Tutorial 6 (AY24/25 Sem 1)¶

In [1]:

# Required imports
import sympy as sym
from ma1522 import Matrix

# Required imports import sympy as sym from ma1522 import Matrix

Question 1¶

(a)¶

Let $\mathbf{u}_1 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$, $\mathbf{u}_2 = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$, $\mathbf{u}_3 = \begin{pmatrix} 0 \\ -1 \\ 3 \end{pmatrix}$. Show that $S = \{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}$ forms a basis for $\mathbb{R}^3$.

In [2]:

S = Matrix.from_str("1 2 -1; 0 2 1; 0 -1 3").T
S

S = Matrix.from_str("1 2 -1; 0 2 1; 0 -1 3").T S

Out[2]:

$\displaystyle \left[\begin{array}{ccc}1 & 0 & 0\\2 & 2 & -1\\-1 & 1 & 3\end{array}\right]$

In [3]:

S.get_linearly_independent_vectors(verbosity=1)

S.get_linearly_independent_vectors(verbosity=1)

Before RREF: [self]

$\displaystyle \left[\begin{array}{ccc}1 & 0 & 0\\2 & 2 & -1\\-1 & 1 & 3\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

Select columns of self corresponding to pivot positions.

Out[3]:

$\displaystyle \left[\begin{array}{ccc}1 & 0 & 0\\2 & 2 & -1\\-1 & 1 & 3\end{array}\right]$

(b)¶

Suppose $\mathbf{w} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$. Find the coordinate vector of $\mathbf{w}$ relative to $S$.

In [4]:

w = Matrix.from_str("1; 1; 1")

w.coords_relative(S, verbosity=2)

w = Matrix.from_str("1; 1; 1") w.coords_relative(S, verbosity=2)

Before RREF: [to | self]

$\displaystyle \left[\begin{array}{ccc|c}1 & 0 & 0 & 1\\2 & 2 & -1 & 1\\-1 & 1 & 3 & 1\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{ccc|c}1 & 0 & 0 & 1\\0 & 1 & 0 & - \frac{1}{7}\\0 & 0 & 1 & \frac{5}{7}\end{array}\right]$

Out[4]:

$\displaystyle \left[\begin{array}{c}1\\- \frac{1}{7}\\\frac{5}{7}\end{array}\right]$

(c)¶

Let $T = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ be another basis for $\mathbb{R}^3$ where $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 5 \\ 4 \end{pmatrix}$, $\mathbf{v}_2 = \begin{pmatrix} -1 \\ 3 \\ 7 \end{pmatrix}$, $\mathbf{v}_3 = \begin{pmatrix} 2 \\ 2 \\ 4 \end{pmatrix}$. Find the transition matrix from $T$ to $S$.

In [5]:

T = Matrix.from_str("1 5 4; -1 3 7; 2 2 4").T

P_T_to_S = T.transition_matrix(S, verbosity=2)
P_T_to_S

T = Matrix.from_str("1 5 4; -1 3 7; 2 2 4").T P_T_to_S = T.transition_matrix(S, verbosity=2) P_T_to_S

Before RREF: [to | self]

$\displaystyle \left[\begin{array}{ccc|ccc}1 & 0 & 0 & 1 & -1 & 2\\2 & 2 & -1 & 5 & 3 & 2\\-1 & 1 & 3 & 4 & 7 & 4\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{ccc|ccc}1 & 0 & 0 & 1 & -1 & 2\\0 & 1 & 0 & 2 & 3 & 0\\0 & 0 & 1 & 1 & 1 & 2\end{array}\right]$

Out[5]:

$\displaystyle \left[\begin{matrix}1 & -1 & 2\\2 & 3 & 0\\1 & 1 & 2\end{matrix}\right]$

(d)¶

Find the transition matrix from $S$ to $T$.

In [6]:

P_S_to_T = S.transition_matrix(T, verbosity=2)
P_S_to_T

P_S_to_T = S.transition_matrix(T, verbosity=2) P_S_to_T

Before RREF: [to | self]

$\displaystyle \left[\begin{array}{ccc|ccc}1 & -1 & 2 & 1 & 0 & 0\\5 & 3 & 2 & 2 & 2 & -1\\4 & 7 & 4 & -1 & 1 & 3\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{ccc|ccc}1 & 0 & 0 & \frac{3}{4} & \frac{1}{2} & - \frac{3}{4}\\0 & 1 & 0 & - \frac{1}{2} & 0 & \frac{1}{2}\\0 & 0 & 1 & - \frac{1}{8} & - \frac{1}{4} & \frac{5}{8}\end{array}\right]$

Out[6]:

$\displaystyle \left[\begin{matrix}\frac{3}{4} & \frac{1}{2} & - \frac{3}{4}\\- \frac{1}{2} & 0 & \frac{1}{2}\\- \frac{1}{8} & - \frac{1}{4} & \frac{5}{8}\end{matrix}\right]$

In [7]:

## Alternatively
P_S_to_T = P_T_to_S.inverse()
P_S_to_T

## Alternatively P_S_to_T = P_T_to_S.inverse() P_S_to_T

Out[7]:

$\displaystyle \left[\begin{array}{ccc}\frac{3}{4} & \frac{1}{2} & - \frac{3}{4}\\- \frac{1}{2} & 0 & \frac{1}{2}\\- \frac{1}{8} & - \frac{1}{4} & \frac{5}{8}\end{array}\right]$

(e)¶

Use the vector $\mathbf{w}$ in Part (b). Find the coordinate vector of $\mathbf{w}$ relative to $T$.

In [8]:

w.coords_relative(T, verbosity=2)

w.coords_relative(T, verbosity=2)

Before RREF: [to | self]

$\displaystyle \left[\begin{array}{ccc|c}1 & -1 & 2 & 1\\5 & 3 & 2 & 1\\4 & 7 & 4 & 1\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{ccc|c}1 & 0 & 0 & \frac{1}{7}\\0 & 1 & 0 & - \frac{1}{7}\\0 & 0 & 1 & \frac{5}{14}\end{array}\right]$

Out[8]:

$\displaystyle \left[\begin{array}{c}\frac{1}{7}\\- \frac{1}{7}\\\frac{5}{14}\end{array}\right]$

Question 3¶

(a)¶

Let $\mathbf{A} = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & -1 & 1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$. Is $\mathbf{b}$ in the column space of $\mathbf{A}$? If it is, express it as a linear combination of the columns of $\mathbf{A}$.

In [9]:

A = Matrix.from_str("1 -1 1; 1 1 -1; -1 -1 1")
b = Matrix.from_str("2; 1; 0")

try:
    b.coords_relative(A, verbosity=2)
except ValueError as e:
    print(f"Error: {e}") # b is not in the column space of A

A = Matrix.from_str("1 -1 1; 1 1 -1; -1 -1 1") b = Matrix.from_str("2; 1; 0") try: b.coords_relative(A, verbosity=2) except ValueError as e: print(f"Error: {e}") # b is not in the column space of A

Before RREF: [to | self]

$\displaystyle \left[\begin{array}{ccc|c}1 & -1 & 1 & 2\\1 & 1 & -1 & 1\\-1 & -1 & 1 & 0\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{ccc|c}1 & 0 & 0 & 0\\0 & 1 & -1 & 0\\0 & 0 & 0 & 1\end{array}\right]$

Error: No solution found due to inconsistent system.

(b)¶

Let $\mathbf{A} = \begin{pmatrix} 1 & 9 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}5 & 1 & -1\end{pmatrix}$. Is $\mathbf{b}$ in the row space of $\mathbf{A}$? If it is, express it as a linear combination of the rows of $\mathbf{A}$.

In [10]:

A = Matrix.from_str("1 9 1; -1 3 1; 1 1 1")
b = Matrix.from_str("5 1 -1")

scalars = b.T.coords_relative(A.T, verbosity=2)

A = Matrix.from_str("1 9 1; -1 3 1; 1 1 1") b = Matrix.from_str("5 1 -1") scalars = b.T.coords_relative(A.T, verbosity=2)

Before RREF: [to | self]

$\displaystyle \left[\begin{array}{ccc|c}1 & -1 & 1 & 5\\9 & 3 & 1 & 1\\1 & 1 & 1 & -1\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{ccc|c}1 & 0 & 0 & 1\\0 & 1 & 0 & -3\\0 & 0 & 1 & 1\end{array}\right]$

In [11]:

scalars # coordinates relative gives the coefficients of the linear combination

scalars # coordinates relative gives the coefficients of the linear combination

Out[11]:

$\displaystyle \left[\begin{array}{c}1\\-3\\1\end{array}\right]$

(c)¶

Let $\mathbf{A} = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 2 & 1 \\ 1 & 2 & 1 & 3 \\ 0 & 1 & 2 & 2 \end{pmatrix}$. Is the row space and column space of $\mathbf{A}$ the whole $\mathbb{R}^4$?

In [12]:

A = Matrix.from_str("1 2 0 1; 0 1 2 1; 1 2 1 3; 0 1 2 2")
A

A = Matrix.from_str("1 2 0 1; 0 1 2 1; 1 2 1 3; 0 1 2 2") A

Out[12]:

$\displaystyle \left[\begin{array}{cccc}1 & 2 & 0 & 1\\0 & 1 & 2 & 1\\1 & 2 & 1 & 3\\0 & 1 & 2 & 2\end{array}\right]$

In [13]:

A.is_same_subspace(verbosity=2) # other defaults to R^n

A.is_same_subspace(verbosity=2) # other defaults to R^n

Check rref(self) does not have zero rows
Before RREF: self

$\displaystyle \left[\begin{array}{cccc}1 & 2 & 0 & 1\\0 & 1 & 2 & 1\\1 & 2 & 1 & 3\\0 & 1 & 2 & 2\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{cccc}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right]$

Out[13]:

True

In [14]:

A.T.is_same_subspace(verbosity=2) # other defaults to R^n

A.T.is_same_subspace(verbosity=2) # other defaults to R^n

Check rref(self) does not have zero rows
Before RREF: self

$\displaystyle \left[\begin{array}{cccc}1 & 0 & 1 & 0\\2 & 1 & 2 & 1\\0 & 2 & 1 & 2\\1 & 1 & 3 & 2\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{cccc}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right]$

Out[14]:

True

Problem 4¶

For each of the following matrices $\mathbf{A}$,

(i) Find a basis for the row space of $\mathbf{A}$.

(ii) Find a basis for the column space of $\mathbf{A}$.

(iii) Find a basis for the nullspace of $\mathbf{A}$.

(iv) Hence determine $\text{rank}(\mathbf{A})$, $\text{nullity}(\mathbf{A})$ and verify the dimension theorem for matrices.

(v) Is $\mathbf{A}$ full rank?

(a)¶

$\mathbf{A} = \begin{pmatrix} 1 & 2 & 5 & 3 \\ 1 & -4 & -1 & -9 \\ -1 & 0 & -3 & 1 \\ 2 & 1 & 7 & 0 \\ 0 & 1 & 1 & 2 \end{pmatrix}$

In [15]:

A = Matrix.from_str("1 2 5 3; 1 -4 -1 -9; -1 0 -3 1; 2 1 7 0; 0 1 1 2")
A

A = Matrix.from_str("1 2 5 3; 1 -4 -1 -9; -1 0 -3 1; 2 1 7 0; 0 1 1 2") A

Out[15]:

$\displaystyle \left[\begin{array}{cccc}1 & 2 & 5 & 3\\1 & -4 & -1 & -9\\-1 & 0 & -3 & 1\\2 & 1 & 7 & 0\\0 & 1 & 1 & 2\end{array}\right]$

In [16]:

# Row space
A.get_linearly_independent_vectors(colspace=False, verbosity=2) 

# Row space A.get_linearly_independent_vectors(colspace=False, verbosity=2)

Before RREF: [self^T]

$\displaystyle \left[\begin{array}{ccccc}1 & 1 & -1 & 2 & 0\\2 & -4 & 0 & 1 & 1\\5 & -1 & -3 & 7 & 1\\3 & -9 & 1 & 0 & 2\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccccc}1 & 0 & - \frac{2}{3} & \frac{3}{2} & \frac{1}{6}\\0 & 1 & - \frac{1}{3} & \frac{1}{2} & - \frac{1}{6}\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Select rows of self corresponding to pivot positions.

Out[16]:

$\displaystyle \left[\begin{array}{cccc}1 & 2 & 5 & 3\\1 & -4 & -1 & -9\end{array}\right]$

In [17]:

# Column space
A.get_linearly_independent_vectors(colspace=True, verbosity=2)

# Column space A.get_linearly_independent_vectors(colspace=True, verbosity=2)

Before RREF: [self]

$\displaystyle \left[\begin{array}{cccc}1 & 2 & 5 & 3\\1 & -4 & -1 & -9\\-1 & 0 & -3 & 1\\2 & 1 & 7 & 0\\0 & 1 & 1 & 2\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc}1 & 0 & 3 & -1\\0 & 1 & 1 & 2\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Select columns of self corresponding to pivot positions.

Out[17]:

$\displaystyle \left[\begin{array}{cc}1 & 2\\1 & -4\\-1 & 0\\2 & 1\\0 & 1\end{array}\right]$

In [18]:

A.nullspace(verbosity=2)

A.nullspace(verbosity=2)

Before RREF: [self]

$\displaystyle \left[\begin{array}{cccc}1 & 2 & 5 & 3\\1 & -4 & -1 & -9\\-1 & 0 & -3 & 1\\2 & 1 & 7 & 0\\0 & 1 & 1 & 2\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc}1 & 0 & 3 & -1\\0 & 1 & 1 & 2\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Out[18]:

$\displaystyle \left[ \left[\begin{matrix}-3\\-1\\1\\0\end{matrix}\right], \ \left[\begin{matrix}1\\-2\\0\\1\end{matrix}\right]\right]$

In [19]:

A.rank(), A.nullity()

A.rank(), A.nullity()

Out[19]:

$\displaystyle \left( 2, \ 2\right)$

(b)¶

$\mathbf{A} = \begin{pmatrix} 1 & 3 & 7 \\ 2 & 1 & 8 \\ 3 & -5 & -1 \\ 2 & -2 & 2 \\ 1 & 1 & 5 \end{pmatrix}$

In [20]:

A = Matrix.from_str("1 3 7; 2 1 8; 3 -5 -1; 2 -2 2; 1 1 5")
A

A = Matrix.from_str("1 3 7; 2 1 8; 3 -5 -1; 2 -2 2; 1 1 5") A

Out[20]:

$\displaystyle \left[\begin{array}{ccc}1 & 3 & 7\\2 & 1 & 8\\3 & -5 & -1\\2 & -2 & 2\\1 & 1 & 5\end{array}\right]$

In [21]:

# Alternatively, you can use `simplify_basis` to get a simple basis
# that is not necessarily a row or column of the original matrix.

rowspace = A.simplify_basis(colspace=False, verbosity=2)
rowspace

# Alternatively, you can use `simplify_basis` to get a simple basis # that is not necessarily a row or column of the original matrix. rowspace = A.simplify_basis(colspace=False, verbosity=2) rowspace

Before RREF: self

$\displaystyle \left[\begin{array}{ccc}1 & 3 & 7\\2 & 1 & 8\\3 & -5 & -1\\2 & -2 & 2\\1 & 1 & 5\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\\0 & 0 & 0\end{array}\right]$

Out[21]:

$\displaystyle \left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]$

In [22]:

colspace = A.simplify_basis(colspace=True, verbosity=2)
colspace

colspace = A.simplify_basis(colspace=True, verbosity=2) colspace

Before RREF: self^T

$\displaystyle \left[\begin{array}{ccccc}1 & 2 & 3 & 2 & 1\\3 & 1 & -5 & -2 & 1\\7 & 8 & -1 & 2 & 5\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{ccccc}1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & \frac{4}{13} & \frac{8}{13}\\0 & 0 & 1 & \frac{6}{13} & - \frac{1}{13}\end{array}\right]$

Out[22]:

$\displaystyle \left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\0 & \frac{4}{13} & \frac{6}{13}\\0 & \frac{8}{13} & - \frac{1}{13}\end{array}\right]$

In [23]:

A.nullspace(verbosity=2) # UserWarning if the nullspace is trivial

A.nullspace(verbosity=2) # UserWarning if the nullspace is trivial

Before RREF: [self]

$\displaystyle \left[\begin{array}{ccc}1 & 3 & 7\\2 & 1 & 8\\3 & -5 & -1\\2 & -2 & 2\\1 & 1 & 5\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

/tmp/ipykernel_2385/40416061.py:1: UserWarning: Only trivial nullspace (0-vector) detected!
  A.nullspace(verbosity=2) # UserWarning if the nullspace is trivial

Out[23]:

$\displaystyle \left[ \right]$

In [24]:

A.rank(), A.nullity()

A.rank(), A.nullity()

Out[24]:

$\displaystyle \left( 3, \ 0\right)$

Question 5¶

Let $W$ be a subspace of $\mathbb{R}^5$ spanned by the following vectors $$\mathbf{u}_1 = \begin{pmatrix} 1 \\ -2 \\ 0 \\ 0 \\ 3 \end{pmatrix}, \quad \mathbf{u}_2 = \begin{pmatrix} 2 \\ -5 \\ -3 \\ -2 \\ 6 \end{pmatrix}, \quad \mathbf{u}_3 = \begin{pmatrix} 0 \\ 5 \\ 15 \\ 10 \\ 0 \end{pmatrix}, \quad \mathbf{u}_4 = \begin{pmatrix} 2 \\ 1 \\ 15 \\ 8 \\ 6 \end{pmatrix}$$

(a)¶

Find a basis for $W$.

In [25]:

W = Matrix.from_str("1 -2 0 0 3; 2 -5 -3 -2 6; 0 5 15 10 0; 2 1 15 8 6").T
W

W = Matrix.from_str("1 -2 0 0 3; 2 -5 -3 -2 6; 0 5 15 10 0; 2 1 15 8 6").T W

Out[25]:

$\displaystyle \left[\begin{array}{cccc}1 & 2 & 0 & 2\\-2 & -5 & 5 & 1\\0 & -3 & 15 & 15\\0 & -2 & 10 & 8\\3 & 6 & 0 & 6\end{array}\right]$

In [26]:

W.get_linearly_independent_vectors(colspace=True, verbosity=2)

W.get_linearly_independent_vectors(colspace=True, verbosity=2)

Before RREF: [self]

$\displaystyle \left[\begin{array}{cccc}1 & 2 & 0 & 2\\-2 & -5 & 5 & 1\\0 & -3 & 15 & 15\\0 & -2 & 10 & 8\\3 & 6 & 0 & 6\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc}1 & 0 & 10 & 0\\0 & 1 & -5 & 0\\0 & 0 & 0 & 1\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1, 3)\right\}$

Select columns of self corresponding to pivot positions.

Out[26]:

$\displaystyle \left[\begin{array}{ccc}1 & 2 & 2\\-2 & -5 & 1\\0 & -3 & 15\\0 & -2 & 8\\3 & 6 & 6\end{array}\right]$

In [27]:

# OR
W.simplify_basis(colspace=True, verbosity=2)

# OR W.simplify_basis(colspace=True, verbosity=2)

Before RREF: self^T

$\displaystyle \left[\begin{array}{ccccc}1 & -2 & 0 & 0 & 3\\2 & -5 & -3 & -2 & 6\\0 & 5 & 15 & 10 & 0\\2 & 1 & 15 & 8 & 6\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{ccccc}1 & 0 & 6 & 0 & 3\\0 & 1 & 3 & 0 & 0\\0 & 0 & 0 & 1 & 0\\0 & 0 & 0 & 0 & 0\end{array}\right]$

Out[27]:

$\displaystyle \left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\6 & 3 & 0\\0 & 0 & 1\\3 & 0 & 0\end{array}\right]$

(c)¶

Extend the basis $W$ found in (a) to a basis for $\mathbb{R}^5$.

In [28]:

W.extend_basis(verbosity=2)  # span_subspace=None extends to R^n by default

W.extend_basis(verbosity=2) # span_subspace=None extends to R^n by default

Before RREF: [self | span_subspace]

$\displaystyle \left[\begin{array}{cccc|ccccc}1 & 2 & 0 & 2 & 1 & 0 & 0 & 0 & 0\\-2 & -5 & 5 & 1 & 0 & 1 & 0 & 0 & 0\\0 & -3 & 15 & 15 & 0 & 0 & 1 & 0 & 0\\0 & -2 & 10 & 8 & 0 & 0 & 0 & 1 & 0\\3 & 6 & 0 & 6 & 0 & 0 & 0 & 0 & 1\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cccc|ccccc}1 & 0 & 10 & 0 & 0 & 0 & - \frac{10}{3} & 6 & \frac{1}{3}\\0 & 1 & -5 & 0 & 0 & 0 & \frac{4}{3} & - \frac{5}{2} & 0\\0 & 0 & 0 & 1 & 0 & 0 & \frac{1}{3} & - \frac{1}{2} & 0\\0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & - \frac{1}{3}\\0 & 0 & 0 & 0 & 0 & 1 & - \frac{1}{3} & 0 & \frac{2}{3}\end{array}\right], \quad\text{pivots} = (0, 1, 3, 4, 5)\right\}$

Select columns of rref([self | span_subspace]) corresponding to pivot positions.

Out[28]:

$\displaystyle \left[\begin{array}{ccccc}1 & 2 & 2 & 1 & 0\\-2 & -5 & 1 & 0 & 1\\0 & -3 & 15 & 0 & 0\\0 & -2 & 8 & 0 & 0\\3 & 6 & 6 & 0 & 0\end{array}\right]$

In [29]:

# Alternatively, use the `simplify_basis` method first for a simpler basis overall
W.simplify_basis(colspace=True, verbosity=0).extend_basis(verbosity=2) 

# Alternatively, use the `simplify_basis` method first for a simpler basis overall W.simplify_basis(colspace=True, verbosity=0).extend_basis(verbosity=2)

Before RREF: [self | span_subspace]

$\displaystyle \left[\begin{array}{ccc|ccccc}1 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & 1 & 0 & 0 & 0\\6 & 3 & 0 & 0 & 0 & 1 & 0 & 0\\0 & 0 & 1 & 0 & 0 & 0 & 1 & 0\\3 & 0 & 0 & 0 & 0 & 0 & 0 & 1\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc|ccccc}1 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{3}\\0 & 1 & 0 & 0 & 0 & \frac{1}{3} & 0 & - \frac{2}{3}\\0 & 0 & 1 & 0 & 0 & 0 & 1 & 0\\0 & 0 & 0 & 1 & 0 & 0 & 0 & - \frac{1}{3}\\0 & 0 & 0 & 0 & 1 & - \frac{1}{3} & 0 & \frac{2}{3}\end{array}\right], \quad\text{pivots} = (0, 1, 2, 3, 4)\right\}$

Select columns of rref([self | span_subspace]) corresponding to pivot positions.

Out[29]:

$\displaystyle \left[\begin{array}{ccccc}1 & 0 & 0 & 1 & 0\\0 & 1 & 0 & 0 & 1\\6 & 3 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0\\3 & 0 & 0 & 0 & 0\end{array}\right]$

Question 6¶

Let $S = \left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \\ 3 \end{pmatrix}, \begin{pmatrix} 2 \\ -1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} -1 \\ 3 \\ 5 \\ 12 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 2 \\ 5 \end{pmatrix}, \begin{pmatrix} 3 \\ -1 \\ 1 \\ 4 \end{pmatrix} \right\}$ and $V = \text{span}(S)$. Find a subset $S' \subseteq S$ such that $S'$ forms a basis for $V$.

In [30]:

V = Matrix.from_str("1 0 1 3; 2 -1 0 1; -1 3 5 12; 0 1 2 5; 3 -1 1 4").T
V

V = Matrix.from_str("1 0 1 3; 2 -1 0 1; -1 3 5 12; 0 1 2 5; 3 -1 1 4").T V

Out[30]:

$\displaystyle \left[\begin{array}{ccccc}1 & 2 & -1 & 0 & 3\\0 & -1 & 3 & 1 & -1\\1 & 0 & 5 & 2 & 1\\3 & 1 & 12 & 5 & 4\end{array}\right]$

In [31]:

S_prime = V.get_linearly_independent_vectors(colspace=True, verbosity=2)
S_prime

S_prime = V.get_linearly_independent_vectors(colspace=True, verbosity=2) S_prime

Before RREF: [self]

$\displaystyle \left[\begin{array}{ccccc}1 & 2 & -1 & 0 & 3\\0 & -1 & 3 & 1 & -1\\1 & 0 & 5 & 2 & 1\\3 & 1 & 12 & 5 & 4\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccccc}1 & 0 & 5 & 2 & 1\\0 & 1 & -3 & -1 & 1\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Select columns of self corresponding to pivot positions.

Out[31]:

$\displaystyle \left[\begin{array}{cc}1 & 2\\0 & -1\\1 & 0\\3 & 1\end{array}\right]$