Math 54 - Cheat Sheet - Math 54 Cheat Sheet Vector spaces...

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Math 54 Cheat SheetVector spacesSubspace:Ifuandvare inW, thenu+vare inW, andcuis inWNul(A):Solutions ofAx=0. Row-reduceA.Row(A):Space spanned by the rows ofA: Row-reduceAand choose the rows thatcontain the pivots.Col(A):Space spanned by columns ofA: Row-reduceAand choose the columns ofAthat contain the pivotsRank(A):=Dim(Col(A)) =number of pivotsRank-Nullity theorem:Rank(A) +dim(Nul(A)) =nLinear transformation:T(u+v) =T(u) +T(v),T(cu) =cT(u),wherecis a constant.Tis one-to-one ifT(u) =0)u=0Tis onto ifCol(T) =Rm.Linearly independence:a1v1+a2v2+· · ·+anvn=0)a1=a2=· · ·=an= 0.To show lin. ind, form the matrix of the vectors, and show thatNul(A) ={0}Linear dependence:a1v1+a2v2+· · ·+anvn=0fora1, a2,· · ·, an, not all zero.Span:Set of linear combinations ofv1,· · ·vnBasisBforV:A linearly independent set such thatSpanB=VTo show sthg is a basis, show it is linearly independent and spans.To find a basis from a collection of vectors, form the matrixAof the vectors, and findCol(A).To find a basis for a vector space, take any element of that v.s. and write it as a linearcombination of ’simpler’ vectors. Then show those vectors form a basis.Dimension:Number of elements in a basis.To finddim, find a basis and find num. elts.Theorem:IfVhas a basis of vectors, then every basis ofVmust havenvectors.Basis theorem:IfVis ann-dim v.s., then any lin. ind. set withnelements is a basis,and any set ofnelts. which spansVis a basis.Matrix of a lin. transfTwith respect to basesBandC:For every vectorvinB,evaluateT(v), and express this as a linear combination of vectors inC. Put thecoefficients in a column vector, and then form the matrix of the column vectors youfound!Coordinates:To find[x]B, expressxin terms of the vectors inB.x=PB[x]B, wherePBis the matrix whole columns are the vectors inB.Invertible matrix theorem:IfAis invertible, then:Ais row-equivalent toI,Ahasnpivots,T(x) =Axis one-to-one and onto,Ax=bhas a solution for everyb,ATis invertible,det(A)6= 0, the columns ofAform a basis forRn,Nul(A) ={0},Rank(A) =nabcd-1=1ad-bcd-b-caA|I!hI|A-1iChange of basis:[x]C=PCB[x]B(think ofCas the new, cool basis)[C | B]!I|PCBPCBis the matrix whose columns are[b]C, wherebis inBDiagonalizationDiagonalizability:AisdiagonalizableifA=P DP-1for some diagonalDandinvertibleP.AandBare similar ifA=P BP-1forPinvertibleTheorem:Ais diagonalizable,Ahasnlinearly independenteigenvectorsTheorem:IFAhasndistinct eigenvalues,THENAis diagonalizable, but the oppositeis not always true!!!!Notes:Acan be diagonalizable even if it’s not invertible (Ex:A=O). Not allmatrices are diagonalizable (Ex:1101)Consequence:A=P DP-1)An=P DnP-1How to diagonalize:To find the eigenvalues, calculatedet(A-λI), and find theroots of that.

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Term
Spring
Professor
Chorin

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