Linear Algebra cheat sheet
Vectors
dot product: u∗v=||u|| ∗ ||v|| ∗ cos(φ) = uxvx+uyvy
cross product: u×v=
uyvz−uzvy
uzvx−uxvz
uxvy−uyvx
norms:
kxkp:= p
pPn
i=1 |xi|p
kxk1:= Pn
i=1 |xi| kxk∞= max
i|xi|
enclosed angle:
cosφ =u∗v
||u|| ∗ ||v||
||u|| ∗ ||v|| =q(u2
x+u2
y)(v2
x+v2
y)
Matrices
basic operations
transpose: [AT]ij = [A]ji : ”mirror over main diagonal”
conjungate transpose / adjugate: A∗= (A)T=AT
”transpose and complex conjugate all entries”
(same as transpose for real matrices)
multiply: AN×M∗BR×K=MN×K
invert: a b
c d−1
=1
det(A)d−b
−c a=1
ad−bc d−b
−c a
norm:
kAkp= max
x6=0
kAxkp
kxkp, induced by vector p-norm
kAk2=pλmax(ATA)
kAk1= max
jPm
i=1 |aij|,
kAk∞= max
iPn
j=1 |aij|,
condition: cond(A) = kAk ·
A−1
determinants
det(A) = Pσ∈Snsgn(σ)Qn
i=1 Ai,σi
For 3×3 matrices (Sarrus rule):
arithmetic rules:
det(A·B) = det(A)·det(B)
det(A−1) = det(A)−1
det (rA) = rndet A, for all An×nand scalars r
eigenvalues, eigenvectors, eigenspace
1. Calculate eigenvalues by solving det (A−λI)=0
2. Any vector xthat satisfies (A−λiI)x= 0 is eigenvector for λi.
3. EigA(λi) = {x∈Cn: (A−λi)x= 0}is eigenspace for λi.
definiteness
defined on n×n square matrices:
∀λ∈σ(A).
λ > 0⇐⇒ positive-definite
λ≥0⇐⇒ positive-semidefinite
λ < 0⇐⇒ negative-definite
λ≤0⇐⇒ negative-semidefinite
if none true (positive and negative λexist): indefinite
equivalent: eg. xTAx > 0⇐⇒ positive-definite
rank
Let A be a matrix and f(x) = Ax.
rank(A) = rank(f) = dim(im(f))
= number of linearly independent column vectors of A
= number of non-zero rows in A after applying Gauss
kernel
kern(A) = {x∈Rn:Ax = 0}(the set of vectors mapping to 0)
For nonsingular A this has one element and dim(kern(A)) = 0 (?)
trace
defined on n×n square matrices: tr(A) = a11 +a22 +··· +ann
(sum of the elements on the main diagonal)
span
Let v1,...,vrbe the column vectors of A. Then:
span(A) = {λ1v1+···+λrvr|λ1,...,λr∈R}
spectrum
σ(A) = {λ∈C:λis eigenvalue of A}
properties
square:N×N
symmetric:A=AT
diagonal: 0 except ak k
⇒implies triangular (eigenvalues on main diagonale)
orthogonal
AT=A−1⇒normal and diagonalizable
unitary
Complex analogy to orthogonal: A complex square matrix is unitary
if all column vectors are orthonormal
⇒diagonolizable
⇒cond2(A)=1
⇒ |det(A)|= 1
nonsingular
An×nis nonsingular = invertible = regular iff:
There is a matrix B:= A−1such that AB =I=BA
det(A)6= 0
Ax =bhas exactly one solution for each b
The column vectors of Aare linearly independent
rank(A) = n
f(x) = Ax is bijective (?)
⇒det(A)−1=det(A−1)
⇒(A−1)−1=A
⇒(AT)−1= (A−1)T
diagonalizable
An×ncan be diagonalized iff:
it has nlinear independant eigenvectors
all eigenvalues are real and distinct
there is an invertible T, such that:
D:= T−1AT =
λ1
...
λn
A=T−1DT and AT =T D
λ1,...,λnare the eigenvalues of A!
T can be created with eigenvectors of A and is nonsingular!
diagonally dominant matrix
∀i.|aii| ≥ Pj6=i|aij |
⇒nonsingular
Hermitian
A square matrix Awhere A∗=A(equal to its adjugate)
A real matrix is Hermitian iff symmetric
⇒ =(det(A)) = 0 (determinante is real)
triangular
A square matrix is right triangular (wlog n = 3):
a11 a12 a13
0a22 a23
0 0 a33
⇒Eigenvalues on main diagonale
idempotent
A square matrix Afor which AA =A.
block matrices
Let B, C be submatrices, and A, D square submatrices. Then:
det A0
C D= det A B
0D= det(A) det(D)
minors
A matrix A has minors Mi,j := remove row i and column j from A
principle minors: {det(upp er left i×imatrix of A) : i..n}
Sylvester’s criterion for hermitian A:
⇒A is positiv-definite iff all principle minors are positive
