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Math 415 Final Exam Solutions: August 5, 2011
- Define each of the first five terms.
(i) (4 points) The list v 1 ,... , vn of vectors is linearly independent.
Whenever there are scalars c 1 ,... , cn with c 1 v 1 + · · · + cnvn = 0, then all ci = 0.
(ii) (4 points) Dimension of a finite-dimensional vector space V.
The dimension of V is the number of elements in a basis of V.
(iii) (4 points) Orthogonal complement S⊥^ of a subspace S of a vector space V.
S⊥^ = {v ∈ V : (v, s) = 0 for all s ∈ S}.
(iv) (4 points) A Hermitian matrix A = [ajk]. (You must explain your nota- tion.)
AH^ = A. Here, AH^ = [aij ]>.
(v) (4 points) An eigenvalue of a linear operator L : V → V.
A number λ for which there exists a nonzero vector v with Av = λv.
(vi) (5 points) What are the columns of the transition matrix P from a basis X to a basis Y?
If X = x 1 ,... , xn, then the jth column of P consists of the coordinates of xj with respect to Y.
- Consider the homogeneous linear system
x 1 + x 2 + x 3 − x 4 = 0 2 x 1 + 2x 2 + 3x 3 + x 4 = 0 x 1 − x 2 + 2x 3 + 3x 4 + x 5 = 0
i) (10 points) Find a basis of the row space of this system.
The row reduced echelon form for the coefficient matrix is
E =
A basis of the row space consists of the rows of E.
ii) (15 points) Find a basis of the solution space of this system.
Using E, we see that the free variables are x 4 and x 5 , and
x 1 = 32 x 4 − 12 x 5 x 2 = 13 x 4 + 12 x 5 x 3 = − 3 x 4
Thus, a basis for the solution space is
( 32 , 13 , −3)>, (− 12 , 12 , 0)>.
- Let A be an n × n matrix, where n ≥ 2.
(i) (10 points) If A is singular, prove that adj(A) is singular.
We know that A adj(A) = det(A)I. Since A is singular, det(A) = 0, and so A adj(A) = 0. If A = 0, then adj(A) = 0; if A 6 = 0, then adj(A) is singular: otherwise, multiply on the right by adj(A)−^1 and get A = 0.
(ii) (15 points) If A is nonsingular, find det(adjA); justify your answer.
Since A adj(A) = det(A)I, we have
det(A adj(A)) = det(A) det(adj(A)) = det(A)n.
Therefore, det(adj(A)) = det(A)n−^1.
- Let V and W be vector spaces, let U = u 1 ,... , un be a basis of V , and let w 1 ,... , wn be a list of (not necessarily distinct) vectors in W. Prove that there is a unique linear transformation L : V → W with L(ui) = wi for all i. You must show that the L you construct is single-valued.
Each v ∈ V has coordinates wrt U : there are unique scalars c 1 ,... , cn with v = c 1 u 1 + · · · + cnun. Define L by
L(v) = c 1 w 1 + · · · + cnwn.
Note that uniqueness of the coordinates shows that L is a single-valued function. Obviously, L(ui) = wi for all i, and so it remains to prove that L is a linear transformation. If v′^ = c′ 1 u 1 + · · · + c′ nun, then v + v′^ = (c 1 + c′ 1 )u 1 + · · · + (cn + c′ n)un, and so L(v + v′) = (c 1 + c′ 1 )w 1 + · · · + (cn + c′ n)wn. On the other hand, L(v) + L(v′) = [c 1 w 1 + · · · + cnwn] + [c′ 1 w 1 + · · · + c′ nwn] = (c 1 + c′ 1 )u 1 + · · · + (cn + c′ n)un. Finally, if α is a scalar, then
L(αv) = L(αc 1 u 1 + · · · + αcnun) = αc 1 w 1 + · · · + αcnwn = α(c 1 w 1 + · · · + cnwn) = αL(v).
- Find the inverse of A =
You can either use adj(A) or Gaussian elimination. Either way,
A−^1 = 14
- Let A =
[ √
3 2
1 2 − (^12)
√ 3 2
]
(i) (10 points) Prove that there is no matrix P with real entries such that P −^1 AP is diagonal.
The characteristic polynomial of A is p(λ) = λ^2 −
3 λ+1, and the quadratic formula gives the eigenvalues to be 12 (
3 ± i). Now if P AP −^1 = D, where D is diagonal, then D has real entries and its diagonal entries would have to be the eigenvalues of A, for similar matrices have the same eigenvalues. But the eigenvalues of A are not real.
(ii) (15 points Find a matrix Q with complex entries such that Q−^1 AQ is diagonal.
If Y = y 1 , y 2 is a basis of eigenvalues of R 2 , then the matrix of A wrt to Y is a diagonal matrix with the eigenvalues of A on the diagonal. Thus, a matrix Q is a transition matrix from Y to the standard basis. An eigenvector of A belonging to
3 + i is (i, 1)>, and an eigenvector of A belonging to
3 − i is (−i, 1)>. Thus,
Q =
[
i −i 1 1
]
is such a matrix (since eigenvectors are determined up to nonzero scalar multi- ples, Q is not unique: we can multiply the first column by any nonzero complex number α and the second column by any nonzero complex number β).
