Linear Algebra Practice Final

Transcription

1 . Let (a) First, Linear Algebra Practice Final Summer 3 3 A = rref([a ) = 5 so if we let x 5 = t, then x 4 = t, x 3 =, x = t, and x = t, so that t t x = t = t t whence ker A = span(,,,, ) and a basis for ker A is β = (b) To find a particular solution, s, we consider rref([a b) = rref = Make it easy on yourself and let x 5 =. Then x = 3, x =, x 3 = and x 4 = 3, so one solution is this one: 3 s = 3 Consequently, 3 K = s + ker A = 3 + t t R

2 (c) Well, one solution is s above, and another may be gotten by picking t =, say, to get 3 s = 3 + = It s easy to verify that these are indeed solutions of the original system.. We know that S is too large, because it contains 5 elements while R 3 [x is a 4-dimensional space. So we will have to reduce it, the question is just by how much. Well, it s easier to deal with matrices, so why don t we represent the vectors in S in the standard basis ρ = {, x, x, x 3 } for R 3 [x, and put the resulting column vectors in a matrix: A = [[ + x + x + 3x 3 ρ [4 + x + 4x + 6x 3 ρ [6 + 3x + 8x + x 3 ρ [ + x + 5x 3 ρ [4 + x + 9x 3 ρ = Row-reducing A gives rref(a) = This tells us that the first three columns and the last are linearly independent, so that the first three and the last polynomials in S are linearly independent. These will form a basis for span S, since they are linearly independent and span S: β = { + x + x + 3x 3, 4 + x + 4x + 6x 3, 6 + 3x + 8x + x 3, 4 + x + 9x 3 } 3. The problem with A is that c A (x) = det(a xi ) = det eigenvalue, and it has algebraic multiplicity, whereas [ x x, so λ = is the only dim(e ) = dim(ker(a I)) = dim(ker(a)) = [ x The problem with B is that c B (x) = det(b xi ) = det x + doesn t factor. That alone is enough, but you ll also note that B consequently doesn t even have eigenvalues, and so doesn t have eigenvectors, which are needed in any basis that will diagonalize it. 4. Let Then A = 3 4 c A (x) = det(a xi 3 ) 3 x = det 4 x x = x 3 + 8x x + 6 = ( )(x 4)(x )

3 so the eigenvalues of A are λ = 4 and λ =, with algebraic multiplicities and, respectively. Let s find the dimensions of E 4 and E : first, = rref(a 4I) E 4 = ker(a 4I) = ker {}}{ = ker 3 has dimension, because the rank of this matrix is, and = rref(a I) {}} { E = ker(a I) = ker = ker has dimension, because the rank of this matrix is. Consequently, the dimensions of the eigenspaces equal the algebraic multiplicities of the corresponding eigenvalues, so A is diagonalizable. To find a basis β for V that diagonalizes A, we merely have to find bases for E 4 and E. This is easy once we look at rref(a 4I) and rref(a I). In the first case, letting z = t, we get that x = t and y = t, so that E 4 = span and similarly with E, letting y = s and z = t gives that x = s t, so that x s t y = z s t = s + t whence E = span, The vectors in these spans are linearly independent, and thus form bases for E 4 and E, respectively. Joining all three into one basis gives a diagonalizing basis β =,, To see that this is true, we compute: 3 4 T A (b ) = Ab = 4 = 8 = 4 = 4b 4 3 T A (b ) = Ab = 4 = = = b 3 T A (b 3 ) = Ab 3 = 4 = = = b 3 3

4 so that 4 Λ = [T A β = 5. Expanding along the 4th row (you could also expand along the first column), we get that f(x) = det x 9 = x det(a 4,)+det(A 4, ) det(a 4,3 )+9 det(a 4,4 ) det(a 4,5 ) 4 Note that the terms det(a 4,j ) are constants, so when we take the derivative of f we treat them as such, and therefore: 3 4 f (x) = det(a 4, ) = det = 3 4 = Let β = (, x, x ) and β = ( + x + x, + x, x ). Then M β,β = [[ + x + x β [ + x β [ x β = and therefore M β,β = M β,β = Finally, which means. (a) (AB) T ij = (AB) ji = [x 5x β = M β,β [x 5x β = x 5x = ( + x + x ) + 9( + x) ( x ) A jk B ki = k= B ki A jk = k= = 9 (B T ) ik (A T ) kj = (B T A T ) ij (b) If A is invertible, then there is a matrix A such that AA = A A = I, whence by part (a) of this problem (A ) T A T = (AA ) T = I T = I = I T = (A A) T = A T (A ) T. This shows that A T is invertible, and (A ) T = (A T ). 8. Let A = [ 3. [ x (a) c A (x) = det(a xi ) = det 3 x 3x + = (x )(x ). Thus λ = and λ = are the eigenvalues of A. Let s find the dimensions of their corresponding eigenspaces: [ [ E = ker(a I ) = ker = ker k= 4

5 has dimension, and E = ker(a I ) = ker [ [ = ker also has dimension. Thus, since the dimensions of the eigenspaces equal the algebraic multiplicities of the corresponding eigenvalues, A is diagonalizable. (b) First, to find bases for E and E, we parametrize the free, or inner, variable y: for E this means letting y = t and so x = t, while for E this means y = t and x = t, so that a basis for E is (, ), and a basis for E is (, ). Joining these two bases gives a basis for R that diagonalizes A: β = {[, [ } Then an easy calculation shows that Ab = b and Ab = b, so [ Λ = [T A β Finally, with ρ the standard basis for R, [ [ M = M β,ρ = ρ [ = ρ [ so that M = [ and an easy verification shows that A = MΛM. 9. det A = 3, which can be computed by using Gaussian elimination or by cofactor expansion.. If A R 5 5 and det(a) = 5, then det( A) = ( ) 5 det A = 3 5 = 6. 5