Math 54 Cheat Sheet
Vector spaces
Subspace: If uand vare in W, then u+vare in W, and cuis in W
Nul(A): Solutions of Ax=0. Row-reduce A.
Row(A): Space spanned by the rows of A: Row-reduce Aand choose
the rows that contain the pivots.
Col(A): Space spanned by columns of A: Row-reduce Aand choose the
columns of Athat contain the pivots
Rank(A): = Dim(Col(A)) = number of pivots
Rank-Nullity theorem: Rank(A) + dim(Nul (A)) = n, where Ais
m×n
Linear transformation: T(u+v) = T(u) + T(v),T(cu) = cT (u),
where cis a number.
Tis one-to-one if T(u) = 0⇒u=0
Tis onto if Col(T) = Rm.
Linearly independence:
a1v1+a2v2+···+anvn=0⇒a1=a2=···=an= 0.
To show lin. ind, form the matrix of the vectors, and show that
Nul(A) = {0}
Linear dependence: a1v1+a2v2+···+anvn=0for
a1, a2,···, an, not all zero.
Span: Set of linear combinations of v1,···vn
Basis Bfor V: A linearly independent set such that Span (B) = V
To show sthg is a basis, show it is linearly independent and spans.
To find a basis from a collection of vectors, form the matrix Aof the
vectors, and find Col(A).
To find a basis for a vector space, take any element of that v.s. and
express it as a linear combination of ’simpler’ vectors. Then show those
vectors form a basis.
Dimension: Number of elements in a basis.
To find dim, find a basis and find num. elts.
Theorem: If Vhas a basis of vectors, then every basis of Vmust have n
vectors.
Basis theorem: If Vis an n−dim v.s., then any lin. ind. set with n
elements is a basis, and any set of nelts. which spans Vis a basis.
Matrix of a lin. transf Twith respect to bases Band C: For every vector
vin B, evaluate T(v), and express T(v)as a linear combination of
vectors in C. Put the coefficients in a column vector, and then form the
matrix of the column vectors you found!
Coordinates: To find [x]B, express xin terms of the vectors in B.
x=PB[x]B, where PBis the matrix whole columns are the vectors in
B.
Invertible matrix theorem: If Ais invertible, then: Ais row-equivalent
to I,Ahas npivots, T(x) = Axis one-to-one and onto, Ax=bhas a
unique solution for every b,ATis invertible, det(A)6= 0, the columns
of Aform a basis for Rn,Nul(A) = {0},Rank (A) = n
a b
c d−1
=1
ad−bc d−b
−c a
A|I→I|A−1
Change of basis: [x]C=PC←B [x]B(think of Cas the new, cool basis)
[C | B]→[I|PC←B]
PC←B is the matrix whose columns are [b]C, where bis in B
Diagonalization
Diagonalizability: Ais diagonalizable if A=P DP −1for some
diagonal Dand invertible P.
Aand Bare similar if A=P BP −1for Pinvertible
Theorem: Ais diagonalizable ⇔Ahas nlinearly independent
eigenvectors
Theorem: IF Ahas ndistinct eigenvalues, THEN Ais diagonalizable,
but the opposite is not always true!!!!
Notes: Acan be diagonalizable even if it’s not invertible (Ex:
A=0 0
0 0). Not all matrices are diagonalizable (Ex: 1 1
0 1)
Consequence: A=P DP −1⇒An=P DnP−1
How to diagonalize: To find the eigenvalues, calculate det(A−λI), and
find the roots of that.
To find the eigenvectors, for each λfind a basis for Nul (A−λI),
which you do by row-reducing
Rational roots theorem: If p(λ)=0has a rational root r=a
b, then a
divides the constant term of p, and bdivides the leading coefficient.
Use this to guess zeros of p. Once you have a zero that works, use long
division! Then A=P DP −1, where D= diagonal matrix of
eigenvalues, P= matrix of eigenvectors
Complex eigenvalues If λ=a+bi, and vis an eigenvector, then
A=P CP −1, where P=Re(v)I m(v),C=a b
−b a
Cis a scaling of pdet(A)followed by a rotation by θ, where:
1
√det(A)C=cos(θ) sin(θ)
−sin(θ) cos(θ)
Orthogonality
u,vorthogonal if u·v= 0.
kuk=√u·u
{u1···un}is orthogonal if ui·uj= 0 if i6=j, orthonormal if
ui·ui= 1
W⊥: Set of vwhich are orthogonal to every win W.
If {u1···un}is an orthogonal basis, then:
y=c1u1+···cnun⇒cj=y·uj
uj·uj
Orthogonal matrix Qhas orthonormal columns!
Consequence:QTQ=I,QQT=Orthogonal projection on Col(Q).
kQxk=kxk
(Qx)·(Qy) = x·y
Orthogonal projection: If {u1···uk}is a basis for W, then orthogonal
projection of yon Wis: ˆ
y=y·u1
u1u1u1+···+y·u1
ukukuk
y−ˆ
yis orthogonal to ˆ
y, shortest distance btw yand Wis ky−ˆyk
Gram-Schmidt: Start with B={u1,···un}. Let:
v1=u1
v2=u2−u2·v1
v1·v1v1
v3=u3−u3·v1
v1·v1v1−u3·v2
v2·v2v2
Then {v1···vn}is an orthogonal basis for Span(B), and if
wi=vi
kvik, then {w1···wn}is an orthonormal basis for Span(B).
QR-factorization: To find Q, apply G-S to columns of A. Then
R=QTA
Least-squares: To solve Ax=bin the least squares-way, solve
ATAx=ATb.
Least squares solution makes kAx−bksmallest.
ˆ
x=R−1QTb, where A=QR.
Inner product spaces f·g=Rb
af(t)g(t)dt. G-S applies with this inner
product as well.
Cauchy-Schwarz: |u·v|≤kukkvk
Triangle inequality: ku+vk≤kuk+kvk
Symmetric matrices (A=AT)
Has nreal eigenvalues, always diagonalizable, orthogonally
diagonalizable (A=P DP T,Pis an orthogonal matrix, equivalent to
symmetry!).
Theorem: If Ais symmetric, then any two eigenvectors from different
eigenspaces are orthogonal.
How to orthogonally diagonalize: First diagonalize, then apply G-S on
each eigenspace and normalize. Then P= matrix of (orthonormal)
eigenvectors, D= matrix of eigenvalues.
Quadratic forms: To find the matrix, put the x2
i-coefficients on the
diagonal, and evenly distribute the other terms. For example, if the
x1x2−term is 6, then the (1,2)th and (2,1)th entry of Ais 3.
Then orthogonally diagonalize A=P DP T.
Then let y=PTx, then the quadratic form becomes
λ1y2
1+···+λny2
n, where λiare the eigenvalues.
Spectral decomposition: λ1u1u1T+λ2u2u2T+···+λnununT
Second-order and Higher-order differential
equations
Homogeneous solutions: Auxiliary equation: Replace equation by
polynomial, so y000 becomes r3etc. Then find the zeros (use the rational
roots theorem and long division, see the ‘Diagonalization-section).
’Simple zeros’ give you ert, Repeated zeros (multiplicity m) give you
Aert +Btert +···Z tm−1ert, Complex zeros r=a+bi give you
Aeat cos(bt) + Beat sin(bt).
Undetermined coefficients: y(t) = y0(t) + yp(t), where y0solves the
hom. eqn. (equation = 0), and ypis a particular solution. To find yp:
If the inhom. term is C tmert, then:
yp=ts(Amtm···+A1t+ 1)ert , where if ris a root of aux with
multiplicity m, then s=m, and if ris not a root, then s= 0.
If the inhom term is Ctmeat sin(βt), then: yp=ts(Amtm···+
A1t+ 1)eat cos(βt) + ts(Bmtm···+B1t+ 1)er t sin(βt), where
s=m, if a+bi is also a root of aux with multiplicity m(s= 0 if not).
cos always goes with sin and vice-versa, also, you have to look at
a+bi as one entity.
Variation of parameters: First, make sure the leading coefficient
(usually the coeff. of y00) is = 1.. Then y=y0+ypas above. Now
suppose yp(t) = v1(t)y1(t) + v2(t)y2(t), where y1and y2are your
hom. solutions. Then y1y2
y0
1y0
2v0
1
v0
2=0
f(t). Invert the matrix and
solve for v0
1and v0
2, and integrate to get v1and v2, and finally use:
yp(t) = v1(t)y1(t) + v2(t)y2(t).
Useful formulas: a b
c d−1
=1
ad−bc d−b
−c a
Rsec(t) = ln |sec(t)+ tan(t)|,Rtan(t) = ln |sec(t)|,
Rtan2(t) = tan(x)−x,Rln(t) = tln(t)−t
Linear independence: f, g, h are linearly independent if
af(t) + bg (t) + ch(t)=0⇒a=b=c= 0. To show linear
