Linear Algebra (Math 311) – Final

Linear Algebra (Math 311) – Final
Problem 1. [30 points] Below you see a matrix A and its reduced rowechelon
form B.
⎛
1
⎜ 0
A = ⎜ 2
⎝ 2
2 3
1 3
3 3
3 −1
−1
1
−3
1
⎞ ⎛
2 1 0 0 −6
−5⎟
⎜
⎟ ⎜0
1 0 4
9 ⎟ ∼ B = ⎜
0 ⎠ ⎝
75
4
−
−1 0 7 −1 −3
47
0 0 1 −1
0 0 0 0
⎞
⎟
4 ⎟
9 ⎟
4 ⎟
0 ⎠
0 0 0 0 0
In answering the following questions, provide at least a short, decisive argument.
1. What is the rank of A?
2. Find linearly independent rows of A.
3. Find linearly independent columns of A.
4. Find a basis for the orthogonal complement of the row space of A.
5. Delete some rows and columns until you find a non–singular square
matrix of maximal size. Justify that your minor is non–singular and of
maximal size.
6. How does the null space {x ∈ R 5 | Ax = 0} and the nullity relate to
any of the above?
Solution: 1. The rank of A is 3, as one may see from the number of
leading 1’s in B.
2. We need to find three linearly independent rows, as the dimension of
the row–space is 3. We may start with the first two rows because they are
linearly independent, none is a multiple of the other one. You may see that
the third row is twice the first minus the second row. We cannot use the first
three rows. We try to use the first, second, and fourth row, and to check
linear independence, we look at the first three coordinates (if you want, you
may also use all coordinates):
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 2 3 1 2 3 1 2 3
⎝0
1 3 ⎠ ∼ ⎝0
1 3 ⎠ ∼ ⎝0
1 3 ⎠
2 3 −1 0 −1 −7 0 0 −4
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From the last matrix in the computation we see that the matrix is nonsingular,
and we conclude that the first, second, and fourth row are linearly
independent.
3. The first three columns of A are linearly independent because that is
where B has the leading 1’s.
4. The orthogonal complement to the row–space is the null space of the
matrix. The last two ccordinates are free. We obtain a basis by setting either
x4 = 1 and x5 = 0 or x4 = 0 and x5 = 1. With these choices we get the basis
⎧⎛
⎞
6
⎪⎨ ⎜
⎜−4
⎟
⎜ 1 ⎟
⎝
⎪⎩
1 ⎠
0
,
⎛ ⎞⎫
−75
⎜ 47 ⎟⎪⎬
⎟
⎜ −9 ⎟
⎝ 0 ⎠
⎪⎭
4
5. We solved part 5. as part 2. of the problem. Delete the last two
columns of the matrix, as well as row three and five. The result is a nonsingular
matrix. There is not larger nonsingular minor, because is size will be
the rank of the matrix, and that is 3.
6. As stated in 4. the nullspace of the matrix is the orthogonal complement
of its rowspace. The nullity is its dimension, and it is the number of
columns minus the rank of the matrix. The nullity is 5 − 3 = 2.
Problem 2. [10 points] Consider the vector space P2 of polynomials of degree
at most 2 with the basis consisting of the vectors v1 = 1, v2 = 1 + t, and
v3 = 1 + t + t 2 . Differentiation defines a linear map D : P2 → P2. Find the
matrix of D with respect to the given basis.
Solution: We apply D to the basis elements, making sure that the results
are given in terms of the basis again. We find: D(v1) = 0, D(v2) = 1 = v1,
and D(v2) = 1 + 2t = −1 + 2(1 + t) = −v1 + 2v2. Accordingly, the matrix
for D is ⎛ ⎞
0 1 −1
⎝0
0 2 ⎠
0 0 0
Problem 3. [20 points] Let P2 be as in Problem 2. Set
S = {1 − t 2 } and T = {1 + t, 1 − t 2 , t + t 2 , 1 + 2t, 1 + t + t 2 }
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1. Is it possible to add elements to S until you have a basis of P2? Explain
why! If possible, do it.
2. Is it possible to omit elements of T until you have a basis of P2? Explain
why! If possible, do it.
Solution: Observe that P2 is of dimension 3, so we are looking for bases
consisting of three elements.
1. Because S is linearly independent, we can add elements, keeping the
set linearly independent, until the set spans and we have a basis. Here is a
set that works, obviously:
{1 − t 2 , 1, t}
2. Note that T spans P2 and contains a basis because
1 = (1 + t + t 2 ) − (t + t 2 )
t = (1 + 2t) − (1 + t)
t 2 = (1 + t + t 2 ) − (1 + t)
To cut down T to a basis, we use the first element, it is nonzero. We keep
the second element, because the first two elements are linearly independent,
none of the elements is a multiple of the other one. Note that the third
element is the difference of the first two, so discard it. It is easy to see that
the fourth element is independent of the first two, so we keep it. By now we
have three linearly independent elements selected from T , and they form a
basis:
{1 + t, 1 − t 2 , 1 + 2t, }
Problem 4. [15 points] Find the determinants of the matrices
⎛ ⎞
⎛ ⎞
1 2 3 1
3 0 1
⎜
A = ⎝1
0 4⎠
and B = ⎜0
1 2 5 ⎟
⎝1
2 1 1⎠
2 1 5
0 7 1 2
Solution: Using the standard formula for 3 × 3 matrices, we find
det A = 1 − 12 = −11.
We may apply elementary row operations (recalling how they affect the determinate)
to bring the matrix into triangular form, and then get the determinante
as the product of the diagonal elements. Here is one way, without
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additional explanations for the steps. We use the notation | · | to indicate
that we find the determinant of the enclosed expression.
1
2 3 1
1
0 3 1
0
0 2 0
det B = 0
1 2 5
1
2 1 1
= 0
1 2 5
1
0 1 1
= 0
1 2 5
0
2 0
1
0 1 1
=
1
2 5
= 66
0
7 1 2
0
7 1 2
0
7 1 2
7 1 2
6 3
Problem 5. [15 points] Let A = .
−2 1
1. Find the eigen vectors and values for A.
2. If possible, find a matrix P so that P −1 AP is diagonal. If A cannot be
diagonalized, explain why.
Solution: We calculate:
6 − λ 3
det
= (6 − λ)(1 − λ) + 6 = λ
−2 1 − λ
2 − 7λ + 12 = (λ − 3)(λ − 4).
1
3
An eigen vector for λ = 3 is , and an eigen vector for λ = 4 is .
−1
−2
As matrix P we use the matrix with the eigen vectors as its columns:
1 3
P = .
−1 −2
Then
P −1 AP =
−2 −3
1 1
provides the diagonaliztion.
6 3
−2 1
1 3
=
−1 −2
3 0
0 4
Problem 6. [25 points] Let V be a vector space, and 〈·, ·〉 an inner product
on V . The inner product defines a norm on V .
1. Define the concept of a norm on a vector space.
2. Tell, by which formula an inner product defines a norm.
3. Prove that the formula that you gave in the previous item does indeed
give you a norm.
4

Solution: A norm on a vector space is a map || · || : V → R, so that
1. ||x|| ≥ 0 for all x ∈ V , and ||x|| = 0 if and only if x = 0.
2. ||ax|| = |a| · ||x|| for all x ∈ V and a ∈ R.
3. ||x + y|| ≤ ||x|| + ||y|| for all x and y in V
An inner product defines a norm by setting
||x|| = 〈x, x〉
We verify that this expression defines a norm. By definition of the square
root, ||x|| = 〈x, x〉 ≥ 0. From the positive definiteness of the inner product
we deduce that
||x|| = 0 ⇐⇒ 〈x, x〉 = 0 ⇐⇒ 〈x, x〉 = 0 ⇐⇒ x = 0.
Using the linearity properties of the inner product and basic rules of algebra
we find:
Finally,
||ax|| = 〈ax, ax〉 = a 2 〈x, x〉 = |a| 〈x, x〉 = |a| · ||x||.
||x + y|| 2 = 〈x + y, x + y〉 = 〈x, x〉 + 2〈x, y〉 + 〈y, y〉
≤ ||x|| 2 + 2||x|| · ||y|| + ||y|| 2 = (||x|| + ||y||) 2
The inequality is a direct consequence of Cauchy–Schwarz’ inequality. The
triangle inequality is obtained by taking square roots of the outermost terms.
Problem 7. [20 points] Consider a vector space V of dimension 2 with basis
{x, y}. On V , define an inner product 〈·, ·〉 by setting 〈x, x〉 = 〈y, y〉 = 3 and
〈x, y〉 = 1. We denote the associated norm by || · ||.
1. Find ||x − y|| and the cosine of the angle between x and y.
2. Orthonormalize the basis.
Solution: We calculate:
||x − y|| = 〈x − y, x − y〉 = √ 3 + 3 − 2 = 2.
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Let α denote the angle between x and y, then
cosα =
〈x, y〉 1
=
||x|| · ||y|| 3 .
Let us first orthogonalize the vectors, resulting in
v1 = x and v2 = y −
〈y, x〉 1
x = y −
〈x, x〉 3 x.
The norms of these vectors are
||v1|| = √
3 and ||v2|| = 〈y − 1
1
x, y − x〉 = 3 +
3 3 1
1 2
− 2 · = 2
3 3 3 .
Normalizing the vectors v1 and v2 we find the orthonormal basis consisting
of
u1 = 1
√ x and u2 =
3 1
3
y −
2 2
1
3 x
.
Problem 8. [15 points] Show, if the columns of C are linearly independent,
then C t C is invertible.
Solution: Suppose the size of C is m × n. Let x ∈ R n . Then Cx is a
linear combination of the columns of C, and because the columns of C are
linearly independent, the only solution of Cx = 0 is the trivial solution x = 0.
Suppose C t C is singular, then there exists some non–zero vector x ∈ R n so
that C t Cx = 0. It follows that x t C t Cx = (Cx) t Cx = 0, and this means that
Cx = 0. This is a contradiction, and it follows that C t C is non–singular,
hence invertible.
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