Tutorial 4 (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 2¶

Let $u_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \\ 3 \end{pmatrix}$, $u_2 = \begin{pmatrix} 3 \\ -1 \\ 5 \\ 2 \end{pmatrix}$, and $u_3 = \begin{pmatrix} -1 \\ 0 \\ 2 \\ 1 \end{pmatrix}$.

(a)¶

If possible, express each of the following vectors as a linear combination of $u_1$, $u_2$, $u_3$.

(i) $\begin{pmatrix} 2 \\ 3 \\ -7 \\ 3 \end{pmatrix}$

(ii) $\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$

(iii) $\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}$

(iv) $\begin{pmatrix} -4 \\ 6 \\ -13 \\ 4 \end{pmatrix}$

In [2]:

U = Matrix.from_str("2 1 0 3; 3 -1 5 2; -1 0 2 1").T
U

U = Matrix.from_str("2 1 0 3; 3 -1 5 2; -1 0 2 1").T U

Out[2]:

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

In [3]:

U.column_constraints(verbosity=1)

U.column_constraints(verbosity=1)

Before RREF: [self | vec]

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

After RREF

$\displaystyle \left[\begin{array}{ccc|c}1 & 0 & 0 & - \frac{x_{2}}{5} - \frac{x_{3}}{5} + \frac{2 x_{4}}{5}\\0 & 1 & 0 & - \frac{6 x_{2}}{5} - \frac{x_{3}}{5} + \frac{2 x_{4}}{5}\\0 & 0 & 1 & 3 x_{2} + x_{3} - x_{4}\\0 & 0 & 0 & x_{1} + 7 x_{2} + 2 x_{3} - 3 x_{4}\end{array}\right]$

For the system to be consistent, the following constraints must be satisfied.

$\displaystyle x_{1} + 7 x_{2} + 2 x_{3} - 3 x_{4} = 0$

Out[3]:

$\displaystyle \left[\begin{array}{ccc|c}1 & 0 & 0 & - \frac{x_{2}}{5} - \frac{x_{3}}{5} + \frac{2 x_{4}}{5}\\0 & 1 & 0 & - \frac{6 x_{2}}{5} - \frac{x_{3}}{5} + \frac{2 x_{4}}{5}\\0 & 0 & 1 & 3 x_{2} + x_{3} - x_{4}\\0 & 0 & 0 & x_{1} + 7 x_{2} + 2 x_{3} - 3 x_{4}\end{array}\right]$

In [4]:

# Alternative Method based on RREF
rhs = Matrix.from_str("2 3 -7 3; 0 0 0 0; 1 1 1 1; -4 6 -13 4").T
rhs = Matrix(rhs, aug_pos=range(3))
rhs

# Alternative Method based on RREF rhs = Matrix.from_str("2 3 -7 3; 0 0 0 0; 1 1 1 1; -4 6 -13 4").T rhs = Matrix(rhs, aug_pos=range(3)) rhs

Out[4]:

$\displaystyle \left[\begin{array}{c|c|c|c}2 & 0 & 1 & -4\\3 & 0 & 1 & 6\\-7 & 0 & 1 & -13\\3 & 0 & 1 & 4\end{array}\right]$

In [5]:

aug_mat = U.row_join(rhs)
aug_mat

aug_mat = U.row_join(rhs) aug_mat

Out[5]:

$\displaystyle \left[\begin{array}{ccc|c|c|c|c}2 & 3 & -1 & 2 & 0 & 1 & -4\\1 & -1 & 0 & 3 & 0 & 1 & 6\\0 & 5 & 2 & -7 & 0 & 1 & -13\\3 & 2 & 1 & 3 & 0 & 1 & 4\end{array}\right]$

In [6]:

aug_mat.rref()

aug_mat.rref()

Out[6]:

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

(b)¶

Is it possible to find 2 vectors $v_1$ and $v_2$ such that they are not a multiple of each other, and both are not a linear combination of $u_1$, $u_2$, $u_3$?

In [7]:

E = U.extend_basis(verbosity=2)
E

E = U.extend_basis(verbosity=2) E

Before RREF: [self | span_subspace]

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

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc|c|c|c|c}1 & 0 & 0 & 0 & - \frac{1}{5} & - \frac{1}{5} & \frac{2}{5}\\0 & 1 & 0 & 0 & - \frac{6}{5} & - \frac{1}{5} & \frac{2}{5}\\0 & 0 & 1 & 0 & 3 & 1 & -1\\0 & 0 & 0 & 1 & 7 & 2 & -3\end{array}\right], \quad\text{pivots} = (0, 1, 2, 3)\right\}$

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

Out[7]:

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

In [8]:

v_1 = E.select_cols(-1) # Not in subspace spanned by U
v_2 = v_1 + U.select_cols(0) # Not a multiple of v_1 or in subspace spanned by U
v_1, v_2

v_1 = E.select_cols(-1) # Not in subspace spanned by U v_2 = v_1 + U.select_cols(0) # Not a multiple of v_1 or in subspace spanned by U v_1, v_2

Out[8]:

$\displaystyle \left( \left[\begin{array}{c}1\\0\\0\\0\end{array}\right], \ \left[\begin{array}{c}3\\1\\0\\3\end{array}\right]\right)$

Question 3¶

Let $V = \left\{ \begin{pmatrix} x \\ y \\ z \end{pmatrix} \middle| x - y - z = 0 \right\}$ be a subset of $\mathbb{R}^3$.

(a)¶

Let $S = \left\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 5 \\ 2 \\ 3 \end{pmatrix} \right\}$. Show that $\text{span}(S) = V$.

In [9]:

V = Matrix([[1, -1, -1]]).nullspace()
V

V = Matrix([[1, -1, -1]]).nullspace() V

Out[9]:

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

In [10]:

S = Matrix.from_str("1 1 0; 5 2 3").T

S.is_same_subspace(Matrix.from_list(V), verbosity=2)

S = Matrix.from_str("1 1 0; 5 2 3").T S.is_same_subspace(Matrix.from_list(V), verbosity=2)

Check if span(self) is subspace of span(other), and vice versa.
Check if span(self) is subspace of span(other)

Before RREF: [other | self]

$\displaystyle \left[\begin{array}{cc|cc}1 & 1 & 1 & 5\\1 & 0 & 1 & 2\\0 & 1 & 0 & 3\end{array}\right]$

After RREF:

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

Span(self) is a subspace of span(other).

Check if span(self) is subspace of span(other)

Before RREF: [other | self]

$\displaystyle \left[\begin{array}{cc|cc}1 & 5 & 1 & 1\\1 & 2 & 1 & 0\\0 & 3 & 0 & 1\end{array}\right]$

After RREF:

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

Span(self) is a subspace of span(other).

Out[10]:

True

(b)¶

Let $T = S \cup \left\{\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\right\}$. Show that $\text{span}(T) = \mathbb{R}^3$.

In [11]:

S

S

Out[11]:

$\displaystyle \left[\begin{array}{cc}1 & 5\\1 & 2\\0 & 3\end{array}\right]$

In [12]:

T = S.row_join(Matrix.from_str("0 0 1").T)
T.rm_aug_line()

# If no `other` subspace is provided, it is assumed to be the entire R^n
T.is_same_subspace(verbosity=2)

T = S.row_join(Matrix.from_str("0 0 1").T) T.rm_aug_line() # If no `other` subspace is provided, it is assumed to be the entire R^n T.is_same_subspace(verbosity=2)

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

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

After RREF:

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

Out[12]:

True

Question 4¶

Which of the following sets $S$ spans $\mathbb{R}^4$?

(i)¶

$S = \left\{\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \\ 0 \end{pmatrix}\right\}$.

In [13]:

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

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

Out[13]:

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

In [14]:

S.is_same_subspace(verbosity=2)

S.is_same_subspace(verbosity=2)

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

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

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

(ii)¶

$S = \left\{\begin{pmatrix} 1 \\ 2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}\right\}$

In [15]:

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

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

Out[15]:

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

In [16]:

S.is_same_subspace(verbosity=2)

S.is_same_subspace(verbosity=2)

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

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

After RREF:

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

Out[16]:

False

(iii)¶

$S = \left\{\begin{pmatrix} 6 \\ 4 \\ -2 \\ 4 \end{pmatrix}, \begin{pmatrix} 2 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 3 \\ 2 \\ -1 \\ 2 \end{pmatrix}, \begin{pmatrix} 5 \\ 6 \\ -3 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 4 \\ -2 \\ -1 \end{pmatrix}\right\}$.

In [17]:

S = Matrix.from_str("6 4 -2 4; 2 0 0 1; 3 2 -1 2; 5 6 -3 2; 0 4 -2 -1").T
S

S = Matrix.from_str("6 4 -2 4; 2 0 0 1; 3 2 -1 2; 5 6 -3 2; 0 4 -2 -1").T S

Out[17]:

$\displaystyle \left[\begin{array}{ccccc}6 & 2 & 3 & 5 & 0\\4 & 0 & 2 & 6 & 4\\-2 & 0 & -1 & -3 & -2\\4 & 1 & 2 & 2 & -1\end{array}\right]$

In [18]:

S.is_same_subspace(verbosity=2)

S.is_same_subspace(verbosity=2)

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

$\displaystyle \left[\begin{array}{ccccc}6 & 2 & 3 & 5 & 0\\4 & 0 & 2 & 6 & 4\\-2 & 0 & -1 & -3 & -2\\4 & 1 & 2 & 2 & -1\end{array}\right]$

After RREF:

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

Out[18]:

False

(iv)¶

$S = \left\{\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix}\right\}$.

In [19]:

S = Matrix.from_str("1 1 0 0; 1 2 -1 1; 0 0 1 1; 2 1 2 1; 1 2 3 4").T
S

S = Matrix.from_str("1 1 0 0; 1 2 -1 1; 0 0 1 1; 2 1 2 1; 1 2 3 4").T S

Out[19]:

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

In [20]:

S.is_same_subspace(verbosity=2)

S.is_same_subspace(verbosity=2)

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

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

After RREF:

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

Out[20]:

True

Question 5¶

Determine whether $\text{span}\{u_1, u_2, u_3\} \subseteq \text{span}\{v_1, v_2\}$ and/or $\text{span}\{v_1, v_2\} \subseteq \text{span}\{u_1, u_2, u_3\}$ if

(a)¶

$u_1 = \begin{pmatrix} 2 \\ -2 \\ 0 \end{pmatrix}$, $u_2 = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix}$, $u_3 = \begin{pmatrix} 0 \\ 0 \\ 9 \end{pmatrix}$, $v_1 = \begin{pmatrix} 1 \\ -1 \\ -5 \end{pmatrix}$, $v_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$.

In [21]:

U = Matrix.from_str("2 -2 0; -1 1 -1; 0 0 9").T
V = Matrix.from_str("1 -1 -5; 0 1 1").T
U, V

U = Matrix.from_str("2 -2 0; -1 1 -1; 0 0 9").T V = Matrix.from_str("1 -1 -5; 0 1 1").T U, V

Out[21]:

$\displaystyle \left( \left[\begin{array}{ccc}2 & -1 & 0\\-2 & 1 & 0\\0 & -1 & 9\end{array}\right], \ \left[\begin{array}{cc}1 & 0\\-1 & 1\\-5 & 1\end{array}\right]\right)$

In [22]:

U.is_same_subspace(V, verbosity=2)

U.is_same_subspace(V, verbosity=2)

Check if span(self) is subspace of span(other), and vice versa.
Check if span(self) is subspace of span(other)

Before RREF: [other | self]

$\displaystyle \left[\begin{array}{cc|ccc}1 & 0 & 2 & -1 & 0\\-1 & 1 & -2 & 1 & 0\\-5 & 1 & 0 & -1 & 9\end{array}\right]$

After RREF:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cc|ccc}1 & 0 & 0 & \frac{1}{5} & - \frac{9}{5}\\0 & 1 & 0 & 0 & 0\\0 & 0 & 1 & - \frac{3}{5} & \frac{9}{10}\end{array}\right], \quad\text{pivots} = (0, 1, 2)\right\}$

Span(self) is not a subspace of span(other).

Out[22]:

False

(b)¶

$u_1 = \begin{pmatrix} 1 \\ 6 \\ 4 \end{pmatrix}$, $u_2 = \begin{pmatrix} 2 \\ 4 \\ -1 \end{pmatrix}$, $u_3 = \begin{pmatrix} -1 \\ 2 \\ 5 \end{pmatrix}$, $v_1 = \begin{pmatrix} 1 \\ -2 \\ -5 \end{pmatrix}$, $v_2 = \begin{pmatrix} 0 \\ 8 \\ 9 \end{pmatrix}$.

In [23]:

U = Matrix.from_str("1 6 4; 2 4 -1; -1 2 5").T
V = Matrix.from_str("1 -2 -5; 0 8 9").T
U, V

U = Matrix.from_str("1 6 4; 2 4 -1; -1 2 5").T V = Matrix.from_str("1 -2 -5; 0 8 9").T U, V

Out[23]:

$\displaystyle \left( \left[\begin{array}{ccc}1 & 2 & -1\\6 & 4 & 2\\4 & -1 & 5\end{array}\right], \ \left[\begin{array}{cc}1 & 0\\-2 & 8\\-5 & 9\end{array}\right]\right)$

In [24]:

U.is_same_subspace(V, verbosity=2)

U.is_same_subspace(V, verbosity=2)

Check if span(self) is subspace of span(other), and vice versa.
Check if span(self) is subspace of span(other)

Before RREF: [other | self]

$\displaystyle \left[\begin{array}{cc|ccc}1 & 0 & 1 & 2 & -1\\-2 & 8 & 6 & 4 & 2\\-5 & 9 & 4 & -1 & 5\end{array}\right]$

After RREF:

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

Span(self) is a subspace of span(other).

Check if span(self) is subspace of span(other)

Before RREF: [other | self]

$\displaystyle \left[\begin{array}{ccc|cc}1 & 2 & -1 & 1 & 0\\6 & 4 & 2 & -2 & 8\\4 & -1 & 5 & -5 & 9\end{array}\right]$

After RREF:

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

Span(self) is a subspace of span(other).

Out[24]:

True