Tutorial 5 (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¶

For each of the following sets of vectors $S$,

(i) Determine if $S$ is linearly independent.

(ii) If $S$ is linearly dependent, express one of the vectors in $S$ as a linear combination of the others.

(a)¶

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

In [2]:

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

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

Out[2]:

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

In [3]:

S.is_linearly_independent(verbosity=2)

S.is_linearly_independent(verbosity=2)

Before RREF: self

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

After RREF:

$\displaystyle \left[\begin{array}{cccc}1 & 0 & 0 & \frac{9}{2}\\0 & 1 & 0 & \frac{15}{2}\\0 & 0 & 1 & -3\end{array}\right]$

Check if Number of columns (4) == Number of pivot columns (3)

Out[3]:

False

Based on the RREF, the last column can be expressed as a linear combination of the first three columns

(b)¶

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

In [4]:

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

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

Out[4]:

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

In [5]:

S.is_linearly_independent(verbosity=2)

S.is_linearly_independent(verbosity=2)

Before RREF: self

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

After RREF:

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

Check if Number of columns (2) == Number of pivot columns (2)

Out[5]:

True

(c)¶

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

In [6]:

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

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

Out[6]:

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

In [7]:

S.is_linearly_independent(verbosity=2)

S.is_linearly_independent(verbosity=2)

Before RREF: self

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

After RREF:

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

Check if Number of columns (3) == Number of pivot columns (2)

Out[7]:

False

(d)¶

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

In [8]:

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

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

Out[8]:

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

In [9]:

S.is_linearly_independent(verbosity=2)

S.is_linearly_independent(verbosity=2)

Before RREF: self

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

After RREF:

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

Check if Number of columns (3) == Number of pivot columns (3)

Out[9]:

True

Question 2¶

Suppose $u$, $v$, $w$ are linearly independent vectors in $\mathbb{R}^n$. Determine which of the sets $S_1$ to $S_5$ are linearly independent.

(d)¶

$S_4 = \{u, u + v, u + v + w\}$.

In [10]:

# Tip: Use `sympy.solve` to help you find non-trivial solutions

a, b, c = sym.symbols("a b c")
u, v, w = sym.symbols("u v w")

v1 = u
v2 = u + v
v3 = u + v + w

# Solve the system of equations to find the coefficients a, b, c
sols = sym.solve(a*v1 + b*v2 + c*v3, (a, b, c))
sols # only trivial solution exists -> the set is linearly independent

# Tip: Use `sympy.solve` to help you find non-trivial solutions a, b, c = sym.symbols("a b c") u, v, w = sym.symbols("u v w") v1 = u v2 = u + v v3 = u + v + w # Solve the system of equations to find the coefficients a, b, c sols = sym.solve(a*v1 + b*v2 + c*v3, (a, b, c)) sols # only trivial solution exists -> the set is linearly independent

Out[10]:

$\displaystyle \left\{ a : 0, \ b : 0, \ c : 0\right\}$

(e)¶

$S_5 = \{u + v, v + w, u + w, u + v + w\}$.

In [11]:

a, b, c, d = sym.symbols("a b c d")
u, v, w = sym.symbols("u v w")

v1 = u + v
v2 = v + w
v3 = u + w
v4 = u + v + w

# Solve the system of equations to find the coefficients a, b, c, d
sols = sym.solve(a*v1 + b*v2 + c*v3 + d*v4, (a, b, c, d))
sols # non-trivial solution exists -> the set is linearly dependent

a, b, c, d = sym.symbols("a b c d") u, v, w = sym.symbols("u v w") v1 = u + v v2 = v + w v3 = u + w v4 = u + v + w # Solve the system of equations to find the coefficients a, b, c, d sols = sym.solve(a*v1 + b*v2 + c*v3 + d*v4, (a, b, c, d)) sols # non-trivial solution exists -> the set is linearly dependent

Out[11]:

$\displaystyle \left[ \left( - \frac{d}{2}, \ - \frac{d}{2}, \ - \frac{d}{2}, \ d\right)\right]$

Question 3¶

For each of the following subspaces $V$, write down a basis for $V$.

(a)¶

$V = \left\{\begin{pmatrix} a + b \\ a + c \\ c + d \\ b + d \end{pmatrix} \middle| \, a, b, c, d \in \mathbb{R}\right\}$.

In [12]:

V = Matrix.from_str("a+b; a+c; c+d; b+d")
vecs = V.sep_unk()
vecs

V = Matrix.from_str("a+b; a+c; c+d; b+d") vecs = V.sep_unk() vecs

Out[12]:

$\displaystyle \left\{ a : \left[\begin{array}{c}1\\1\\0\\0\end{array}\right], \ b : \left[\begin{array}{c}1\\0\\0\\1\end{array}\right], \ c : \left[\begin{array}{c}0\\1\\1\\0\end{array}\right], \ d : \left[\begin{array}{c}0\\0\\1\\1\end{array}\right]\right\}$

In [13]:

V = Matrix.from_list([*vecs.values()])
V

V = Matrix.from_list([*vecs.values()]) V

Out[13]:

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

In [14]:

# Tip: Use pivots in the RREF to get linearly independent vectors
V.get_linearly_independent_vectors(verbosity=1)

# Tip: Use pivots in the RREF to get linearly independent vectors V.get_linearly_independent_vectors(verbosity=1)

Before RREF: [self]

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

After RREF:

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

Select columns of self corresponding to pivot positions.

Out[14]:

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

(b)¶

$V = \text{span}\left\{\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} 0 \\ 3 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\right\}$.

In [15]:

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

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

Out[15]:

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

In [16]:

V.get_linearly_independent_vectors(verbosity=1)

V.get_linearly_independent_vectors(verbosity=1)

Before RREF: [self]

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

After RREF:

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

Select columns of self corresponding to pivot positions.

Out[16]:

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

(c)¶

$V$ is the solution space of the following homogeneous linear system $$\begin{cases} a_1 && &+& a_3 &+& a_4 &-& a_5 &=& 0 \\ && a_2 &+& a_3 &+& 2a_4 &+& a_5 &=& 0 \\ a_1 &+& a_2 &+& 2a_3 &+& a_4 &-& 2a_5 &=& 0 \end{cases}$$

In [17]:

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

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

Out[17]:

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

In [18]:

V = mat.nullspace()
V

V = mat.nullspace() V

Out[18]:

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

In [19]:

Matrix.from_list(V).get_linearly_independent_vectors(verbosity=1)

Matrix.from_list(V).get_linearly_independent_vectors(verbosity=1)

Before RREF: [self]

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

After RREF:

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

Select columns of self corresponding to pivot positions.

Out[19]:

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

Question 4¶

For what values of $a$ will $u_1 = \begin{pmatrix} a \\ 1 \\ -1 \end{pmatrix}$, $u_2 = \begin{pmatrix} -1 \\ a \\ 1 \end{pmatrix}$, $u_3 = \begin{pmatrix} 1 \\ -1 \\ a \end{pmatrix}$ form a basis for $\mathbb{R}^3$?

In [20]:

a = sym.symbols("a", real=True) # remember to set real=True for real numbers only
U = Matrix([[a, 1, -1], [-1, a, 1], [1, -1, a]]).T
U

a = sym.symbols("a", real=True) # remember to set real=True for real numbers only U = Matrix([[a, 1, -1], [-1, a, 1], [1, -1, a]]).T U

Out[20]:

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

In [21]:

sym.solve(U.det()) # basis for R^3 if a != 0

sym.solve(U.det()) # basis for R^3 if a != 0

Out[21]:

$\displaystyle \left[ 0\right]$

Question 5¶

Let $U$ and $V$ be subspaces of $\mathbb{R}^n$. We define the sum $U + V$ to be the set of vectors $\{u + v \mid u \in U \text{ and } v \in V\}$.

Suppose $U = \text{span}\left\{\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 2 \\ 1 \end{pmatrix}\right\}$, $V = \text{span}\left\{\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 2 \\ -1 \end{pmatrix}\right\}$.

(b)¶

Show that $U + V$ is a subspace of $\mathbb{R}^n$ by showing that it can be written as a span of a set. What is the dimension?

In [22]:

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

U, V

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

Out[22]:

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

In [23]:

U.row_join(V).get_linearly_independent_vectors(verbosity=1)

U.row_join(V).get_linearly_independent_vectors(verbosity=1)

Before RREF: [self]

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

After RREF:

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

Select columns of self corresponding to pivot positions.

Out[23]:

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

(e)¶

Show that $U \cap V$ a subspace by showing that it can be written as a span of a set. What is the dimension?

In [24]:

U.intersect_subspace(V, verbosity=2) # intersection of U and V

U.intersect_subspace(V, verbosity=2) # intersection of U and V

A linear system whose solution space is the subspace of self. Null(self^T)^T

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

A linear system whose solution space is the subspace of other. Null(other^T)^T

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

Before RREF: [self ; other]

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

After RREF:

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

Out[24]:

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