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 onetoone 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: Rowreduce 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: Rowreduce 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. Rowreduce A. Null(A) = R^n, dim = nrank, Basis of Null = rref(A), FV, pivots, special solutions 
LeftNull(A) 
Solutions of A'x = 0. Rowreduce A'. LeftNull(A) = R^m, dim = mrank, 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 
GramSchmidt 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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