Complex IP
<u,v>= conjugate of <v,u>
Basis =
maximal LI set
minimal spanning set
any u in v can be written as a sum of the vectors
Nullspace/kernel
Solution space to Ax=0
Rowspace
A^ty= c has a solution
Columnspace
Ax=b has a solution
Solving differential equations
-write as a matrix
-diagonalise A=MDM
-x= Mw
-w dot = Dw
-solve e
-use x=Mw to get back to x
