2013-04-24, 11:24 PM
So, test on a few chapters tomorrow. Going to kind of use this as a checklist for what I know and what the topics are.
Coordinate systems - good
[x]B
x = P_B[x]_B (and therefore) P^-1_B[x]=[x]_B
p(t) = a0 + a1*t + a2*t^2 + ... + an*t^2 then R^n+1 = [a0;...;an]
Dimension of a vector space - good
Find basis. If no basis (no free variables), dim = 0.
dim Col A + dim Nul A = n (n is number of colums)
Nul A found by : reduced echelon form equal to zero. Use equations from pivot variables.
Rank - good
rank A + dim Nul A = n
rank A = dim Col A
For row equivalent matrices : Use echelon form to find rank A, basis for Col A uses the original matrix.
Basis for Row A uses echelon (NOT reduced). Just rows with pivots.
Basis Nul A found same as before.
Change of Basis - okay. straightforward but the language confuses me.
P = [ [d1]A [d2]A [d3]A ] then [x]A = P[x]D
/*if D = {d1,d2,d3} and F = {f1,f2,f3} and f1 = 2d1 - d2 + d3, f2 = 3d2 + d3, f3 = -3d1 + 2d3
P(D<-F) is just [d1 d2 d3]
Finding [x]D for x = f1 - 2f2 + 2f2 is simply P(D<-F)(dot)x because the coefficients of x create a vector.*/
[c1 c2 | b1 b2] ~ [I | P(C<-B)]
P(B<-C) = P(C<-B)^-1
Applications to Markov chains - meh fractions suck
state vectors
word problems
columns add to 1
xk+1 = Pxk
Steady-State vector : (P - I)x = 0
Probability vector: divide steady state by sum of its entries.
Eigenvectors and Eigenvalues - sehr gut
A-λIx = 0
if above is linearly dependent (free variables) then λ is an eigenvalue. Corresponding eigenvector solve and take vector, you know what I mean. Same stuff you've been doing.
Basically a bunch of preliminary stuff for sections coming up
Characteristic equation - good
solve det(A-λI) = 0 to find the characteristic equation for A; shows all possible values of λ. Expand for polynomial.
Let A be an n x n matrix. A is invertible iff:
a. The number 0 is not an eigenvalue of A
b. the determinant of A is not zero.
detAB = detA*detB
detA^T=detA
if A is triangular, determinant is the diagonal
row replacement doesn't change det A. row switching does. (-1)^r where r is number of row switches.
Diagonalization - good
If A = PDP^-1 then A^k = PD^kP^-1
Diagonalizing an n x n matrix A:
1. Find eigenvalues
2. Find n linearly independent eigenvectors of A
3. Construct P from vectors in step 2
4. Construct D from the corresponding eigenvalues
(check: AP = PD)
Eigenvectors and Linear Transformations - wat
mapping sounds confusing but is easy
D = {d1, d1} and B = {b1,b2} are bases for vector spaces V and W
let T : V -> W be a linear transformation with T(d1) = 3b1 - 3b2, T(d2) = -2b1 + 5b2
Matrix for T relative to D and B is simply [3 -3;-2 5]
Discrete Dynamical Systems - semi good
for λ < 0, as k -> ∞, λ^k -> 0
origin is an attractor when all λ < 1
origin is a repeller when all λ > 1
origin is saddle point when some λ < 1 and some λ > 1
probably gonna yolo questions on this
Inner product, Length, and Orthogonality - good
basically calc III with matrices
dot product : multiply and add. gives scalar
unit vector in direction of given vector : divide everything by length
Distance between two vectors : subtract them and find distance of that vector
Orthogonal vectors : dot product = 0
two vectors are orthogonal iff : ||U + V||^2 = ||u||^2 + ||v||^2
ok I think that's everything I guess if anyone feels like reading through all of it and wants to comment on something wrong or just whatever then go ahead but this is mostly for me and yes this is a run-on sentence dealwithit
Coordinate systems - good
[x]B
x = P_B[x]_B (and therefore) P^-1_B[x]=[x]_B
p(t) = a0 + a1*t + a2*t^2 + ... + an*t^2 then R^n+1 = [a0;...;an]
Dimension of a vector space - good
Find basis. If no basis (no free variables), dim = 0.
dim Col A + dim Nul A = n (n is number of colums)
Nul A found by : reduced echelon form equal to zero. Use equations from pivot variables.
Rank - good
rank A + dim Nul A = n
rank A = dim Col A
For row equivalent matrices : Use echelon form to find rank A, basis for Col A uses the original matrix.
Basis for Row A uses echelon (NOT reduced). Just rows with pivots.
Basis Nul A found same as before.
Change of Basis - okay. straightforward but the language confuses me.
P = [ [d1]A [d2]A [d3]A ] then [x]A = P[x]D
/*if D = {d1,d2,d3} and F = {f1,f2,f3} and f1 = 2d1 - d2 + d3, f2 = 3d2 + d3, f3 = -3d1 + 2d3
P(D<-F) is just [d1 d2 d3]
Finding [x]D for x = f1 - 2f2 + 2f2 is simply P(D<-F)(dot)x because the coefficients of x create a vector.*/
[c1 c2 | b1 b2] ~ [I | P(C<-B)]
P(B<-C) = P(C<-B)^-1
Applications to Markov chains - meh fractions suck
state vectors
word problems
columns add to 1
xk+1 = Pxk
Steady-State vector : (P - I)x = 0
Probability vector: divide steady state by sum of its entries.
Eigenvectors and Eigenvalues - sehr gut
A-λIx = 0
if above is linearly dependent (free variables) then λ is an eigenvalue. Corresponding eigenvector solve and take vector, you know what I mean. Same stuff you've been doing.
Basically a bunch of preliminary stuff for sections coming up
Characteristic equation - good
solve det(A-λI) = 0 to find the characteristic equation for A; shows all possible values of λ. Expand for polynomial.
Let A be an n x n matrix. A is invertible iff:
a. The number 0 is not an eigenvalue of A
b. the determinant of A is not zero.
detAB = detA*detB
detA^T=detA
if A is triangular, determinant is the diagonal
row replacement doesn't change det A. row switching does. (-1)^r where r is number of row switches.
Diagonalization - good
If A = PDP^-1 then A^k = PD^kP^-1
Diagonalizing an n x n matrix A:
1. Find eigenvalues
2. Find n linearly independent eigenvectors of A
3. Construct P from vectors in step 2
4. Construct D from the corresponding eigenvalues
(check: AP = PD)
Eigenvectors and Linear Transformations - wat
mapping sounds confusing but is easy
D = {d1, d1} and B = {b1,b2} are bases for vector spaces V and W
let T : V -> W be a linear transformation with T(d1) = 3b1 - 3b2, T(d2) = -2b1 + 5b2
Matrix for T relative to D and B is simply [3 -3;-2 5]
Discrete Dynamical Systems - semi good
for λ < 0, as k -> ∞, λ^k -> 0
origin is an attractor when all λ < 1
origin is a repeller when all λ > 1
origin is saddle point when some λ < 1 and some λ > 1
probably gonna yolo questions on this
Inner product, Length, and Orthogonality - good
basically calc III with matrices
dot product : multiply and add. gives scalar
unit vector in direction of given vector : divide everything by length
Distance between two vectors : subtract them and find distance of that vector
Orthogonal vectors : dot product = 0
two vectors are orthogonal iff : ||U + V||^2 = ||u||^2 + ||v||^2
ok I think that's everything I guess if anyone feels like reading through all of it and wants to comment on something wrong or just whatever then go ahead but this is mostly for me and yes this is a run-on sentence dealwithit

