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Math 51 Exam 2 Solutions — May 22, 2007
- (12 points) (a) Find the eigenvalues of the following matrix.
A =
(6 points) To find the eigenvalues, we must solve the equation det(A − λI) = 0 for λ.
det(A − λI) =
1 − λ − 1 1 1 − 1 − λ 1 − 1 1 1 − λ
To evaluate the determinant, we expand it along a row or column. Here, we show it expanded along the top row: (1 − λ)
∣∣−^1 1 − λ^1 −^1 λ
∣∣ (^) −^11 − 1 1 − λ
∣∣ (^) −^11 −^1 1 −^ λ
= (1 − λ)(λ^2 − 2) + (2 − λ) − λ = −λ^3 + λ^2 + 2λ − 2 + 2 − λ − λ = −λ^3 + λ^2 Setting this equal to 0 and factoring yields λ^2 (1 − λ) = 0 from which we conclude that λ = 0 or 1. So the eigenvalues are 0 or 1. (b) Compute the eigenspace of the largest eigenvalue you found in part (a). (3 points) The largest eigenvalue is λ = 1. The eigenspace E 1 is precisely the nullspace of the matrix A − I. Let’s compute N (A − I) by row-reducing:
A − I =
which is row-reduced. There are two pivots, and their equations are x 1 − x 3 = 0 ⇒ x 1 = x 3 and x 2 − x 3 = 0 ⇒ x 2 = x 3
So N (A − I) =
−→x =
x 1 x 2 x 3
x 3 x 3 x 3
= span
(c) Is there a nonzero vector −→v ∈ R^3 such that A−→v = −→v? Briefly explain. (3 points) Yes. There is a nonzero vector −→v such that A−→v = −→v exactly when and only when 1 is an eigenvalue of A. In part a), we found that 1 is an eigenvalue of A.
- (12 points) Let A =
[ 1 3 − 3
]
(a) Find a basis for the orthogonal complement of the null space N (A). (5 points) Since N (A)⊥^ = C(AT^ ), we will find a basis for C(AT^ ).
AT^ =
By definition, the two columns of AT^ span C(AT^ ). These vectors are linearly independent, since they are clearly not scalar multiples of each other. Thus a basis for N (A)⊥^ = C(AT^ ) is (^)
(Alternatively, rref(AT^ ) =
(b) Given the matrix A above, fill in the blanks. (3 points) From part (a), we can see that N (A) is a subspace of R^3. (Vectors in N (A) have 3 components.) Thus the nullspace and its complement together form R^3 , so dim(N (A)) + dim(N (A)⊥) = 3 The columns of A have 2 components, so C(A) is a subspace of R^2. Thus dim(C(A)) + dim(C(A)⊥) = 2 Since A has 3 columns, the Rank-Nullity theorem says dim(N (A)) + dim(C(A)) = 3
(c) Using your answers to parts (a) and (b), give the dimensions of the four subspaces.
(4 points) From part (a), N (A)⊥^ has 2 basis vectors, so dim(N (A)⊥) = 2. The rest then follow from part (b). dim(N (A)) = 1 dim(N (A)⊥) = 2
dim(C(A)) = 2 dim(C(A)⊥) = 0
- (10 points) Find the inverse of the given matrix.
The inverse matrix is (^)
- (12 points) Let V be the 2-dimensional subspace of R^3 with basis
B = {−→v 1 , −→v 2 } =
(a) Use the Gram-Schmidt process to find an orthonormal basis {−w→ 1 , −w→ 2 } for V. (6 points)
−w→ 1 =^ −→v^1 ||−→v 1 || =
−→y 2 = −→v 2 − projV 1 (−→v 2 ) = −→v 2 − (−→v 2 · −w→ 1 )−w→ 1
=
−w→ 2 =^ −→y^2 ||−→y 2 || =^
√^1
(b) −→v 1 and −→v 2 are sketched in V below. Draw the vector −w→ 2 in the figure. (3 points) (^) −→ v 2 √ 41 ≈ 6. 4
−→v 1 3
−w→ 2 1
line spanned by −→v 1
(c) Find the distance from −→v 2 to the line spanned by −→v 1. (You may use your work from parts (a) and (b)). (3 points) The distance is the length of the shortest path from the line spanned by −→v 1 to the tip of −→v 2. This path is perpendicular to the line and is represented by −→y 2 , so distance = ||−→y 2 || = √ 5.
- (14 points) Mark each statement below as true or false by circling T or F. No justification is necessary. (2 points each)
T F If A, B, and C are matrices such that AB = AC, then B = C.
The matrix A is not necessarily invertible! Take a matrix with nontrivial nullspace (like A from question 3), and let B and C be one column matrices which are different vectors in the nullspace. Then AB = 0 = AC, but B 6 = C. T F If A is a 2 × 2 matrix with det(A) = 0, then one column is a multiple of the other. det(A) = 0 if and only if the columns of A are linearly dependent. In the 2 × 2 case, this means one column is a scalar multiple of the other.
T F If A and B are n × n matrices with det(A) = 2 and det(B) = 3, then det(A + B) = 5.
Determinant is multiplicative (det(AB) = det(A) det(B)), but not additive. For a specific example, take A = B = [ 1 00 1 ]. Then det A = det B = 1. But A + B = [ 2 00 2 ], so det(A + B) = 4 6 = 1 + 1.
T F Suppose that A is an n×n matrix with rref(A) = In. It follows that A is diagonalizable.
Diagonalizability does not mean that the reduced row echelon form of a matrix is diagonalizable or diagonal. The matrix [ 1 10 1 ] row reduces to the identity. But it has only a single eigenvalue 1 with eigenspace spanned by [ 10 ]. This does not form an eigenbasis for R^2 , and hence the matrix is not diagonalizable.
T F The components of a function f : R → Rm^ are vectors.
Each component fi(x) is a scalar.
T F (^) (x,ylim)→(0,0)^ x
(^2) y (^3) − 2 xy xy =^ −^2
(x,ylim)→(0,0)^ x
(^2) y (^3) − 2 xy xy =^ (x,ylim)→(0,0)
xy (xy^2 − 2) xy = 0^ −^ 2 =^ −^2
T F The function f (x, y) =
{ (^) y (^4) −x 4 x^2 +y^2 when(x, y)^6 = (0,^ 0) 2 when(x, y) = (0, 0) is continuous. By a factorization of the numerator of y x^42 −+xy^42 similar to the previous problem, lim(x,y)→(0,0) f (x, y) = lim(x,y)→(0,0) y x^42 −+xy^42 = 0. But f (0, 0) = 2, so f is not continuous.
- (7 points) Match the graph of the surface with one of the contour maps.
Write the letter of the contour map corresponding to the graph of the given function in the box: f (x, y) = x^2 + y 42 f (x, y) = e^1 −x^2 +y^2 b d f (x, y) = e^1 −x^2 −y^2 f (x, y) = ln |y − x^2 | a c
