Tutorial 11 (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:

(i) Determine whether the following are linear transformations.

(ii) Write down the standard matrix for each of the linear transformations.

(iii) Find a basis for the range for each of the linear transformations.

(iv) Find a basis for the kernel for each of the linear transformations.

(a)¶

$T_1 : \mathbb{R}^2 \to \mathbb{R}^2$ such that $T_1\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + y \\ y - x \end{pmatrix}$ for $\begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^2$.

In [2]:

in_vec = Matrix.from_str("x; y")
out_vec = Matrix.from_str("x+y; y-x")

in_vec, out_vec

in_vec = Matrix.from_str("x; y") out_vec = Matrix.from_str("x+y; y-x") in_vec, out_vec

Out[2]:

$\displaystyle \left( \left[\begin{array}{c}x\\y\end{array}\right], \ \left[\begin{array}{c}x + y\\- x + y\end{array}\right]\right)$

In [3]:

mat = in_vec.standard_matrix(out_vec, matrices=1)
mat # Double brackets indicates its a list of 1 matrix, i.e. unique solution

mat = in_vec.standard_matrix(out_vec, matrices=1) mat # Double brackets indicates its a list of 1 matrix, i.e. unique solution

Out[3]:

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

In [4]:

mat[0].columnspace()

mat[0].columnspace()

Out[4]:

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

In [5]:

mat[0].nullspace()

mat[0].nullspace()

Out[5]:

$\displaystyle \left[ \right]$

(b)¶

$T_2 : \mathbb{R}^2 \to \mathbb{R}^2$ such that $T_2\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2^x \\ 0 \end{pmatrix}$ for $\begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^2$.

In [6]:

in_vec = Matrix.from_str("x; y")
out_vec = Matrix.from_str("2**x; 0")

in_vec, out_vec

in_vec = Matrix.from_str("x; y") out_vec = Matrix.from_str("2**x; 0") in_vec, out_vec

Out[6]:

$\displaystyle \left( \left[\begin{array}{c}x\\y\end{array}\right], \ \left[\begin{array}{c}2^{x}\\0\end{array}\right]\right)$

In [7]:

try: 
    in_vec.standard_matrix(out_vec, matrices=1)
except ValueError as e:
    print(f"Error: {e}")

try: in_vec.standard_matrix(out_vec, matrices=1) except ValueError as e: print(f"Error: {e}")

Error: No solution found for the standard matrix. This may indicate that the transformation is not linear.

(c)¶

$T_3 : \mathbb{R}^2 \to \mathbb{R}^3$ such that $T_3\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + y \\ 0 \\ 0 \end{pmatrix}$ for $\begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^2$.

In [8]:

in_vec = Matrix.from_str("x; y")
out_vec = Matrix.from_str("x+y; 0; 0")

in_vec, out_vec

in_vec = Matrix.from_str("x; y") out_vec = Matrix.from_str("x+y; 0; 0") in_vec, out_vec

Out[8]:

$\displaystyle \left( \left[\begin{array}{c}x\\y\end{array}\right], \ \left[\begin{array}{c}x + y\\0\\0\end{array}\right]\right)$

In [9]:

mat = in_vec.standard_matrix(out_vec, matrices=1)
mat

mat = in_vec.standard_matrix(out_vec, matrices=1) mat

Out[9]:

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

In [10]:

mat[0].columnspace()

mat[0].columnspace()

Out[10]:

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

In [11]:

mat[0].nullspace()

mat[0].nullspace()

Out[11]:

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

(d)¶

$T_4 : \mathbb{R}^3 \to \mathbb{R}^3$ such that $T_4\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ y - x \\ y - z \end{pmatrix}$ for $\begin{pmatrix} x \ y \ z \end{pmatrix} \in \mathbb{R}^3$.

In [12]:

in_vec = Matrix.from_str("x; y; z")
out_vec = Matrix.from_str("1; y-x; y-z")

in_vec, out_vec

in_vec = Matrix.from_str("x; y; z") out_vec = Matrix.from_str("1; y-x; y-z") in_vec, out_vec

Out[12]:

$\displaystyle \left( \left[\begin{array}{c}x\\y\\z\end{array}\right], \ \left[\begin{array}{c}1\\- x + y\\y - z\end{array}\right]\right)$

In [13]:

try:
    mat = in_vec.standard_matrix(out_vec, matrices=1)
except ValueError as e:
    print(f"Error: {e}")

try: mat = in_vec.standard_matrix(out_vec, matrices=1) except ValueError as e: print(f"Error: {e}")

Error: No solution found for the standard matrix. This may indicate that the transformation is not linear.

(e)¶

$T_5 : \mathbb{R}^5 \to \mathbb{R}$ such that $T_5\begin{pmatrix} x_1 \\ \vdots \\ x_5 \end{pmatrix} = x_3 + 2x_4 - x_5$ for $\begin{pmatrix} x_1 \\ \vdots \\ x_5 \end{pmatrix} \in \mathbb{R}^5$.

In [14]:

in_vec = Matrix.create_unk_matrix(5, 1, "x", is_real=True)
out_vec = Matrix.from_str("x3 + 2*x4 - x5", col_sep="?", is_real=True)

in_vec, out_vec

in_vec = Matrix.create_unk_matrix(5, 1, "x", is_real=True) out_vec = Matrix.from_str("x3 + 2*x4 - x5", col_sep="?", is_real=True) in_vec, out_vec

Out[14]:

$\displaystyle \left( \left[\begin{array}{c}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{array}\right], \ \left[\begin{array}{c}x_{3} + 2 x_{4} - x_{5}\end{array}\right]\right)$

In [15]:

in_vec.standard_matrix(out_vec, matrices=1)

in_vec.standard_matrix(out_vec, matrices=1)

Out[15]:

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

In [16]:

mat[0].columnspace()

mat[0].columnspace()

Out[16]:

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

In [17]:

mat[0].nullspace()

mat[0].nullspace()

Out[17]:

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

(f)¶

$T_6 : \mathbb{R}^n \to \mathbb{R}$ such that $T_6(\mathbf{x}) = \mathbf{x} \cdot \mathbf{x}$ for $\mathbf{x} \in \mathbb{R}^n$.

In [18]:

x = Matrix.create_unk_matrix(3, 1) # set n=3
x, x.dot(x)

x = Matrix.create_unk_matrix(3, 1) # set n=3 x, x.dot(x)

Out[18]:

$\displaystyle \left( \left[\begin{array}{c}x\\y\\z\end{array}\right], \ x^{2} + y^{2} + z^{2}\right)$

In [19]:

# Wrap `x.dot(x)` in a matrix so that it can be 
# used with `standard_matrix`
try:
    mat = x.standard_matrix(Matrix([x.dot(x)]), matrices=1)
except ValueError as e:
    print(f"Error: {e}")

# Wrap `x.dot(x)` in a matrix so that it can be # used with `standard_matrix` try: mat = x.standard_matrix(Matrix([x.dot(x)]), matrices=1) except ValueError as e: print(f"Error: {e}")

Error: No solution found for the standard matrix. This may indicate that the transformation is not linear.

Question 2¶

Let $F : \mathbb{R}^3 \to \mathbb{R}^3$ and $G : \mathbb{R}^3 \to \mathbb{R}^3$ be linear transformations such that

$$F\left( \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \right) = \begin{pmatrix} x_1 - 2x_2 \\ x_1 + x_2 - 3x_3 \\ 5x_2 - x_3 \end{pmatrix} \quad \text{and} \quad G\left( \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \right) = \begin{pmatrix} x_3 - x_1 \\ x_2 + 5x_1 \\ x_1 + x_2 + x_3 \end{pmatrix}$$

and let $\mathbf{A}_F$ and $\mathbf{B}_G$ be the standard matrix of $F$ and $G$, respectively.

(a)¶

Find $\mathbf{A}_F$ and $\mathbf{B}_G$.

In [20]:

F_in = Matrix.from_str("x1; x2; x3")
F_out = Matrix.from_str("x1 - 2*x2; x1 + x2 - 3*x3; 5*x2 - x3", col_sep="?")

F_in, F_out

F_in = Matrix.from_str("x1; x2; x3") F_out = Matrix.from_str("x1 - 2*x2; x1 + x2 - 3*x3; 5*x2 - x3", col_sep="?") F_in, F_out

Out[20]:

$\displaystyle \left( \left[\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right], \ \left[\begin{array}{c}x_{1} - 2 x_{2}\\x_{1} + x_{2} - 3 x_{3}\\5 x_{2} - x_{3}\end{array}\right]\right)$

In [21]:

A_F = F_in.standard_matrix(F_out, matrices=1)[0]
A_F

A_F = F_in.standard_matrix(F_out, matrices=1)[0] A_F

Out[21]:

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

In [22]:

G_in = Matrix.from_str("x1; x2; x3")
G_out = Matrix.from_str("x3 - x1; x2 + 5*x1; x1 + x2 + x3", col_sep="?")

G_in, G_out

G_in = Matrix.from_str("x1; x2; x3") G_out = Matrix.from_str("x3 - x1; x2 + 5*x1; x1 + x2 + x3", col_sep="?") G_in, G_out

Out[22]:

$\displaystyle \left( \left[\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right], \ \left[\begin{array}{c}- x_{1} + x_{3}\\5 x_{1} + x_{2}\\x_{1} + x_{2} + x_{3}\end{array}\right]\right)$

In [23]:

B_G = G_in.standard_matrix(G_out, matrices=1)[0]
B_G

B_G = G_in.standard_matrix(G_out, matrices=1)[0] B_G

Out[23]:

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

(c)¶

Write down the formula for $F(G(\mathbf{x}))$ and find its standard matrix.

In [24]:

# Composition of function is the composition of their standard matrices
A_F @ B_G

# Composition of function is the composition of their standard matrices A_F @ B_G

Out[24]:

$\displaystyle \left[\begin{array}{ccc}-11 & -2 & 1\\1 & -2 & -2\\24 & 4 & -1\end{array}\right]$

(d)¶

Find a linear transformation $H : \mathbb{R}^3 \to \mathbb{R}^3$ such that $H(G(\mathbf{x})) = \mathbf{x}$, for all $\mathbf{x} \in \mathbb{R}^3$.

In [25]:

H = B_G.inverse(verbosity=2)
H

H = B_G.inverse(verbosity=2) H

Left inverse found!
Right inverse found!
Before RREF: [self | eye]

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

After RREF:

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

Out[25]:

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

Question 3¶

For each of the following linear transformations, (i) determine whether there is enough information for us to find the formula of $T$; and (ii) find the formula and the standard matrix for $T$ if possible.

(a)¶

$T : \mathbb{R}^3 \to \mathbb{R}^4$ such that $$T\left( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right) = \begin{pmatrix} 1 \\ 3 \\ 0 \\ 1 \end{pmatrix}, \quad T\left( \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \right) = \begin{pmatrix} 2 \\ 2 \\ -1 \\ 4 \end{pmatrix}, \quad T\left( \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 0 \\ 4 \\ 1 \\ 6 \end{pmatrix}$$

In [26]:

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

T_in, T_out

T_in = Matrix.eye(3) T_out = Matrix.from_str("1 3 0 1; 2 2 -1 4; 0 4 1 6").T T_in, T_out

Out[26]:

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

In [27]:

mat = T_in.standard_matrix(T_out, matrices=1)
mat

mat = T_in.standard_matrix(T_out, matrices=1) mat

Out[27]:

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

(b)¶

$T : \mathbb{R}^2 \to \mathbb{R}^2$ such that $$T\left( \begin{pmatrix} 1 \\ -1 \end{pmatrix} \right) = \begin{pmatrix} 2 \\ 0 \end{pmatrix}, \quad T\left( \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 0 \\ 2 \end{pmatrix}, \quad T\left( \begin{pmatrix} 2 \\ 0 \end{pmatrix} \right) = \begin{pmatrix} 2 \\ 2 \end{pmatrix}$$

In [28]:

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

T_in, T_out

T_in = Matrix.from_str("1 -1; 1 1; 2 0").T T_out = Matrix.from_str("2 0; 0 2; 2 2").T T_in, T_out

Out[28]:

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

In [29]:

mat = T_in.standard_matrix(T_out, matrices=1)
mat

mat = T_in.standard_matrix(T_out, matrices=1) mat

Out[29]:

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

(c)¶

$T : \mathbb{R}^3 \to \mathbb{R}$ such that $$T\left( \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \right) = -1, \quad T\left( \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} \right) = 1, \quad T\left( \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \right) = 0$$

In [30]:

T_in = Matrix.from_str("1 -1 0; 0 1 -1; -1 0 1").T
T_out = Matrix.from_str("-1 1 0")

T_in, T_out

T_in = Matrix.from_str("1 -1 0; 0 1 -1; -1 0 1").T T_out = Matrix.from_str("-1 1 0") T_in, T_out

Out[30]:

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

In [31]:

mat = T_in.standard_matrix(T_out, matrices=1)
mat # free parameters in the solution means its not unique

mat = T_in.standard_matrix(T_out, matrices=1) mat # free parameters in the solution means its not unique

Out[31]:

$\displaystyle \left[ \left[\begin{array}{ccc}x_{1,3} & x_{1,3} + 1 & x_{1,3}\end{array}\right]\right]$