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1
(Notes, books, and calculators are not authorized)
Show all your work in the blank space you are given on the exam sheet. Always justify your
answer. Answers with no justification will not be graded. P2 is the vector space over R of
polynomials of degree at most two.
Question 1: Consider the subspace U of R4 spanned by v1 = (1, 1, 1, 1), v2 = (1, 1, 2, 4),
v3 = (1, 2, −4, −3). Use the Gram-Schmidt process to compute an orthogonal basis of U .
Apply the Gram-Schmidt process.
w1 = v1 = (1, 1, 1, 1)
w 2 = v2 −
w3 = v3 −
hv2 , w1 i
w1 = (1, 1, 2, 4) − 2(1, 1, 1, 1) = (−1, −1, 0, 2)
kw1 k2
hv3 , w1 i
hv3 , w2 i
3
1 3
w1 −
w2 = (1, 2, −4, −3)+(1, 1, 1, 1)+ (−1, −1, 0, 2) = ( , , −3, 1).
2
2
kw1 k
kw2 k
2
2 2
Question 2: Let V = C 0 ([−π, π]; R) be the vector space over R composed of Rthe functions
π
from [−π, π] to R that are continuous. Consider the inner product hf, gi = π1 −π f (t)g(t)dt
√
and√the associated norm k · k. We recall that k1/ 2k = k sin(t)k = k√cos(t)k = 1 and B =
{1/ 2, sin(t),
√ cos(t)} are orthonormal. Let U
√ = span(B). Let u1 (t) = 1/ 2 + 2 sin(t) + 4 cos(t),
u2 (t) = 2/ 2 − 3 sin(t) + cos(t), u3 (t) = 2/ 2 + sin(t) − cos(t) in V . (a) Compute < u1 , u2 >,
< u2 , u3 >, < u2 , u3 > and compute ku1 k, ku2 k, ku3 k.
We compute hui , uj i for i 6= j, i, j ∈ {1, 2, 3}.
√
hu1 , u2 i = 2k1/ 2k − 6k sin k + 4k cos k = 2 − 6 + 4 = 0,
√
hu1 , u3 i = 2k1/ 2k + 2k sin k − 4k cos k = 2 + 2 − 4 = 0,
√
hu2 , u3 i = 4k1/ 2k − 3k sin k − 1k cos k = 4 − 3 − 1 = 0.
This proves the statement. Clearly
ku1 k =
√
21,
ku2 k =
√
14,
ku3 k =
√
6.
2
Quiz 8, November 6, 2014
3 − λ
PA (λ) := det(A − λI) = 2
3
2
−4
0 −1
and B =
.
−6
1 0
−4 = (3 − λ)(−6 − λ) + 8 = λ2 + 3λ − 10
−6 − λ
Question 3: Find the eigenvalues of A =
i.e., PA (λ) = λ2 + 3λ − 10 = (λ − 2)(λ + 5). The eigenvalues of A are λ1 = 2 and λ1 = −5.
−λ −1 = λ2 + 1
PB (λ) := det(B − λI) = 1 −λ
i.e., PB (λ) = λ2 + 1 = (λ + i)(λ − i). The eigenvalues of B are λ1 = i and λ1 = −i.
Question 4: The following linearly 
independent vectors e1 := (1, −1, 0), e2 := (1, 0, 1), and
4 1 −1
e3 := (1, 2, 1) are eigenvectors of A = 2 5 −2. Find an invertible matrix P and a diagonal
1 1 2
matrix D so that A = P DP −1 .
The matrix is clearly diagonalizable since there exist three linearly independent eigenvectors. Let
us compute Ae1 , Ae2 and Ae3 to get the associated eigenvalues. (A second possibility would
be to compute the characteristic polynomial of A, but that would be too long and unnecessarily
complicated since we are already given the eigenvectors.)
Ae1 = (3, −3, 0) = 3e1
Ae2 = (3, 0, 3) = 3e2
Ae3 = (5, 10, 5) = 5e3 .
This means that e1 and e2 are associated with the eigenvalue 3 (the multiplicity of this eigenvalue
is two) and e3 is associated with the eigenvalue 5. Let us now set




3 0 0
1 1 1
D = 0 3 0 , P = −1 0 2 .
0 0 5
0 1 1
Then A = P DP −1 .