linear algebra midterm study
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Apr 3, 2024
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Multiple choice
Let T : R^n -> R^n be an isomorphism. Which of the following is always true?
B) [T] is invertible
Let {u1, u2, u3} be a basis for vector space V. Which of the following is also a basis for V?
C) {u1 −u2, u3 −u2, u1 +u3}
Let n ≥ 1 be an integer, and consider the mapping T_n : P2 -> R defined as T_n(a
+ bx + cx^2) = a + b^n − c^n. The mapping T_n is linear
A) only for n = 1
The eigenspace of the matrix [-4, 15; -2, 7] for eigenvalue λ = 2 is:
B) Span{[5; 2]}
The eigenvalues of the matrix [2, -1; -1, 2] are:
D) λ = 1 and λ = 3
The eigenvalues of the matrix [1, 2; 2, 1] are:
A) λ = −1 and λ = 3
The eigenvalues of the matrix [-2, -1; -1, -2] are:
D) λ = −1 and λ = −3
The eigenvalues of the matrix [3, 3; 5, 1] are:
D) λ = −2 and λ = 6
If A is a 4 x 4 matrix with det A = 10, then det(A + I) is:
D) Cannot be determined
If A is a 4×4 matrix with det A = 2, then det(cof A) is:
D)8
If A is a 4 × 4 matrix with det A = −2, then det(cof A) is:
D) −8
The determinant of the matrix [0, 0, -3; -2, -1, 3; 3, -4, -4] is:
C) −33
The determinant of the matrix [2, 0, 0; -1, 3, -2; 3, 3, 1] is:
B) 18
The determinant of the matrix [3, 2, 0, 0; -3, 1, 0, 0; 0, 0, -4, 2; 0, 0, 4, -1] is:
A)-36
The determinant of the matrix [3, 0, 1; 0, 2, -1; 0, 1, 1] is:
B) 9
Question - Gram-Schmidt
Use the Gram-Schmidt procedure to produce an orthonormal basis for the sub-
space spanned by
B = {[1; -1; 1], [1; 0; 1], [1; 1; 2]}
Do not change the order of the vectors.
Denote the given basis vectors by w1, w2, and w3 respectively. Step 1: Let v1 = w1.
Step 2: perp w2 = [1; 0; 1] - 2/3[1; -1; 1] = -1/3[1; 2; 1] so take v2 = [1; 2; 1]
Step 3: perp w3 = [1; 1; 2] - 2/3[1; -1; 1] - 5/6[1; 2; 1] = 1/2[-1; 0; 1] so take v3 = [-1; 0; 1]
Normalizing we get that an orthonormal basis for this subspace is
{1/sqrt(3) [1; -1; 1], 1/sqrt(6) [1; 2; 1], 1/sqrt(2) [-1; 0; 1]}
(b) Find the coordinates of the vector x = 2 with respect to the orthonormal basis obtained in part (a).
[x]_B = [x1 * u1; x2 * u2; x3 * u3] = [0; sqrt(6); 0]
Question - Gram schmid again
Use the Gram-Schmidt procedure to produce an orthonormal basis for the sub- space spanned by
{[1; -2; 1], [0; 1; 0], [0; 0; 1]}
Do not change the order of the vectors.
v1 = [1; −2; 1], S1 = Span{v1}.
perp_S1 = [0; 1; 0] = [0; 1; 0] - (-2/6) [1; -2; 1] = [1/3; 1/3; 1/3], choose v2 = [1; 1; 1], S2 = Span{v1 ,v2 }
perp_S2 [0; 0; 1] = [0; 0; 1] - 1/6[1; -2; 1] - 1/3[1; 1; 1] = [-1/2; 0; 1/2], choose v3 = [-1; 0; 1]
Then an orthonormal basis for the subspace is {1/sqrt(6)[1; -2; 1], 1/sqrt(3)[1; 1; 1], 1/sqrt(2)
[-1; 0; 1]}
(b) Find the coordinates of the vector x = [-2; 3; 0] with respect to the orthonormal basis obtained in part (a).
X = [-2; 3; 0] = (-8/sqrt(6)) 1/sqrt(6)[1; -2; 1] + (1/sqrt(3)) 1/sqrt(3)[1; 1; 1] + (2/sqrt(2)) 1/sqrt(2)[-1; 0; 1]
So [X]_B = [-8/sqrt(6); 1/sqrt(3); 2/sqrt(2)]
Question - linear mapping, bases
4. Let T : R^2 -> R^2 be the linear mapping defined by T[x; y] = [2y; x+y] and consider the bases B1 = {[0; 1], [1; -1]) and B2 = {[-1; 1], [1; 1]}
(a) Find the matrix M_B2B1 of T with respect to the given bases.
M_B2B1 = [C_B2(T[0; 1]) C_B2(T[1; -1)]
= [C_B2([2; 1]) C_B2([-2; 0)]
= [C_B2(-1/2[-1; 1] + 3/2[1; 1]) C_B2(1[-1; 1] - 1[1; 1])
= [[-1/2; 3/2], [1; 1]] = [-½, 1; 3/2, -1]
(b) Find C_B2 (T (u)) where u = [1; 1] using the results in part (a).
u = [1; 1] = 2[0; 1] + [1; -1] so C_B1(u) = [2; 1]
So C_B2(T(u)) = M_B2B1 C_B1 (u) = [-1/2, 1; 3/2, -1] [2; 1] = [0, 2]
Question - linearly dependent or independent
5. (a) Is the set {x^2 + x + 1, x^2 − 1, −x + 1} linearly dependent or independent in P^2? Justify your answer.
Let a(x^2+x+1) + b(x^2 - 1) + c(-x + 1) = 0x^2+0x+0. Then
[1, 1, 0; 1, 0, -1; 1, -1, 1] [a; b; c] = [0; 0; 0]
[1, 1, 0; 1, 0, -1; 1, -1, 1] row reduces to [1, 0, 0; 0, 1, 0; 0, 0, 1] therefore a = b = c = 0.
So yes, the set is linearly independent.
(b) Consider the vector space M_2,2. Is the matrix A = [2, 1; 1, 4] in the set
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