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Triangular or diagonal matrix
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1/diagonal entries
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Permuted matrix
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P transpose
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Other
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rref ( [A eye()] )
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Way 1
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A*B full multiplication
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Way 2
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[row A]*B
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Way 3
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[col A]*B
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Way 3
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B11*col(A1)+B21*col(A2)
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Find entry 2,3
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[row A2]*[columnB3] = 1 number
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Rank 1 matrix
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[a11*rowB1; a21*rowB1;a31*rowB1] + ...
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Angle
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cos(theta) = (v*w)/(||v||*||w||)
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Outer Product
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[column1]*[1 # #] find numbers that work
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Linear Independent
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Linearly independent if rref(A) ----> #pivots = #row
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Linear transformation (x and y given)
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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.
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formula
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A'*A*xhat=A'*
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Step 1
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rref ( [A'*A A'*b] )
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Step 2
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xhat = last column of rref
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Step 3
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bhat = A*xhat --> bhat is the vector spaned A closest to v and the projection of the vector onto subspace
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Step 4
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be = b - bhat --> be is the vector perpendicular
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Step 4
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error vector/distance = norm (be) (1/sqr of components of b swuares)
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For regression
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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
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Echelon form
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Leading entries in every row are farther to the right than the row above. To do = elimination steps
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Reduced Echelon form (rref)
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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
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Ax=b with LU
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L = identity but a21 = -λ1, a31 = -λ2, a32 = -λ3. U =
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Echelon form
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Leading entries in every row are farther to the right than the row above. To do = elimination steps
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Reduced Echelon form (rref)
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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
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Ax=b with LU
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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)
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Ax=b with CR
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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)
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v
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eigenvector
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λ
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eigenvalue
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Finding λ
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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
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Finding v
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rref ( [A - λ*eye ] ) and find FV, pivots, and ss
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Diagonalization
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A = P*D*P^(-1), where P = [eigenvectors] , D = diag(λ)
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When can we diagonalize*
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Only when: square, real λ, and if repeated λ - look rref ( [A - λ*eye ] ) and only 1 pivot.
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A = Q*D*Q'
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Q = special solutions form rref ( [A - λ*eye ] ) for every λ, and then doing norm(q1) for all of them. D = diag(λs)
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Is λ an eigenvalue
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Do rref ( [A - λ*eye ] ) and has to be only 1 pivot (linearly dependent)
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Positive definite
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λs all positive
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Semipositive definite
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λs all positive and at least a 0
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Indefinite
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λ at least one is negative
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Subspace
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If u and v are in W , then u + v are in W , and cu is in W
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Basis B for V
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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).
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Row(A)
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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
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Col(A)
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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
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Null(A) / Vector in Null
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Solutions of Ax = 0. Row-reduce A. Null(A) = R^n, dim = n-rank, Basis of Null = rref(A), FV, pivots, special solutions
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LeftNull(A)
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Solutions of A'x = 0. Row-reduce A'. LeftNull(A) = R^m, dim = m-rank, Basis of LeftNull = rref(A'), FV, pivots, special solutions
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Rank(A)
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number of pivots
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Is v in Null
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do A*v and it needs to equal to vector 0
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find v in ColA
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same vectors as in matrix
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Is v in col space of B
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is B*x=v consistent? do rref([B v]) and see if consistent
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A
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q1 = A(:,1)
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Q = q1
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xhat =(q1'*A(:,2))/(q1'*q1)
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ahat = Q*xhat
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q2 = A(:,2) - ahat
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Q(:,2) = q2
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Q(:,1) = 1/(q'1*q1)*q1
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Q(:,2) = 1/(q'2*q2)*q2
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Q = [ Q(:,1) Q(:,2) ]
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R = Q'*A
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if 3x3 keep going
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v and u are othogonal
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if v*u = 0
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W⊥:
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Set of v which are orthogonal to every w in W.
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Orthogonal projection:
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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||
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Basis of W⊥:
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basis of Null(Mw)
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Equalities between basis
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(RowA)' = NullA and vice versa. (ColA)'=LeftNullA and vice versa
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