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 rowsof A: Row-reduce Aand choose the rowsthat
contain the pivots.
Col(A): Space spanned by columns of A: Row-reduce Aand choose the columns of A
that 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.
Toshow lin. ind,form the matrix of the vectors, and show that N ul(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
Toshow sthg is a basis, show it is linearly independent and spans.
Tofind a basis from a collection of vectors, form the matrix Aof the vectors, and find
Col(A).
Tofind 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 vectorsform a basis.
Dimension: Number of elements in a basis.
Tofind dim, find a basis and find num. elts.
Theorem: If Vhas a basis of vectors, then everybasis of Vmust have nvectors.
Basis theorem: If Vis an n−dim v.s.,then any lin. ind. set with nelements 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,
evaluateT(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: Tofind [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.
Invertiblematrix theorem: If Ais invertible, then: Ais row-equivalent to I,Ahas n
pivots, 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→hI|A−1i
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=PD P −1for some diagonal Dand
invertibleP.
Aand Bare similar if A=PB P −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’snot invertible (Ex: A=0 0
0 0). Not
all matrices are diagonalizable (Ex: 1 1
0 1)
Consequence: A=PD P −1⇒An=P DnP−1
How to diagonalize: Tofind the eigenvalues, calculate det(A−λI), and find the
roots of that.
Tofind the eigenvectors, for each λfind a basis for N ul(A−λI), which you do by
row-reducing
Rational roots theorem: If p(λ)=0hasa rational root r=a
b, then adivides
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=PD P −1, where D= diagonal matrix of eigenvalues, P= matrix of
eigenvectors
Complex eigenvaluesIf λ=a+bi, and vis an eigenvector, then
A=PC P −1, where P=Re(v)Im(v),C=a b
−b a
Cis a scaling of pdet(A)followed by a rotation by θ, where:
1
pdet(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· · · ukis 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
vi
,
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: Tosolve 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=PD P T,Pis an orthogonal matrix, equivalent to symmetry!).
Theorem: If Ais symmetric, then any two eigenvectorsfrom 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: Tofind 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=PD P 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 +· · · Ztm−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 Ctmert , 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)ert sin(β t),
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.
Variationof 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
2"v0
1
v0
2#=0
f(t). Invert the matrix and solve for v0
1
and 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
dependence, do it directly. Toshow linear independence, form the Wronskian:
f
W(t) = f(t)g(t)
f0(t)g0(t)(for 2functions),
f
W(t) =
f(t)g(t)h(t)
f0(t)g0(t)h0(t)
f00(t)g00(t)h00 (t)
(for 3functions). Then pick a point
t0where det(f
W(t0)) is easy to evaluate. If det 6= 0, then f, g, h are linearly
independent! Try to look for simplifications before you differentiate.
Fundamental solution set: If f, g, h are solutions and linearly independent.
Largest interval of existence:First make sure the leading coefficient equals to 1. Then
look at the domain of each term. For each domain, consider the part of the interval which
contains the initial condition. Finally, intersect the intervals and change any bracketsto
parentheses. Harmonic oscillator: my00 +by0+ky = 0 (m=inertia, b=
damping, k=stiffness)
Systems of differential equations
Tosolve x0=Ax:x(t) = Aeλ1tv1+B eλ2tv2+eλ3tv3(λi
are your eigenvalues,viare your eigenvectors)
Fundamental matrix: Matrix whose columns are the solutions, without the constants
(the columns are solutions and linearly independent)
Complex eigenvaluesIf λ=α+iβ , and v=a+ib. Then:
x(t) = Aeαt cos(βt)a−eαt sin(βt)b+
Beαt sin(βt)a+eαt cos(βt)b
Notes: Youonly need to consider one complex eigenvalue. For real eigenvalues, use the
formula above. Also, 1
a+bi =a−bi
a2+b2
Generalized eigenvectorsIf you only find one eigenvector v(even though there are
supposed to be 2), then solve the following equation for u:(A−λI)(u) = v
(one solution is enough).
Then: x(t) = Aeλtv+Bteλtv+eλt u
