1. WORKING WITH BASIC DEFINITIONS Assume: {x1, ..., xn} is a basis for X, T:X->Y is a linear transformation. Consider any y=Tx in the image of T. This x, a vector in X, can be written as some linear combination a1*x1+...+an*xn. Then y=T(a1*x1+...+an*xn)=a1*Tx1+...+an*Txn, so y can be written as a linear combination of {Tx1,...,Txn}. This proves any vector in the image of T is in the span of {Tx1,...,Txn}. The set {Tx1,..,Txn} is not necessarily a basis for the image of T, because it might not be linearly independent. For example, any time dim(Y) [ 0 0 0 4 3 ] -> [ 0 0 0 4 3 ] = U [ 0 4 -6 6 7 ] [ 0 0 0 4 3 ] [ 0 0 0 0 0 ] The answers below aren't the only correct answers. Basis for NS(A): Columns 1, 3, and 5 don't have pivots, so they correspond to free variables. So there ought to be three vectors in a basis for NS(U) (=NS(A)), and you can choose them so that =<1,0,0>, <0,1,0>, <0,0,1>. Along with the equations (2)x2+(-3)x3+(1)x4+(2)x5=0, (4)x4+(3)x5=0, these choices allow you to solve "Ux=0" by back-substitution. Answer: {<1, 0, 0, 0, 0>, <0, 3/2, 1, 0, 0>, <0, -5/8, 0, -3/4, 1>} Basis for RS(A): RS(A)=RS(U), and the non-zero rows of U are linearly independent. Read off the basis {<0, 2, -3, 1, 2>, <0, 0, 0, 4, 3>}. Basis for CS(A): CS(A) is NOT CS(U), but the columns of A which have pivots in U form a basis for CS(A). So read off columns 2 and 4: get the basis {<2, -2, 4>, <1, 3, 6>}. rk(A) = dim RS(A) = dim CS(A) = #columns(A)-dim NS(A) = 2. 3. BASES FOR THE SPAN OF A COLLECTION OF VECTORS (a) Put the vectors into a matrix as rows, and find a basis for RS as above. [ -2 4 1 2 ] [ -2 4 1 2 ] [ -2 4 1 2 ] [ 4 2 3 -1 ] -> [ 0 10 5 3 ] -> [ 0 10 5 3 ] [ 2 6 4 1 ] [ 0 10 5 3 ] [ 0 0 0 0 ] => {<-2, 4, 1, 2>, <0, 10, 5, 3>} (b) Use the Gram-Schmidt method, as given at the end of Hill sec. 4.4. w1 = <1,1,1> w2 = <1,2,2> - [(<1,2,2> . <1,1,1>)/(<1,1,1> . <1,1,1>)] <1,1,1> = <-2/3,1/3,1/3>. Actually, it's fine to use w2=<-2,1,1> since we normalize later. w3 = <1,0,1> - [(<1,0,1> . <1,1,1>)/(<1,1,1> . <1,1,1>)] <1,1,1> - [(<1,0,1> . <-2,1,1>)/(<-2,1,1> . <-2,1,1>)] <-2,1,1> = <1,0,1> - <2/3, 2/3, 2/3> - <1/3, -1/6, -1/6> = <0, -1/2, 1/2>. Again, it's fine to use w3=<0, -1, 1>. Now normalize to get {(1/sqrt(3))<1,1,1>, (1/sqrt(6))<-2,1,1>, (1/sqrt(2))<0,-1,1>} (c) Apply the method of (a) to the coordinate vectors (wrt standard basis). [ 2 -3 4 -5 ] [ 2 -3 4 -5 ] [ 2 -3 4 -5 ] [ 4 -1 15 -2 ] -> [ 0 5 7 8 ] -> [ 0 5 7 8 ] [ 2 2 11 3 ] [ 0 5 7 8 ] [ 0 0 0 0 ] So the polynomials with coordinate vectors <2, -3, 4, -5> & <0, 5, 7, 8> form a basis: { 2 - 3x + 4x^2 -5x^3, 5x + 7x^2 + 8x^3 } (d) Use the Gram-Schmidt method, but to find f.g or ||f||=sqrt(f.f), use the formula f.g = integral of f(x)g(x) from -1 to 1. h1(x) = (1 - x) (1+x).(1-x) = ... = 4/3 (1-x).(1-x) = ... = 8/3 h2(x) = (1 + x) - [(4/3)/(8/3)] (1-x) = 1/2 + 3/2 x As in (b), it's fine to scale: let's use h2(x)=1+3x. (1+x+x^2).(1- x) = ... = 2 (1+x+x^2).(1+3x) = ... = 14/3 (1+3x).(1+3x) = ... = 8 h3(x) = (1+x+x^2) - [(2)/(8/3)] (1-x) - [(14/3)/(8)] (1+3x) = -1/3 + x^2 Again, it's fine to use h3(x) = 1-3x^2 (1-3x^2).(1-3x^2) = ... = 8/5 Normalize to get { sqrt(3/8)(1-x), sqrt(1/8)(1+3x), sqrt(5/8)(1-3x^2) } (Simplifying expressions involving sqrt(8) may be a good idea...) 4. CHANGE OF BASIS AND TRANSITION MATRICES (a) Row-reduce [C|B] to get [I|P], then compute [x]_C = P [x]_B. (1) Columns of "B" and "C" are the vectors in B, C. [2 4 | 1 1] -> [ 2 4 | 1 1 ] -> [ 2 0 | 1/2 -1/2 ] -> [3 2 | 1 0] -> [ 0 -4 | -1/2 -3/2 ] -> [ 0 -4 | -1/2 -3/2 ] -> [1 0 | 1/4 -1/4 ] => P = [ 1/4 -1/4 ] [0 1 | 1/8 3/8 ] => [ 1/8 3/8 ] => [x]_C = <-3/4, 5/8> (2) Columns of "B", "C" are coordinate vectors of the members of B, C. Since C is the std. basis, the matrix "C" would be I already. So you can just write down [ 1 1 0 ] => [5] P = [ 1 1 1 ] => [x]_C = [3] [ 1 0 1 ] => [0] (b) (1) [x]_B: solve this linear system for the coefficients [x]_B= (i.e., x=a<1,1>+b<1,-1>). Then find [x]_C using the given P. [ 1 1 | 4 ] -> [ 1 1 | 4 ] => a=3 [ 1 -1 | 2 ] -> [ 0 -2 | -2 ] => b=1 => [x]_B = <3,1> => [x]_C = <1.6, 1.2> (2) The inverse of the matrix below! Use elimination or 2x2 inverse... The determinant is .6*.6+.2*.2=0.4 so P^{-1} is [ 1.5 +0.5 ] [ -0.5 1.5 ] (3) The columns of the transition matrix from (2) are [c1]_B and [c2]_B. You can also figure these out by translating coordinates from [c1]_C=<1,0> and [c2]_C=<0,1>. So c1 = 1.5<1,1>-0.5<1,-1> = <1, 2> and c2 = 0.5<1,1>+1.5<1,-1> = <2,-1> 5. LINEAR TRANSFORMATIONS (a) Domain: R^2 Range : R^2 Linear: Yes [2 0] [0 1] (b) T(x, y, z) = (xy,zx) Domain: R^3 Range : R^2 Linear: No! (c) Domain: P_3 Range : P_2 Linear: Yes [0 1 0 0] [0 0 2 0] [0 0 0 3] (d) T(x, y, z) is "the reflection of (x,y,z) through the xy-plane, rotated by 30 degrees around the z-axis." (In other words, T is the transformation which rotates the xy-plane by 30 degrees and reflects the z-axis.) Domain: R^3 Range : R^3 Linear: Yes [c -s 0] [s c 0] [0 0 -1] where c=cos(30deg)=1/2, s=sin(30deg)=sqrt(3)/2. 6. LEAST SQUARES Solve for [a0; a1] such that y=a0*1 + a1*x is the best-fit line. A = [1, 1; 1, 2; 1, 3; 1, 4; 1, 5] b = [5; 4; 1; 1; -1] [a0; a1] = (A^T A)^-1 (A^T b) = [6.5; -1.5] Of course that line hides a fairly large calculation. You could potentially make it easier by solving for an equation of the form b0(1)+b1(x-3), which would make the columns of A orthogonal... For the quadratic regression, look for equation a0x^0+a1x^1+a2x^2: use A = [1, 1, 1; 1, 2, 4; 1, 3, 9; 1, 4, 16; 1, 5, 25]. [a0; a1; a2] = [7, -27/14; 1/14], approximately [7; -1.93; 0.071] Perpendicular projection problem: Form a matrix A with c1 and c2 as columns: A = [1, 2; 1, 1; -1, 2; -1, 0]. Then P=A(A^T A)^-1 A^T, and p=Pv. Again, this line hides a significant calculation. You can make it much easier! Just ORTHOGONALIZE the vectors first: do the first half of the Gram-Schmidt process, and scale the vectors to avoid fractions if desired. * You won't change the span of the vectors, so you won't change the answer. * You can calculate A^T*A more easily because only the diagonal entries can be nonzero * You can invert A^T*A more easily because it's diagonal! These advantages aren't so pronounced when you only have two vectors. [ 21 14 7 -7 ] [56] 1 [ 14 11 -2 -8 ] 1 [29] P = -- [ 7 -2 29 11 ] , p = -- [52] 35 [ -7 -8 11 9 ] 35 [-2] 7. DETERMINANTS | -1 1 0 1 2 1 0 | | -1 1 0 1 2 1 | | -1 1 1 2 1 | | 4 -4 0 2 1 -1 0 | | 4 -4 0 2 1 -1 | | 4 -4 2 1 -1 | | 1 2 1 1 1 1 1 | ( )= -1 | 2 8 0 0 0 | | 2 8 0 0 0 0 0 | = +1 | 2 8 0 0 0 0 | | 0 -3 0 0 0 | | 0 -3 0 0 0 0 0 | | 0 -3 0 0 0 0 | | 1 5 0 0 1 | | 1 5 0 0 0 1 0 | | 1 5 0 0 0 1 | ( ) | 0 -7 1 2 -2 2 0 | | 0 -7 1 2 -2 2 | | -1 1 2 1 | | 1 2 1 | | 1 2 | | 4 2 1 -1 | | 2 1 -1 | = +6 | 2 1 | = 6(1*1-2*2) = -18 = +3 | 2 0 0 0 | = +6 ( ) ( ) ( ) | 0 0 1 | | 1 0 0 1 | Trace is the sum of the diagonal entries: -3. Finally, the identity det(AB)=det(A)det(B) and the fact that the first matrix in the long product is singular tell you that the last determinant is zero! 8. EIGENVALUE PROBLEMS det(lambda I-A) = (lambda-2)[(lambda+6)(lambda-11)-(-12)(6)] = (lambda-2)(lambda-2)(lambda-3) => eigenvalues are 2 (multiplicity of 2), 3. Use elimination to find NS(lambda I-A). [8 -12 1 ] [8 -12 1 ] => By method from #2, get the basis 2I-A = [6 - 9 0 ] -> [0 0 -3/4] => {<3, 2, 0>} [0 0 0 ] [0 0 0 ] => for the eigenspace W_2 [9 -12 1 ] [9 -12 1 ] [9 -12 1] => Get the basis 3I-A = [6 - 8 0 ] -> [0 0 -2/3] -> [0 0 1] => {<4,3,0>} [0 0 1 ] [0 0 1 ] [0 0 0] => for the eigenspace W_3. Whenever Av=lambda v, (I-A)v=(1-lambda)v, and vice-versa. So I-A should have eigenvalues -1 associated to <3,2,0>, -2 associated to <4,3,0>. Since 0 is not an eigenvalue of A, A is invertible. 9. DIAGONALIZABILITY The matrix from the previous problem doesn't have enough independent eigenvectors to be diagonalizable. 2x2 matrix: det(lambda I-A)=(lambda-2)(lambda-1). The eigenvectors are <1,1> for lambda=2, <0,1> for lambda=1. [2 0] [1 0] [2 0] [1 0]^-1 [1 1] = [1 1] [0 1] [1 1] A S Lambda S ^-1 So A^12 = S Lambda^12 S^-1 = [1 0] [4096 0] [ 1 0] = [4096 0] [1 1] [0 1] [-1 1] = [4095 1] The last matrix is symmetric, so it's possible to diagonalize it as Q Lambda Q^T, where Q is an orthogonal matrix and Lambda is a real diagonal matrix. 10. PROOFS Let A be a matrix satisfying A^2-5A-6=0. If Av=lambda*v, then 0=(A^2-5A-6)v=lambda^2 v - 5lambda v - 6 v. Assuming v is nonzero (an eigenvector is nonzero by definition!) this implies lambda^2-5lambda-6=0. Since (lambda-6)(lambda+1)=0, we can conclude lambda=6 or lambda=-1. If A is an mxn matrix, and x is a vector in R^n, then (A^T(Ax)).x = ((Ax)).(A(x)) = ||Ax||^2. This is never negative; if it is zero then Ax=0; if A is invertible this only happens when x=0. (See Hill, 4.15.)