Tutorial 07 (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 $a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b$ be a linear equation. Express this linear system as $\mathbf{a} \cdot \mathbf{x} = b$ for some (column) vectors $\mathbf{a}$ and $\mathbf{x}$.

In [2]:

# set n = 4
a = Matrix.create_unk_matrix(r=4, symbol="a")
x = Matrix.create_unk_matrix(r=4, symbol="x")
a, x

# set n = 4 a = Matrix.create_unk_matrix(r=4, symbol="a") x = Matrix.create_unk_matrix(r=4, symbol="x") a, x

Out[2]:

$\displaystyle \left( \left[\begin{array}{c}a_{1}\\a_{2}\\a_{3}\\a_{4}\end{array}\right], \ \left[\begin{array}{c}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{array}\right]\right)$

(b)¶

Find the solution set of the linear system $$\begin{cases} x_1 &+& 3x_2 &-& 2x_3 && &=& 0 \\ 2x_1 &+& 6x_2 &-& 5x_3 &-& 2x_4 &=& 0 \\ && && 5x_3 &+& 10x_4 &=& 0 \end{cases}$$

In [3]:

mat = Matrix.from_str("1 3 -2 0; 2 6 -5 -2; 0 0 5 10")
aug = Matrix.zeros(rows=3, cols=1)
mat, aug

mat = Matrix.from_str("1 3 -2 0; 2 6 -5 -2; 0 0 5 10") aug = Matrix.zeros(rows=3, cols=1) mat, aug

Out[3]:

$\displaystyle \left( \left[\begin{array}{cccc}1 & 3 & -2 & 0\\2 & 6 & -5 & -2\\0 & 0 & 5 & 10\end{array}\right], \ \left[\begin{array}{c}0\\0\\0\end{array}\right]\right)$

In [4]:

sol = mat.solve(aug)
sol

sol = mat.solve(aug) sol

Out[4]:

$\displaystyle \left[ \left[\begin{array}{c}- 3 x - 4 z\\x\\- 2 z\\z\end{array}\right]\right]$

In [5]:

sol.sep_unk()

sol.sep_unk()

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
Cell In[5], line 1
----> 1 sol.sep_unk()

AttributeError: 'list' object has no attribute 'sep_unk'

In [6]:

# Alternatively, use nullspace to find the solution set
mat.nullspace(verbosity=1) 

# Alternatively, use nullspace to find the solution set mat.nullspace(verbosity=1)

Before RREF: [self]

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

After RREF:

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

Out[6]:

$\displaystyle \left[ \left[\begin{matrix}-3\\1\\0\\0\end{matrix}\right], \ \left[\begin{matrix}-4\\0\\-2\\1\end{matrix}\right]\right]$

(c)¶

Find a nonzero vector $\mathbf{v} \in \mathbb{R}^4$ such that $\mathbf{a}_1 \cdot \mathbf{v} = 0$, $\mathbf{a}_2 \cdot \mathbf{v} = 0$, and $\mathbf{a}_3 \cdot \mathbf{v} = 0$, where $$\mathbf{a}_1 = \begin{pmatrix} 1 \\ 3 \\ -2 \\ 0 \end{pmatrix}, \quad \mathbf{a}_2 = \begin{pmatrix} 2 \\ 6 \\ -5 \\ -2 \end{pmatrix}, \quad \mathbf{a}_3 = \begin{pmatrix} 0 \\ 0 \\ 5 \\ 10 \end{pmatrix}$$

In [7]:

A = Matrix.from_str("1 3 -2 0; 2 6 -5 -2; 0 0 5 10").T
v = Matrix.create_unk_matrix(r=4, symbol="v")
A, v

A = Matrix.from_str("1 3 -2 0; 2 6 -5 -2; 0 0 5 10").T v = Matrix.create_unk_matrix(r=4, symbol="v") A, v

Out[7]:

$\displaystyle \left( \left[\begin{array}{ccc}1 & 2 & 0\\3 & 6 & 0\\-2 & -5 & 5\\0 & -2 & 10\end{array}\right], \ \left[\begin{array}{c}v_{1}\\v_{2}\\v_{3}\\v_{4}\end{array}\right]\right)$

In [8]:

# Fast method to find v without showing the steps
sol = sym.solve(A.T @ v, v)
v.subs(sol)

# Fast method to find v without showing the steps sol = sym.solve(A.T @ v, v) v.subs(sol)

Out[8]:

$\displaystyle \left[\begin{array}{c}- 3 v_{2} - 4 v_{4}\\v_{2}\\- 2 v_{4}\\v_{4}\end{array}\right]$

In [9]:

# Alternative method using `orthogonal_complement`
mat = A.orthogonal_complement(verbosity=1)
mat

# Alternative method using `orthogonal_complement` mat = A.orthogonal_complement(verbosity=1) mat

Before RREF: [self]

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

After RREF:

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

Out[9]:

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

In [10]:

free_params = Matrix.create_unk_matrix(r=2)
mat @ free_params

free_params = Matrix.create_unk_matrix(r=2) mat @ free_params

Out[10]:

$\displaystyle \left[\begin{array}{c}- 3 x - 4 y\\x\\- 2 y\\y\end{array}\right]$

Question 3¶

Let $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \\ \ -1 \end{pmatrix}$, $\mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$, and $\mathbf{V} = \begin{pmatrix} \mathbf{v}_1 & \mathbf{v}_2 \end{pmatrix}$.

(a)¶

Compute $\mathbf{v}_1 \cdot \mathbf{v}_1$, $\mathbf{v}_1 \cdot \mathbf{v}_2$, $\mathbf{v}_2 \cdot \mathbf{v}_1$ and $\mathbf{v}_2 \cdot \mathbf{v}_2$.

In [11]:

v1 = Matrix.from_str("1; 2; -1")
v2 = Matrix.from_str("1; 0; 1")
v1, v2

v1 = Matrix.from_str("1; 2; -1") v2 = Matrix.from_str("1; 0; 1") v1, v2

Out[11]:

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

In [12]:

# use `dot` method for explicit dot product
v1.dot(v1), v1.dot(v2), v2.dot(v1), v2.dot(v2)

# use `dot` method for explicit dot product v1.dot(v1), v1.dot(v2), v2.dot(v1), v2.dot(v2)

Out[12]:

$\displaystyle \left( 6, \ 0, \ 0, \ 2\right)$

(b)¶

Compute $\mathbf{V}^T \mathbf{V}$. What do the entries of $\mathbf{V}^T \mathbf{V}$ represent?

In [13]:

V = v1.row_join(v2)
V.T @ V

V = v1.row_join(v2) V.T @ V

Out[13]:

$\displaystyle \left[\begin{array}{cc}6 & 0\\0 & 2\end{array}\right]$

Question 4¶

Let $W$ be a subspace of $\mathbb{R}^n$. The orthogonal complement of $W$, denoted as $W^{\perp}$, is defined to be $$W^{\perp} := \{ \mathbf{v} \in \mathbb{R}^n : \mathbf{v} \cdot \mathbf{w} = 0 \text{ for all } \mathbf{w} \in W \}$$

Let $\mathbf{w}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}$, $\mathbf{w}_2 = \begin{pmatrix} 1 \\ 2 \\ -1 \\ -2 \\ 0 \end{pmatrix}$, and $\mathbf{w}_3 = \begin{pmatrix} 1 \\ -1 \\ 1 \\ -1 \\ 0 \end{pmatrix}$, and $W = \text{span}\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3\}$.

(a)¶

Show that $S = \{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3\}$ is linearly independent.

In [14]:

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

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

Out[14]:

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

In [15]:

S.is_linearly_independent(verbosity=2)

S.is_linearly_independent(verbosity=2)

Before RREF: self

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

After RREF:

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

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

Out[15]:

True

(b)¶

Show that $S$ is orthogonal.

In [16]:

S.is_vec_orthogonal(verbosity=1)

S.is_vec_orthogonal(verbosity=1)

Check if [self^T @ self] is a diagonal matrix

$\displaystyle \left[\begin{array}{ccc}5 & 0 & 0\\0 & 10 & 0\\0 & 0 & 4\end{array}\right]$

Out[16]:

True

(c)¶

Show that $W^{\perp}$ is a subspace of $\mathbb{R}^5$ by showing that it is a span of a set. What is the dimension? (Hint: See Question 1.)

In [17]:

W_perp = S.orthogonal_complement(verbosity=1)
W_perp

W_perp = S.orthogonal_complement(verbosity=1) W_perp

Before RREF: [self]

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

After RREF:

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

Out[17]:

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

(d)¶

Obtain an orthonormal set $T$ by normalizing $\mathbf{w}_1$, $\mathbf{w}_2$, $\mathbf{w}_3$.

In [18]:

# `normalized` is a mutable method, so use `copy` to avoid modifying S
T = S.copy().normalized()
T

# `normalized` is a mutable method, so use `copy` to avoid modifying S T = S.copy().normalized() T

Out[18]:

$\displaystyle \left[\begin{array}{ccc}\frac{\sqrt{5}}{5} & \frac{\sqrt{10}}{10} & \frac{1}{2}\\\frac{\sqrt{5}}{5} & \frac{\sqrt{10}}{5} & - \frac{1}{2}\\\frac{\sqrt{5}}{5} & - \frac{\sqrt{10}}{10} & \frac{1}{2}\\\frac{\sqrt{5}}{5} & - \frac{\sqrt{10}}{5} & - \frac{1}{2}\\\frac{\sqrt{5}}{5} & 0 & 0\end{array}\right]$

In [19]:

# Use ScalarFactor so that it is easier to read the matrix
T = S.copy().normalized(factor=True)
T

# Use ScalarFactor so that it is easier to read the matrix T = S.copy().normalized(factor=True) T

Out[19]:

$\left[\begin{array}{ccc}1 & \frac{1}{10} & 1\\1 & \frac{1}{5} & -1\\1 & - \frac{1}{10} & 1\\1 & - \frac{1}{5} & -1\\1 & 0 & 0\end{array}\right]\left[\begin{array}{ccc}\frac{1}{\sqrt{5}} & 0 & 0\\0 & \sqrt{10} & 0\\0 & 0 & \frac{1}{2}\end{array}\right]$

(e)¶

Let $\mathbf{v} = \begin{pmatrix} 2 \\ 0 \\ 1 \\ 1 \\ -1 \end{pmatrix}$. Find the projection of $\mathbf{v}$ onto $W$.

In [20]:

v = Matrix.from_str("2; 0; 1; 1; -1")
W = S # W is the subspace spanned by S
# `proj_comp` solves least squares solution under the hood and 
# does not require W to be a set of orthonormal column vectors
v_proj = v.proj_comp(W, verbosity=1)

v = Matrix.from_str("2; 0; 1; 1; -1") W = S # W is the subspace spanned by S # `proj_comp` solves least squares solution under the hood and # does not require W to be a set of orthonormal column vectors v_proj = v.proj_comp(W, verbosity=1)

self.T @ self is invertible. The lest squares solution is unique.

$\displaystyle \mathbf{x} = \left(\mathbf{A}^\top \mathbf{A}\right)^{-1} \mathbf{A}^\top \mathbf{b}$

$\displaystyle \left[\begin{array}{c}\frac{3}{5}\\- \frac{1}{10}\\\frac{1}{2}\end{array}\right]$

Projected component: Au

$\displaystyle \left[\begin{array}{c}1\\- \frac{1}{10}\\\frac{6}{5}\\\frac{3}{10}\\\frac{3}{5}\end{array}\right]$

Normal component: b - b_proj

$\displaystyle \left[\begin{array}{c}1\\\frac{1}{10}\\- \frac{1}{5}\\\frac{7}{10}\\- \frac{8}{5}\end{array}\right]$

(f)¶

Let $\mathbf{v}_W$ be the projection of $\mathbf{v}$ onto $W$. Show that $\mathbf{v} - \mathbf{v}_W$ is in $W^{\perp}$.

In [21]:

v_norm = v - v_proj
aug_mat = W_perp.row_join(v_norm)
aug_mat

v_norm = v - v_proj aug_mat = W_perp.row_join(v_norm) aug_mat

Out[21]:

$\displaystyle \left[\begin{array}{cc|c}2 & \frac{1}{4} & 1\\-1 & - \frac{1}{2} & \frac{1}{10}\\-2 & - \frac{3}{4} & - \frac{1}{5}\\1 & 0 & \frac{7}{10}\\0 & 1 & - \frac{8}{5}\end{array}\right]$

In [22]:

aug_mat.rref() # consistent system, so v_norm is in W_perp

aug_mat.rref() # consistent system, so v_norm is in W_perp

Out[22]:

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{cc|c}1 & 0 & \frac{7}{10}\\0 & 1 & - \frac{8}{5}\\0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Question 5¶

Let $S = \{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \mathbf{u}_4\}$ where $$\mathbf{u}_1 = \begin{pmatrix} 1 \\ 2 \\ 2 \\ -1 \end{pmatrix}, \quad \mathbf{u}_2 = \begin{pmatrix} 1 \\ 1 \\ -1 \\ 1 \end{pmatrix}, \quad \mathbf{u}_3 = \begin{pmatrix} -1 \\ 1 \\ -1 \\ -1 \end{pmatrix}, \quad \mathbf{u}_4 = \begin{pmatrix} -2 \\ 1 \\ 1 \\ 2 \end{pmatrix}$$

(a)¶

Check that $S$ is an orthogonal basis for $\mathbb{R}^4$.

In [23]:

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

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

Out[23]:

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

In [24]:

S.is_vec_orthogonal(verbosity=1)

S.is_vec_orthogonal(verbosity=1)

Check if [self^T @ self] is a diagonal matrix

$\displaystyle \left[\begin{array}{cccc}10 & 0 & 0 & 0\\0 & 4 & 0 & 0\\0 & 0 & 4 & 0\\0 & 0 & 0 & 10\end{array}\right]$

Out[24]:

True

(b)¶

Is it possible to find a nonzero vector $\mathbf{w}$ in $\mathbb{R}^4$ such that $S \cup {\mathbf{w}}$ is an orthogonal set?

In [25]:

w = Matrix.orthogonal_complement(S, verbosity=1) # No orthogonal complement found

w = Matrix.orthogonal_complement(S, verbosity=1) # No orthogonal complement found

Before RREF: [self]

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

After RREF:

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

/home/runner/work/linear-algebra/linear-algebra/src/ma1522/symbolic.py:2870: UserWarning: Only trivial nullspace (0-vector) detected!
  return Matrix.from_list(self.transpose().nullspace(verbosity))

(c)¶

Obtain an orthonormal set $T$ by normalizing $\mathbf{u}_1$, $\mathbf{u}_2$, $\mathbf{u}_3$, $\mathbf{u}_4$.

In [26]:

T = S.copy().normalized(factor=True)
T

T = S.copy().normalized(factor=True) T

Out[26]:

$\left[\begin{array}{cccc}1 & 1 & -1 & -2\\2 & 1 & 1 & 1\\2 & -1 & -1 & 1\\-1 & 1 & -1 & 2\end{array}\right]\left[\begin{array}{cccc}\frac{1}{\sqrt{10}} & 0 & 0 & 0\\0 & \frac{1}{2} & 0 & 0\\0 & 0 & \frac{1}{2} & 0\\0 & 0 & 0 & \frac{1}{\sqrt{10}}\end{array}\right]$

(d)¶

Let $\mathbf{v} = \begin{pmatrix} 0 \\ 1 \\ 2 \\ 3 \end{pmatrix}$. Find $[\mathbf{v}]_S$ and $[\mathbf{v}]_T$.

In [27]:

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

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

Out[27]:

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

In [28]:

v_S = v.coords_relative(S, verbosity=2)
v_S

v_S = v.coords_relative(S, verbosity=2) v_S

Before RREF: [to | self]

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

After RREF:

$\displaystyle \left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & \frac{3}{10}\\0 & 1 & 0 & 0 & \frac{1}{2}\\0 & 0 & 1 & 0 & -1\\0 & 0 & 0 & 1 & \frac{9}{10}\end{array}\right]$

Out[28]:

$\displaystyle \left[\begin{array}{c}\frac{3}{10}\\\frac{1}{2}\\-1\\\frac{9}{10}\end{array}\right]$

In [29]:

T = T.eval() # Convert the ScalarFactor to a regular matrix
v_T = v.coords_relative(T, verbosity=2)
v_T

T = T.eval() # Convert the ScalarFactor to a regular matrix v_T = v.coords_relative(T, verbosity=2) v_T

Before RREF: [to | self]

$\displaystyle \left[\begin{array}{cccc|c}\frac{\sqrt{10}}{10} & \frac{1}{2} & - \frac{1}{2} & - \frac{\sqrt{10}}{5} & 0\\\frac{\sqrt{10}}{5} & \frac{1}{2} & \frac{1}{2} & \frac{\sqrt{10}}{10} & 1\\\frac{\sqrt{10}}{5} & - \frac{1}{2} & - \frac{1}{2} & \frac{\sqrt{10}}{10} & 2\\- \frac{\sqrt{10}}{10} & \frac{1}{2} & - \frac{1}{2} & \frac{\sqrt{10}}{5} & 3\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & \frac{3 \sqrt{10}}{10}\\0 & 1 & 0 & 0 & 1\\0 & 0 & 1 & 0 & -2\\0 & 0 & 0 & 1 & \frac{9 \sqrt{10}}{10}\end{array}\right]$

Out[29]:

$\displaystyle \left[\begin{array}{c}\frac{3 \sqrt{10}}{10}\\1\\-2\\\frac{9 \sqrt{10}}{10}\end{array}\right]$

(e)¶

Suppose $\mathbf{w}$ is a vector in $\mathbb{R}^4$ such that $[\mathbf{w}]_S = \begin{pmatrix} 1 \\ 2 \\ 1 \\ 1 \end{pmatrix}$. Find $[\mathbf{w}]_T$.

In [30]:

w_S = Matrix.from_str("1; 2; 1; 1")
w_S

w_S = Matrix.from_str("1; 2; 1; 1") w_S

Out[30]:

$\displaystyle \left[\begin{array}{c}1\\2\\1\\1\end{array}\right]$

In [31]:

# Quick and dirty way to find w_T
w = S @ w_S
w

# Quick and dirty way to find w_T w = S @ w_S w

Out[31]:

$\displaystyle \left[\begin{array}{c}0\\6\\0\\2\end{array}\right]$

In [32]:

w_T = w.coords_relative(T, verbosity=2)
w_T

w_T = w.coords_relative(T, verbosity=2) w_T

Before RREF: [to | self]

$\displaystyle \left[\begin{array}{cccc|c}\frac{\sqrt{10}}{10} & \frac{1}{2} & - \frac{1}{2} & - \frac{\sqrt{10}}{5} & 0\\\frac{\sqrt{10}}{5} & \frac{1}{2} & \frac{1}{2} & \frac{\sqrt{10}}{10} & 6\\\frac{\sqrt{10}}{5} & - \frac{1}{2} & - \frac{1}{2} & \frac{\sqrt{10}}{10} & 0\\- \frac{\sqrt{10}}{10} & \frac{1}{2} & - \frac{1}{2} & \frac{\sqrt{10}}{5} & 2\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & \sqrt{10}\\0 & 1 & 0 & 0 & 4\\0 & 0 & 1 & 0 & 2\\0 & 0 & 0 & 1 & \sqrt{10}\end{array}\right]$

Out[32]:

$\displaystyle \left[\begin{array}{c}\sqrt{10}\\4\\2\\\sqrt{10}\end{array}\right]$

In [33]:

# Alternative way via `transition_matrix`

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

# Alternative way via `transition_matrix` S_to_T = S.transition_matrix(T, verbosity=2) S_to_T

Before RREF: [to | self]

$\displaystyle \left[\begin{array}{cccc|cccc}\frac{\sqrt{10}}{10} & \frac{1}{2} & - \frac{1}{2} & - \frac{\sqrt{10}}{5} & 1 & 1 & -1 & -2\\\frac{\sqrt{10}}{5} & \frac{1}{2} & \frac{1}{2} & \frac{\sqrt{10}}{10} & 2 & 1 & 1 & 1\\\frac{\sqrt{10}}{5} & - \frac{1}{2} & - \frac{1}{2} & \frac{\sqrt{10}}{10} & 2 & -1 & -1 & 1\\- \frac{\sqrt{10}}{10} & \frac{1}{2} & - \frac{1}{2} & \frac{\sqrt{10}}{5} & -1 & 1 & -1 & 2\end{array}\right]$

After RREF:

$\displaystyle \left[\begin{array}{cccc|cccc}1 & 0 & 0 & 0 & \sqrt{10} & 0 & 0 & 0\\0 & 1 & 0 & 0 & 0 & 2 & 0 & 0\\0 & 0 & 1 & 0 & 0 & 0 & 2 & 0\\0 & 0 & 0 & 1 & 0 & 0 & 0 & \sqrt{10}\end{array}\right]$

Out[33]:

$\displaystyle \left[\begin{matrix}\sqrt{10} & 0 & 0 & 0\\0 & 2 & 0 & 0\\0 & 0 & 2 & 0\\0 & 0 & 0 & \sqrt{10}\end{matrix}\right]$

In [34]:

w_T = S_to_T @ w_S
w_T

w_T = S_to_T @ w_S w_T

Out[34]:

$\displaystyle \left[\begin{array}{c}\sqrt{10}\\4\\2\\\sqrt{10}\end{array}\right]$