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linear algrebra Cheat Sheet by

linear algebra MTH 301

Inverse of a matrix

Triangular or diagonal matrix
1/diagonal entries
Permuted matrix
P transpose
Other
rref ( [A eye()] )

Multip­lic­ation of Matrix + angle

Way 1
A*B full multip­lic­ation
Way 2
[row A]*B
Way 3
[col A]*B
Way 3
B11*co­l(A­1)+­B21­*co­l(A2)
Find entry 2,3
[row A2]*[c­olu­mnB3] = 1 number
Rank 1 matrix
[a11*r­owB1; a21*ro­wB1­;a3­1*r­owB1] + ...
Angle
cos(theta) = (v*w)/(||v||*||w||)
Outer Product
[colum­n1]*[1 # #] find numbers that work

Linear Transf­orm­ation and dependency

Linear Indepe­ndent
Linearly indepe­ndent if rref(A) ----> #pivots = #row
Linear transf­orm­ation (x and y given)
T (u + v) = T (u) + T (v), T (cu) = cT (u), where c is a number. T is one-to-one if T(u)=0⇒u=0 T is onto if Col(T) = Rm.

Projec­tions or Ax=b is incons­istent

formula
A'*A*x­hat=A'*
Step 1
rref ( [A'*A A'*b] )
Step 2
xhat = last column of rref
Step 3
bhat = A*xhat --> bhat is the vector spaned A closest to v and the projection of the vector onto subspace
Step 4
be = b - bhat --> be is the vector perpen­dicular
Step 4
error vector­/di­stance = norm (be) (1/sqr of components of b swuares)
For regression
step 1: f(x) = [x][b], step 2: A = [x.^0 ...] and y = given, step 3: do LSE and find xhat which will be a,b,c
 

Ax = b

Echelon form
Leading entries in every row are farther to the right than the row above. To do = elimin­ation steps
Reduced Echelon form (rref)
echelon + columns of leading entries are all 0 except the entry which must be a 1. To do = elimin­ations steps down to right, then left to top
Ax=b with LU
L = identity but a21 = -λ1, a31 = -λ2, a32 = -λ3. U =

Ax = b (A and b specified)

Echelon form
Leading entries in every row are farther to the right than the row above. To do = elimin­ation steps
Reduced Echelon form (rref)
echelon + columns of leading entries are all 0 except the entry which must be a 1. To do = elimin­ations steps down to right, then left to top
Ax=b with LU
L = identity but a21 = -λ1, a31 = -λ2, a32 = -λ3. U = echelon. Then do Ly=b - given (solve for y), then Ux=y (solve for x)
Ax=b with CR
tMaybe not full rank. C = columns of A that have a pivot in R. R = rref form. To find x --> using R to find FV, pivots, and special solutions (if b not 0 do rref([A b])), if one soln is given then add that in gen sol and just do rref(A)

Eigenv­ectors and Eigenv­alues

v
eigenv­ector
λ
eigenvalue
Finding λ
1. Diag or triang = entries of diag. 2. 2x2 do λ = m +- sqrt (m^2 - p), where m = (a11+a­22)/2, and p = a11*a22 - a12*a21
Finding v
rref ( [A - λ*eye ] ) and find FV, pivots, and ss
Diagon­ali­zation
A = P*D*P^­(-1), where P = [eigen­vec­tors] , D = diag(λ)
When can we diagon­alize*
Only when: square, real λ, and if repeated λ - look rref ( [A - λ*eye ] ) and only 1 pivot.
A = Q*D*Q'
Q = special solutions form rref ( [A - λ*eye ] ) for every λ, and then doing norm(q1) for all of them. D = diag(λs)
Is λ an eigenvalue
Do rref ( [A - λ*eye ] ) and has to be only 1 pivot (linearly dependent)
Positive definite
λs all positive
Semipo­sitive definite
λs all positive and at least a 0
Indefinite
λ at least one is negative
 

Vector Spaces and Basis

Subspace
If u and v are in W , then u + v are in W , and cu is in W
Basis B for V
A linearly indepe­ndent set such that Span (B) = V To show sthg is a basis, show it is linearly indepe­ndent (rref(A) has NO FV) and spans(no row of 0's).
Row(A)
Space spanned by the rows of A: Row-reduce A and choose the rows that contain the pivots. Row(A) = R^n, dim = rank, Basis of Row = R in A = CR
Col(A)
Space spanned by columns of A: Row-reduce A and choose the columns of A that contain the pivots. Col(A) = R^m, dim = rank, Basis of Col = C in A = CR
Null(A) / Vector in Null
Solutions of Ax = 0. Row-reduce A. Null(A) = R^n, dim = n-rank, Basis of Null = rref(A), FV, pivots, special solutions
LeftNu­ll(A)
Solutions of A'x = 0. Row-reduce A'. LeftNu­ll(A) = R^m, dim = m-rank, Basis of LeftNull = rref(A'), FV, pivots, special solutions
Rank(A)
number of pivots
Is v in Null
do A*v and it needs to equal to vector 0
find v in ColA
same vectors as in matrix
Is v in col space of B
is B*x=v consis­tent? do rref([B v]) and see if consistent

Gram-S­chmidt steps

A
q1 = A(:,1)
Q = q1
xhat =(q1'*­A(:­,2)­)/(­q1'*q1)
ahat = Q*xhat
q2 = A(:,2) - ahat
Q(:,2) = q2
Q(:,1) = 1/(q'1­*q1)*q1
Q(:,2) = 1/(q'2­*q2)*q2
Q = [ Q(:,1) Q(:,2) ]
R = Q'*A
if 3x3 keep going

Orthog­onality

v and u are othogonal
if v*u = 0
W⊥:
Set of v which are orthogonal to every w in W.
Orthogonal projec­tion:
If {u1 · · · uk } is a basis for W , then orthogonal projection of y on W is: yˆ=(y·­u1/­u1*­u1)­+··­·+(­y·u­1/u­k*uk), and y − yˆ is orthogonal to yˆ, shortest distance btw y and W is ||y−yˆ||
Basis of W⊥:
basis of Null(Mw)
Equalities between basis
(RowA)' = NullA and vice versa. (ColA)­'=L­eft­NullA and vice versa
 

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  • Published: 14th December, 2022

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