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Sage and Linear Algebra Worksheet: FCLA Section B

Robert Beezer

Department of Mathematics and Computer Science
University of Puget Sound

Fall 2019

Section 1 Bases

Five “random” vectors, each with 4 entries, collected into a set S.

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v1 = vector(QQ, [-4, -2,  3, -11])
v2 = vector(QQ, [-2,  7,  3,   9])
v3 = vector(QQ, [ 6, -4, -7,   5])
v4 = vector(QQ, [-1,  0,  3,  -4])
v5 = vector(QQ, [-4,  5, -5,  11])
S = [v1, v2, v3, v4, v5]

Consider the subspace spanned by these five vectors. We will make these vectors the rows of a matrix and row-reduce to see a basis for the space (subspace, or row space, take your pick). This is an application of Theorem BRS.

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A = matrix(S)
A

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A.rref()

Sage does this semi-automatically, tossing zero rows for us.

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W = span(S)
B = W.basis()
B

Demonstration 1.

Construct a random vector, w, in this subspace by choosing scalars for a linear combination of the vectors we used to build W as a span originally.

Then use the three basis vectors in B to recreate the vector w. Question: how many ways can you do this? By Theorem VRRB there should always be exactly one way to create w using a linear combination of a basis of W.

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w = *v1 + *v2 + *v3 + *v4 + *v5
w

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w in W

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*B[0] + *B[1] + *B[2]

Section 2 Nonsingular Matrices

We will obtain a basis of $\mathbb{C}^{10}$ from the columns of a $10\times 10$ nonsingular matrix.

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entries = [[ 1,  1,  1, -1, -2,  4,  2, -3,  1, -6],
           [-2, -1, -2,  2,  4, -7, -4,  5, -1,  7],
           [ 1, -1,  2, -2, -5,  8,  5, -3,  4, -4],
           [-1, -2,  0,  1,  0, -5,  0, -3, -5,  6],
           [ 0, -2,  1, -1, -2,  3,  2,  3,  3,  7],
           [ 1,  0,  1, -1, -2,  4,  2,  0,  2,  0],
           [-1,  0, -1,  1,  3, -1, -2,  7,  5,  1],
           [ 1,  1,  1, -1, -2,  8,  3,  2,  8, -6],
           [ 0,  2, -1,  1,  2, -1, -2,  2,  2, -6],
           [ 1,  3,  0,  0,  1,  3,  0,  0,  3, -8]]
M = matrix(QQ, entries)
M

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not M.is_singular()

A totally random vector with 10 entries:

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v = random_vector(ZZ, 10, x=-9, y=9)
v

Demonstration 2.

By Theorem CNMB, the columns of the matrix are a basis of $\mathbb{C}^{10}\text{.}$ So the vector v should be a linear combination of the columns of the matrix. Verify this fact in three ways.

First, the old-fashioned way, thus exposing Theorem NMUS.
Then, the modern way, with an inverse, since a nonsingular matrix is invertible, thus exposing Theorem SNCM.
Finally, the Sage way, as described below.

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aug = M.augment(v)
aug.rref()

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M.inverse()*v

The Sage way: first create a space with a user basis.

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X = (QQ^10).subspace_with_basis(M.columns())
X

Sage still carries an echelonized basis, in addition to the user-installed basis.

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X.basis()

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X.echelonized_basis()

Now ask for a coordinatization, relative to the basis in X, thus exposing Theorem VRRB.

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X.coordinates(v)

Section 3 Orthonormal Bases

A particularly simple orthonormal basis of $\mathbb{C}^3\text{,}$ collected into the set S.

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v1 = vector(QQ, [1/3, 2/3, 2/3])
v2 = vector(QQ, [2/3, -2/3, 1/3])
v3 = vector(QQ, [2/3, 1/3, -2/3])
S = [v1, v2, v3]

Demonstration 3.

If these vectors are an orthonormal basis, then as the columns of a matrix they should create an orthonormal basis.

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Q = column_matrix(S)
Q

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Q.conjugate_transpose()*Q

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Q.is_unitary()

Demonstration 4.

Build a random vector of size $3$ and find our ways to express the vector as a (unique) linear combination of the basis vectors. Which method is most efficient?

A totally random vector with 3 entries.

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v = random_vector(ZZ, 3, x=-9, y=9)
v

First, the old-fashioned way, thus exposing Theorem NMUS.

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aug = Q.augment(v)
aug.rref()

Now, the modern way, with an inverse, since a nonsingular matrix is invertible, thus exposing Theorem SNCM.

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Q.inverse()*v

The Sage way. Create a space with a “user basis” and ask for a coordinatization, thus exposing Theorem VRRB.

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X = (QQ^3).subspace_with_basis(Q.columns())
X.coordinates(v)

Finally, exploiting the orthonormal basis, and computing scalars for the linear combination with an inner product, thus exposing Theorem COB. (Sage's .inner\_product() does not conjugate the entries of either vector, so we use the more careful .hermitian\_inner\_product() vector method instead.)

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a1 = v1.hermitian_inner_product(v)
a2 = v2.hermitian_inner_product(v)
a3 = v3.hermitian_inner_product(v)
a1, a2, a3