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Math 20F Solutions to Final Exam 11:30 AM March 21, 2003
- Since the matrix is triangular, the eigenvalues are the diagonal entries. Finding eigen- vectors involves finding bases for the null spaces of A − λI. The answers are
λ = 1 v = α(1, 0 , 0)T^ λ = 3 v = β(1, 1 , 0)T^ λ = 4 v = γ(0, 0 , 1)T
where α, β and γ are nonzero scalars. Any values you chose for those scalars are correct.
- (a) The easiest way to construct an orthonormal basis is to note that the vectors are orthogonal and each has length 3. This gives us
u 1 = (1/ 3 , 2 / 3 , 0 , 2 / 3 , 0)T u 2 = (2/ 3 , − 1 / 3 , 0 , 0 , 2 /3)T u 3 = (0, 2 / 3 , 0 , − 2 / 3 , 1 /3)T^.
You could use Gram-Schmidt orthogonalization and obtain the same answer. The orthonormal basis is not unique, so you may have obtained a different correct answer; however, that’s fairly unlikely since this choice is the obvious one.
(b) If the given vector is v, then vT^ u 1 = 15, vT^ u 2 = 9 and vT^ u 3 = 3. Thus the projection of v onto W is 15u 1 + 9u 2 + 3u 3 = p = (11, 9 , 0 , 8 , 7)T^. Thus q = v − p = (− 2 , 0 , 9 , 1 , 2) is in W ⊥^ and so v = p + q is the answer. (There is no other correct answer.)
3. L =
- Recall that the determinant is the product of the eigenvalues and so det(A) = 6. (i) 1, −1, 1/2 and 1/3; det(A−^1 ) = − 1 / 6 (ii) Same as A. (iii) If Av = λv, then p(A)v = p(λ)v. Thus we have 0, 2, 2 and 6; det(A^2 − A) = 0.
- Using the hint, column k of A is wkv. Hence all the columns are multiples of v and so lie in the space spanned by v. We need to show that v is in the span of the columns. Since we are given w 6 = 0 , there is some j with wj 6 = 0. Multiplying column j by the scalar 1/wj , we obtain v.
Math 20F Solutions to Final Exam 11:30 AM March 21, 2003
- The only information about the row space R(M T^ ) and the null space N (M ) of an arbitrary matrix M ∈ Rn×k^ is: (i) dim(R(M )) ≤ n, the number of rows of M. (ii) Since R(M T^ ) = N (M )⊥, dim(R(M T^ ))+dim(N (M )) = k, the number of columns of M.
A: Using the previous observations, we cannot rule this answer out In fact, there is a matrix with the given row and null space: Construct a matrix with that row space and the null space will automatically take care of itself. B: Since (0, 0 ′^0 , 1) and (0, 1 , 1 , 2)T^ are not orthogonal, the answer must be wrong. C: The spaces are orthogonal, but the sum of their dimensions is wrong, so C is incorrect. (The student must have missed a vector in the basis of either the row space or the null space.)
- (i) Let y = Ax. Recall that yT^ y ≥ 0 with equality if and only if y = 0. Now
〈x, x〉 = (Ax)T^ (Ax) = yT^ y ≥ 0 ,
with equality if and only if y = 0. y = Ax = 0 if and only if x = 0 because A is nonsingular.
(ii) 〈x, y〉 = xT^ AT^ Ay = (yT^ AT^ Ax)T^ = (〈y, x〉)T^. Since the inner product is a scalar, it equals its transpose.
(iii) 〈αx + βy, z〉 = (αx + βy)T^ AT^ Az = αxT^ AT^ Az + βyT^ AT^ Az = α〈x, z〉 + β〈y, z〉.
- It doesn’t make sense to write det(A) because A is not a square matrix.
