Review• {v1,, vp} is a basis of V if the vectors span V and are independent.• To obtain a basis for Nul(A), solve Ax = 0:3 6 6 36 12 15 0>RR E F1 2 0 50 0 1 −2x =−2x2− 5x4x22x4x4= x2−2100+ x4−5021Hence,−2100,−5021form a basis for Nul(A).• To obtain a basis for Col(A), take th e pivot col umns of A.1 2 0 42 4 −1 33 6 2 224 8 0 16>1 2 0 40 0 −1 −50 0 0 00 0 0 0Hence,1234,0−120form a basis for Col(A).• Row operations do not preserve the c olumn space.For instance,10>R1↔R201.• On th e other hand: row operations do preserve the nu ll space.Why? Recall why/tha t we can operate on rows to solve systems likeAx = 0!Dimension of Col(A) and Nul(A)Definition 1. T he ra nk of a matrix A is the num ber of its piv ot s.Theorem 2. Let A be an m × n matrix of rank r. Then:• dim Col(A) = rWhy? A basis for Col(A) is given by the pivot columns of A.• dim Nul(A) = n − r is the num ber of free v ariables of AWhy? In our recipe for a basis for Nul(A), each free variable corresponds to an element in thebasis.• dim Col(A) + dim Nul(A) = nWhy? Each of the n columns either contains a pivot or corresponds to a free variable.Armin [email protected] four fundamental subspacesRow space and left null spaceDefinition 3.•The row space of A is the column space of AT.Col(AT) is spanned by the colu mns of ATand these are the rows o f A.• The left null space of A is the null space of AT.Why “left”? A vector x is in Nul(AT) if a nd on ly if ATx = 0.Note thatATx = 0(ATx)T= xTA = 0T.Hence, x is in Nul(AT) if a nd only if xTA = 0.Example 4. Find a basis for Col(A) and Col(AT) whereA =1 2 0 42 4 −1 33 6 2 224 8 0 16.Solution. We k n ow wha t to do for Col(A) from an echelon form of A, and we couldlikewise handleCol(AT) from an echelon form of AT.But wait!Instead of doing twice the work, we only need an echelon form ofA:1 2 0 42 4 −1 33 6 2 224 8 0 16 1 2 0 40 0 −1 −50 0 2 100 0 0 0Armin [email protected]1 2 0 40 0 −1 −50 0 0 00 0 0 0= BHence, the rank of A is 2.A basis forCol(A) is1234,0−120.Recall that Col(A)Col(B). That’s because we performed row operations.However, the row spaces are the same!Col(AT) = Col(BT)The row space is preserved by elementary row operations.In particular: a basis for Col(AT) is given by1204,00−1−5.Theorem 5. (Fundamental Theorem of Linear Algebra, Part I)Let A be an m × n matr i x of rank r.• dim C o l(A) = r (subspace of Rm)• dim C o l(AT) = r (subspace of Rn)• dim N u l(A) = n − r (subspace of Rn) (# of free variables of A)• dim N u l(AT) = m − r (subspace of Rm)In particular:The co l umn and row space always have th e same d imension!In ot her words ,A and AThave the same rank. [i.e. sam e num b er of pivots]Easy to see for a matrix in echelon form2 1 3 00 0 1 20 0 0 7,but not obvious for a rando m matrix.Armin [email protected] transformationsThroughout,V and W are vector spaces.Definition 6. A map T : V → W is a linear transformation ifT (cx + dy) = cT (x) + dT (y)for all x, y in V and all c, d in R.In other words, a linear transformation respects addi tion and scaling:• T (x + y) = T (x) + T (y)• T (cx) = cT (x)It also sends the zero vector in V to the zero vector in W :• T (0) = 0 [because T (0) = T (0 · 0) = 0 · T (0) = 0]Example 7. Let A be an m × n ma tr ix.Then the mapT (x) = Ax is a linear transformation T : Rn→ Rm.Why?Be cause matr ix multiplication is linear:A(cx + dy) = cAx + dAyThe LHS is T (cx + dy) and t he RHS is cT (x) + dT (y).Example 8. Let Pnbe the vector space of all polynomials of degree at most n. Con siderthe mapT : Pn→ Pn− 1given byT (p(t)) =ddtp(t).This map is linear! Why?Be cause differentiation is lin ear:ddt[ap(t) + bq(t)] = addtp(t) + bddtq(t)The LHS is T (ap(t) + bq(t)) an d the RHS is aT (p(t)) + bT (q(t)).Representing linear map s by matricesLet x1,, xnbe a basis for V .A linear mapT : V → W is dete rmined by the values T (x1),, T (xn).Armin [email protected]?Take anyv in V .It can be written asv = c1x1++ cnxnbecause {x1,, xn} is a basis and hence spans V .Hence, by the linearity ofT ,T (v) = T (c1x1++ cnx) = c1T (x1) ++ cnT (xn).Definition 9. (From linear maps to matrices)Let x1,, xnbe a basis for V , and y1,, yma basis for W .The matrix representingT with respect to these bases• has n colum ns (one for each of the xj),• the j-th column has m entri e s a1,j,, am,jdetermined byT (xj) = a1, jy1++ am,jym.Example 10. Let V = R2and W = R3. Let T be the linear map such thatT10=123, T01=407.What is the matrix A re presenting T with respect to th e standard bases?Solution. The standard bases are10x1,01x2for R2,100y1,010y2,001y3for R3.T (x1) =123= 1100+ 2010+ 3001= 1y1+ 2y2+ 3y3A =1 ∗2 ∗3 ∗T (x2) =407= 4y1+ 0y2+ 7y3A =1 42 03 7Armin [email protected](We did not have time yet to discuss the next example in class, but i t will be helpful ifyour discussion section already meets Tuesdays.)Example 11. As in the previous exam ple, l e t V = R2and W = R3. Let T be the (same)linear map such thatT10=123, T01=407.What is the matrix B representing T with respect to the following bases ?11x1,−12x2for R2,111y1,010y2,001y3for R3.Solution. This time :T (x1) = T11= T10+ T01=123+407=5210= 5111− 3010+ 5001can you see it?otherw ise: d o it!B =5 ∗−3 ∗5 ∗T (x2) = T−12= −T10+ 2T01= −123+ 2407=7−211= 7111− 9010+ 4001B =5 7−3 −95 4Tedious, eve n in t his simpl e example! (But we can certainly do it.)A
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