Tutorial 09 (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 father wishes to distribute an amount of money among his three sons Jack, Jim, and John. He wishes to distribute such that the following conditions are all satisfied.

(i) The amount Jack receives plus twice the amount Jim receives is $300.

(ii) The amount Jim receives plus the amount John receives is $300.

(iii) Jack receives $300 more than twice of what John receives.

(a)¶

Is it possible for the following conditions to all be satisfied?

In [2]:

mat = Matrix.from_str("1 2 0; 0 1 1; 1 0 -2")
aug = Matrix.from_str("300; 300; 300")

mat, aug

mat = Matrix.from_str("1 2 0; 0 1 1; 1 0 -2") aug = Matrix.from_str("300; 300; 300") mat, aug

Out[2]:

$\displaystyle \left( \left[\begin{array}{ccc}1 & 2 & 0\\0 & 1 & 1\\1 & 0 & -2\end{array}\right], \ \left[\begin{array}{c}300\\300\\300\end{array}\right]\right)$

In [3]:

aug.is_subspace_of(mat, verbosity=2)

aug.is_subspace_of(mat, verbosity=2)

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

Before RREF: [other | self]

$\displaystyle \left[\begin{array}{ccc|c}1 & 2 & 0 & 300\\0 & 1 & 1 & 300\\1 & 0 & -2 & 300\end{array}\right]$

After RREF:

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

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

Out[3]:

False

(b)¶

If it is not possible, find a least square solution. (Make sure that your least square solution is feasible. For example, one cannot give a negative amount of money to anybody.)

In [4]:

sol = mat.solve_least_squares(aug, verbosity=2)
sol

sol = mat.solve_least_squares(aug, verbosity=2) sol

Before RREF: [self.T @ self | self.T @ rhs]

$\displaystyle \left[\begin{array}{ccc|c}2 & 2 & -2 & 600\\2 & 5 & 1 & 900\\-2 & 1 & 5 & -300\end{array}\right]$

After RREF

$\text{RREF}\left\{\text{rref} = \left[\begin{array}{ccc|c}1 & 0 & -2 & 200\\0 & 1 & 1 & 100\\0 & 0 & 0 & 0\end{array}\right], \quad\text{pivots} = (0, 1)\right\}$

Out[4]:

$\displaystyle \left[\begin{array}{c}2 z + 200\\100 - z\\z\end{array}\right]$

In [5]:

sol.sep_part_gen()

sol.sep_part_gen()

Out[5]:

$\left(\left[\begin{array}{c}200\\100\\0\end{array}\right] + \left[\begin{array}{c}2 z\\- z\\z\end{array}\right]\right)$

Question 2¶

(b)¶

Let $$\mathbf{A} = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix}$$

What is $\mathbf{Q}$ and $\mathbf{R}$? Compare this with the answer in tutorial 8 question 5(a).

In [6]:

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

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

Out[6]:

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

In [7]:

# To get the full QR decomposition (as computed by MATLAB), set `full=True`
A.QRdecomposition(full=True, verbosity=0)

# To get the full QR decomposition (as computed by MATLAB), set `full=True` A.QRdecomposition(full=True, verbosity=0)

Out[7]:

$\left[\begin{array}{cccc}\frac{\sqrt{3}}{3} & 0 & - \frac{\sqrt{6}}{6} & \frac{\sqrt{2}}{2}\\\frac{\sqrt{3}}{3} & 0 & - \frac{\sqrt{6}}{6} & - \frac{\sqrt{2}}{2}\\\frac{\sqrt{3}}{3} & 0 & \frac{\sqrt{6}}{3} & 0\\0 & 1 & 0 & 0\end{array}\right]\left[\begin{array}{ccc}\sqrt{3} & \sqrt{3} & \frac{\sqrt{3}}{3}\\0 & 1 & 1\\0 & 0 & \frac{\sqrt{6}}{3}\\0 & 0 & 0\end{array}\right]$

Question 3 (Cayley-Hamilton Theorem)¶

Consider $$p(X) = X^3 - 4X^2 - X + 4I$$

(a)¶

Compute $p(X)$ for $X = \begin{pmatrix} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 1 \end{pmatrix}$.

In [8]:

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

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

Out[8]:

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

In [9]:

X**3 - 4 * X**2 - X + 4 * Matrix.eye(3)

X**3 - 4 * X**2 - X + 4 * Matrix.eye(3)

Out[9]:

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

(b)¶

Find the characteristic polynomial of $X$.

In [10]:

X.cpoly()

X.cpoly()

Out[10]:

$\displaystyle \left(x - 4\right) \left(x - 1\right) \left(x + 1\right)$

Question 4¶

For each of the following matrices $\mathbf{A}$, determine if $\mathbf{A}$ is diagonalizable. If $\mathbf{A}$ is diagonalizable, find an invertible $\mathbf{P}$ that diagonalizes $\mathbf{A}$ and determine $\mathbf{P}^{-1}\mathbf{A}\mathbf{P}$.

(a)¶

$\mathbf{A} = \begin{pmatrix} 1 & -3 & 3 \\ 3 & -5 & 3 \\ 6 & -6 & 4 \end{pmatrix}$

In [11]:

A = Matrix.from_str("1 -3 3; 3 -5 3; 6 -6 4")
A

A = Matrix.from_str("1 -3 3; 3 -5 3; 6 -6 4") A

Out[11]:

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

In [12]:

sym.Matrix(A).is_diagonalizable()

sym.Matrix(A).is_diagonalizable()

Out[12]:

True

In [13]:

# To check diagaonalization
# Take note that the `reals_only` argument is set to `True` by default 
# in accordance with MA1522 syllabus. This overrides SymPy's default 
# behavior of allowing complex eigenvalues.
A.is_diagonalizable(reals_only=True, verbosity=1)

# To check diagaonalization # Take note that the `reals_only` argument is set to `True` by default # in accordance with MA1522 syllabus. This overrides SymPy's default # behavior of allowing complex eigenvalues. A.is_diagonalizable(reals_only=True, verbosity=1)

Characteristic Polynomial is: 

$\displaystyle \left(x - 4\right) \left(x + 2\right)^{2}$

Check if algebraic multiplicity equals number of eigenvectors.

Eigenvectors are:

$\displaystyle \left\{\text{eigenvalue} = -2, \quad\text{algebraic multiplicity} = 2, \quad\text{eigenspace} = \left[\begin{array}{cc}1 & -1\\1 & 0\\0 & 1\end{array}\right]\right\}$

$\displaystyle \left\{\text{eigenvalue} = 4, \quad\text{algebraic multiplicity} = 1, \quad\text{eigenspace} = \left[\begin{array}{c}\frac{1}{2}\\\frac{1}{2}\\1\end{array}\right]\right\}$

Out[13]:

True

In [14]:

# Full diagaonlisation procedure
A.diagonalize(reals_only=True, verbosity=1)

# Full diagaonlisation procedure A.diagonalize(reals_only=True, verbosity=1)

Characteristic Polynomial

$\displaystyle \left(x - 4\right) \left(x + 2\right)^{2}$

$\displaystyle \text{Before RREF: }4\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{ccc}3 & 3 & -3\\-3 & 9 & -3\\-6 & 6 & 0\end{array}\right]$

After RREF:

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

Eigenvectors:

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

$\displaystyle \text{Before RREF: }-2\mathbb{I} - \mathrm{self}$

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

After RREF:

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

Eigenvectors:

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

Out[14]:

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

(b)¶

$\mathbf{A} = \begin{pmatrix} 9 & 8 & 6 & 3 \\ 0 & -1 & 3 & -4 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}$

In [15]:

A = Matrix.from_str("9 8 6 3; 0 -1 3 -4; 0 0 2 0; 0 0 0 3")
A

A = Matrix.from_str("9 8 6 3; 0 -1 3 -4; 0 0 2 0; 0 0 0 3") A

Out[15]:

$\displaystyle \left[\begin{array}{cccc}9 & 8 & 6 & 3\\0 & -1 & 3 & -4\\0 & 0 & 2 & 0\\0 & 0 & 0 & 3\end{array}\right]$

In [16]:

A.diagonalize(reals_only=True, verbosity=1)

A.diagonalize(reals_only=True, verbosity=1)

Characteristic Polynomial

$\displaystyle \left(x - 9\right) \left(x - 3\right) \left(x - 2\right) \left(x + 1\right)$

$\displaystyle \text{Before RREF: }9\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cccc}0 & -8 & -6 & -3\\0 & 10 & -3 & 4\\0 & 0 & 7 & 0\\0 & 0 & 0 & 6\end{array}\right]$

After RREF:

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

Eigenvectors:

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

$\displaystyle \text{Before RREF: }3\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cccc}-6 & -8 & -6 & -3\\0 & 4 & -3 & 4\\0 & 0 & 1 & 0\\0 & 0 & 0 & 0\end{array}\right]$

After RREF:

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

Eigenvectors:

$\displaystyle \left[ \left[\begin{matrix}\frac{5}{6}\\-1\\0\\1\end{matrix}\right]\right]$

$\displaystyle \text{Before RREF: }2\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cccc}-7 & -8 & -6 & -3\\0 & 3 & -3 & 4\\0 & 0 & 0 & 0\\0 & 0 & 0 & -1\end{array}\right]$

After RREF:

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

Eigenvectors:

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

$\displaystyle \text{Before RREF: }-1\mathbb{I} - \mathrm{self}$

$\displaystyle \left[\begin{array}{cccc}-10 & -8 & -6 & -3\\0 & 0 & -3 & 4\\0 & 0 & -3 & 0\\0 & 0 & 0 & -4\end{array}\right]$

After RREF:

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

Eigenvectors:

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

Out[16]:

$\left[\begin{array}{cccc}-4 & -2 & 5 & 1\\5 & 1 & -6 & 0\\0 & 1 & 0 & 0\\0 & 0 & 6 & 0\end{array}\right]\left[\begin{array}{cccc}-1 & 0 & 0 & 0\\0 & 2 & 0 & 0\\0 & 0 & 3 & 0\\0 & 0 & 0 & 9\end{array}\right]\left[\begin{matrix}0 & \frac{1}{5} & - \frac{1}{5} & \frac{1}{5}\\0 & 0 & 1 & 0\\0 & 0 & 0 & \frac{1}{6}\\1 & \frac{4}{5} & \frac{6}{5} & - \frac{1}{30}\end{matrix}\right]$

(c)¶

$\mathbf{A} = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}$

In [17]:

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

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

Out[17]:

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

In [18]:

# If matrix is not diagonalizable, it will raise a MatrixError
try:
    A.diagonalize(reals_only=True, verbosity=1)
except sym.matrices.matrixbase.MatrixError as e:
    print(f"MatrixError: {e}")

# If matrix is not diagonalizable, it will raise a MatrixError try: A.diagonalize(reals_only=True, verbosity=1) except sym.matrices.matrixbase.MatrixError as e: print(f"MatrixError: {e}")

Characteristic Polynomial

$\displaystyle \left(x - 1\right)^{3}$

$\displaystyle \text{Before RREF: }1\mathbb{I} - \mathrm{self}$

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

After RREF:

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

Eigenvectors:

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

MatrixError: Matrix is not diagonalizable

(d)¶

$\mathbf{A} = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$

In [19]:

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

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

Out[19]:

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

In [20]:

A.diagonalize(reals_only=True, verbosity=1)

A.diagonalize(reals_only=True, verbosity=1)

Characteristic Polynomial

$\displaystyle \left(x - 1\right)^{2} \left(x + 1\right)^{2}$

$\displaystyle \text{Before RREF: }1\mathbb{I} - \mathrm{self}$

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

After RREF:

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

Eigenvectors:

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

$\displaystyle \text{Before RREF: }-1\mathbb{I} - \mathrm{self}$

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

After RREF:

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

Eigenvectors:

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

Out[20]:

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

(e)¶

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

In [21]:

A = Matrix.from_str("-1 1 1; 1 1 -1; -4 2 3")
A

A = Matrix.from_str("-1 1 1; 1 1 -1; -4 2 3") A

Out[21]:

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

In [22]:

# `cpoly` returning a tuple of expressions means that the characteristic 
# polynomial is not irreducible over the reals. As the rhs expression 
# only has non-real roots. This means that the matrix is not 
# diagaonalizable over the reals (with regards to MA1522 syllabus).
A.cpoly()

# `cpoly` returning a tuple of expressions means that the characteristic # polynomial is not irreducible over the reals. As the rhs expression # only has non-real roots. This means that the matrix is not # diagaonalizable over the reals (with regards to MA1522 syllabus). A.cpoly()

Out[22]:

$\displaystyle \left( x - 1, \ x^{2} - 2 x + 2\right)$

In [23]:

Matrix.eye(4).inv()

Matrix.eye(4).inv()

Out[23]:

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

In [24]:

# By setting `reals_only=False`, we allow complex eigenvalues
# and hence the matrix can be diagonalized.
PDP = A.diagonalize(reals_only=False)
PDP

# By setting `reals_only=False`, we allow complex eigenvalues # and hence the matrix can be diagonalized. PDP = A.diagonalize(reals_only=False) PDP

Out[24]:

$\left[\begin{array}{ccc}1 & 1 & 1\\1 & - i & i\\1 & 2 & 2\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 - i & 0\\0 & 0 & 1 + i\end{array}\right]\left[\begin{matrix}2 & 0 & -1\\- \frac{1}{2} - i & \frac{i}{2} & \frac{1}{2} + \frac{i}{2}\\- \frac{1}{2} + i & - \frac{i}{2} & \frac{1}{2} - \frac{i}{2}\end{matrix}\right]$