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MATH 33A LECTURE 2 FINAL EXAM
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Problem 1. (True/False, 2 pts each) Mark your answers by lling in the appropriate bosx next to each question.
(i: T F ) The matrices
[
]
and
[
]
are similar. NO: they have different traces.
(ii: T F ) The matrices
[
]
and
[
]
are similar. NO: They have different determi- nants (iii: T F ) A symmetric matrix A can have i =
− 1 as one of its eigenvalues. NO: all eigen- values are real (iv: T F ) If a 2 × 2 orthogonal matrix B has an eigenvector with eigenvalue 1 , then B is the identity matrix. NO: consider B to be the reection about the x-axis. (v: T F ) No 3 × 2 matrix A can have an inverse. NO: it's not a square matrix. (vi: T F ) If A is 2 × 3 , the range of A may have dimension 0 , 1 or 2. YES. (vii: T F ) If A is 3 × 4 , the kernel of A may have dimension 0 , 1 , 2 , 3 or 4. NO: it can't be dimension 0 because of rank-nullity (note that the image of A may have dimension at most 3 ). (viii: T F ) No three linearly independent vectors span R^4. YES: any basis must have four elements. (ix: T F ) If A and B are similar, then all eigenvectors of A are also eigenvectors of B. NO: otherwise every diagonalizable matrix would have to be diagonal. (x: T F ) If (u, v) are linearly independent and (v, w) are linearly independent, then (u, w) are linearly independent. NO: imagine that u = w (xi: T F ) If A and B are square matrices, then rank(AB) ≤ rank(B). YES: the kernel of AB includes that of B so the conclusion follows from rank-nullity. (xii: T F ) If A is orthogonal, then det A 6 = 0 YES: an orthogonal matrix always has ± 1 as its determinant. (xiii: T F ) If A is diagonalizable, so is A^3. YES: A = SDS−^1 so A^3 = SDS−^1 SDS−^1 SDS−^1 = SD^3 S−^1 and D^3 is diagonal. (xiv: T F ) There is a 4 × 4 matrix with no real eigenvalues. YES: make A have characteristic
polynomial fA(x) = (x^2 + 1)^2. One example is A =
(xv: T F ) There is a 3 × 3 matrix with no real eigenvalues. NO: such an A must have as its characteristic polynomial a polynomial of degree 3 , which must have at least one real root.
Problem 3. (20 pts) Let A =
. (a) Compute rref(A). (b) Find a basis
for the kernel and image of A. (b) Find the rank of A.
Solution. We rst compute the reduced row echelon form.
Thus the image of A has as basis the vectors
and
. Thus^ A^ has rank^2. Com-
puting the kernel gives the relations
x 1 = t + 2s + 3r x 2 = − 2 t − 3 s − 4 r x 3 = t x 4 = s x 5 = r
So the kernel is spanned by the vectors 1 − 2 1 0 0
Problem 4. (20 pts) (a) Assume that B is a 2 × 2 matrix with 0 as an eigenvalue of geometric multiplicity 2. Must B = 0? Give a proof or provide an example of a nonzero B. (c) Assume that B is a 2 × 2 matrix and that B^2 = 0 but B 6 = 0. Find all eigenvalues of B together with their algebraic and geometric multiplicities. (Hint: assume that Bv = λv and then derive an equation for λ; next, can ker B = 0? Lastly, use (a) to rule out the other possible geometric multiplicities).
Solution. (a) If 0 has geometric multiplicity 2 , the associated subspace has dimension 2 and thus must be the entire space. Hence Bv = 0v for all v and so B = 0. (b) Assume that B has an eigenvector v with eigenvalue λ. Then 0 = B^2 v = B(Bv) = Bλv = λ^2 v, so that λ = 0. Thus all eigenvalues of B must be zero. So the characteristic polynomial of B must be a degree 2 polynomial with all of its roots zero, so that fB (x) = x^2. Hence the algebraic multiplicity of the zero eigenvalue is 2. The geometric multiplicity of the zero eigenvalue could thus be 1 or 2. If it were 2 , we would get that B = 0 by part (a). So it must be 1.
Problem 6. Find the volume of the parallelepiped determined by the vectors
Solution. The volume is the absolute value of the determinant
det
Problem 7. (20 pts) (a) Let A be a non-zero matrix so that A^100 = 0. Can A be diagonaliz- able? (Hint: assume that A were diagonalizable and then compute its 100th power). (b) Find a 3 × 3 matrix A so that A has exactly one eigenvalue λ = 0 of algebraic multi- plicity 3 and geometric multiplicity 2.
Solution. (a) If A = SDS−^1 then A^100 = SD^100 S−^1. So if A^100 = 0 it follows that D^100 = 0 so D = 0. (b) One example is
A =
Problem 9. (20 pts) Let
A =
(a) Find fA(λ) and all eigenvalues of A. (b) Find the algebraic multiplicity of each eigen- value. (c) For each eigenvalue, nd a basis for the corresponding eigenspace. (d) Com- pute the geometric multiplicity of each eigenvalue. (e) Is A diagonalizable?
Solution. fA(x) = (x^2 − 1)(x − 3)^2. Thus A has eigenvalues ± 1 each of algebraic mul- tiplicity 1 and 3 of algebraic multiplicity 2. Computing ker(A − 3 I), we nd that the eigenspace corresponding to 3 is spanned by the vector e 3 , and so is one-dimensional. Thus all eigenvalues have geometric multiplicity 1. Since not all geometric multiplicities equal algebraic multiplicities, the matrix is not diagonalizable. Computing ker(A ± I)
gives that the eigenspaces corresponding to ± 1 are spanned by
Problem 10. (20 pts) Let Q be the matrix
Q =
and denote the i, j-th entry of Q by qij. Assume that you are given four vectors v 1 , v 2 , v 3 , v 4 in R^55 , satisfying vi · vj = qij for all i, j. Use this information to: (a) Express the projection of v 1 onto the span of v 3 and v 4 in terms of v 1 , v 2 , v 3 , v 4 (b) Compute the angle between v 1 and v 3 (c) Compute the length of v 1 + v 3 (d) Find an orthonormal basis for the span of v 2 , v 3 , v 4 (express your answer in terms of these vectors). Solution. (a) Note that v 3 · v 4 = 0 so that these vectors are orthogonal. Thus the desired projection is v 3 √ 3
v 4 √ 2
(b) v 1 · v 3 = 1 so that cos θ =
so that θ = cos−^1 (1/
(c) ‖v 1 + v 3 ‖^2 = 6 so the length is
(d) v 2 /
2 , v 3 /
3 , v 4 /(
Problem 12. Let V be the subspace of R^3 , which is the intersection of the planes x + 2y +
2 z = 0 and 4 x − y − z = 0. Find the orthogonal projection of the vector v =
(^) onto
V. (Hint: there are several ways to do this quickly; one involves nding the projection of v onto V ⊥^ rst). Solution. We know that P rojV (v) = v − P rojV ⊥ (v).
Now since the rst plane consists of all vectors perpendicular to w 1 =
(^) while the
second plane consists of all vectors perpendicular to w 2 =
, V ⊥^ has is the span of
these two vectors. Thus
P rojV ⊥ (v) =
v · w 1 ‖w 1 ‖
w 1 ‖w 1 ‖
v · w 2 ‖w 2 ‖
w 2 ‖w 2 ‖
w 1 = w 1 =
It follows that
P rojV (v) =
