Once more, we review questions from a previous exam to prepare ourselves for an upcoming exam.

Transcription

1 Final course review Once more, we review questions from a previous exam to prepare ourselves for an upcoming exam.. Suppose we know that A is an m by n matrix of rank r, Ax = has no solution, and Ax = has exactly one solution. a) Whatcan we say about m, n and r? Theproduct Ax is a vector in threedimensions, so m =. The fact that Ax = has no solution tells us that the column space is not all of R. In addition, we know that the column space contains, so r is not zero: r <. The fact that Ax = has exactly one solution tells us that the nullspace of A contains only the zero vector and so n = r. Hence n <. b) Writedownanexampleof amatrix A that fits this description. The vector must be in the column space, so we ll make it a column of A. The simplest way to answer this question is to stop here. A =. In this solution, n = r = and m =. To find a [ solution ] in which n = r =, add a second column. Make sure that is not in the column space: A =. There are many other correct answers to this question.

2 c) Crossoutall statementsthat arefalseabout any matrixwith thegiven properties(whichare r = n, m = ). i. deta T A = detaa T ii. A T A is invertible iii. AA T is positive definite Onegoodapproachto thisproblemisto use our sample matrixto test each statement. i. Ifweleavethisparttolast, wecanquickly answerit(false)using what we learn while answering the following twoparts. ii. The matrix A T A is invertible if r = n; i.e. if the columns of A are independent. The nullspace ofour A contains only the zero vector, so this statement is true. For each of our sample matrices, A T A equals the identity and so is invertible. Note that this means deta T A =. iii. We know that m = and r <, so AA T will be a by matrix with rank less than ; it can t be positive definite. (It is true that for any matrix A with real valued entries, AA T is positive semidefinte.) For our test matrices, AA T has atleastone row that s all zeros, so is an eigenvalue(andis not positive). Note also that detaa T = and so statement (i) must be false. (However,if A and B aresquarematricesthendet BA = detab = deta detb.) d) Prove that A T y = c has atleast one solution for every right hand side c, andin facthas infinitely many solutions for every c. We know A T is an n by m matrix with m = and rank r = n < m. If A T has full row rank, the equation A T y = c is always solvable. We have n rows and rank r = n, so A T has full row rank. Therefore A T y = c has a solution for everyvector c. The solvable system A T y = c will have infinitely many solutions if the nullspace of A T haspositivedimension. Weknowdim(N(A T )) = m r >, so A T y = c has infinitely many solutions for every c.. Suppose the columns of A are v, v and v. a) Solve Ax = v v + v. This is just the column method of multiplying matrices from the first lecture. Choose x =.

3 b) True or false: if v v + v =, then the solution to (a) is not unique. Explainyour answer. True. Any scalar multiple of x = will be a solution. Another wayof answeringthisisto notethat A T has a nontrivialnullspace, andwe can always add any vectorinthe nullspaceto a solution x toget adifferent solution. c) Supposev, v and v are orthonormal(forget about(b)). What combination of v and v is closest to v? If we imagine the right triangle out from the origin formed by av + bv and v, thepythagorean theorem tells us thatv + v = is the closest point to v in theplane spannedby v and v.. Suppose we have the Markov matrix... A = Note that the sum of the first two columns of A equals twice the third column of A. a) Whatare the eigenvalues of A? Zero is an eigenvalue because the columns of A are dependent. (A is singular.) One is an eigenvalue because A is amarkov matrix. The third eigenvalue is. because the trace of A is.8. So λ =,,.. b) Let u k = A k u(). If u() =,what is lim k u k? We llstartby computing u k andthen findthesteady state. Thismeans finding a general expression of the form: u k = c λ k x + c λ k x + c λ k x. When we plug in the eigenvalues we found inpart(a),this becomes u k = + c x + c (.) k x. We see that as k approaches infinity, c x is the only term that does not go to zero. The key eigenvector in any Markov process is the one with eigenvalue one.

4 To find x, solve (A I)x = : x =....6 Thebestwayto solvethis mightbeby elimination. However,because the first two columns look like multiples of and the third column looks like a multiple of, we might get lucky and guess x =. This gives us u = c. We know that in a Markov process, the sum of the entries of u k is the samefor all k. The sum of the entries of u() is, so c = and u =.. Find a twoby twomatrix that: a) projectsonto the line spanned by a =. aa T Theformula for this matrixis P =. This gives us ata 6/5 /5 P =. /5 9/5 (To test this answer, we canquickly check that detp =.) b) has eigenvalues λ = and λ = and eigenvectors x = and x =. Herethe formula we needis A = SΛS. A = / / = / / A =. Iftime permits, we can check this by computing the products Ax i. c) hasrealentries and cannot be factored as B T B for any B. We know that B T B will always be symmetric, so any [ asymmetric ] ma trixhas this property. For example,we could choose.

5 d) is not symmetric, but has orthogonal eigenvectors. We know that symmetric matrices have orthogonal eigenvectors, but so do other types of matrices (e.g. skew symmetric and orthogonal) when we allow complex eignevectors. Two possible answers are: (skew symmetric) cos θ sin θ (orthogonal). sin θ cos θ 5. Applying the least squaresmethod tothe system c = = b d ĉ / gives the best fit vector =. dˆ a) Whatis the projection p of b = onto the column space of A =? We know that / times the first column minus times [ the ] second column is the closest point P in the column space to, so the answeris / ĉ A = = 8/. dˆ 5/ b) Drawthe straight line problemthatcorresponds tothis system. Plotting the entries of the second column of A against the entries of b weget the threepoints shown infigure. The best fit line is ĉ+ dt ˆ. c) Findadifferent vector b = R sothattheleast squaressolution is ĉ =. dˆ ĉ Weknowthat istheprojectionofbontothe column space, soto dˆ getazeroprojectionweneedto find avector orthogonal to the columns. 5

6 P (, ) (, ) P b = t P (, ) Figure: Threedatapoints and their bestfit line t. We could get the answer b = by inspection, or we could use the crossproduct of the columns to find avaluefor b. Thank you for taking this course! 6

7 MIT OpenCourseWare 8.6SC Linear Algebra Fall For information about citing these materials or our Terms of Use, visit: