Trial and Error – Physicspupil

November 27, 2023November 27, 2023

202311270959 Exercise 1.4.33

Simplify:

$(AB)^{-1}(AC^{-1})(D^{-1}C^{-1})^{-1}D^{-1}$ .

_{Extracted from H. Anton & C. Rorres. (2010). Elementary Linear Algebra Application Version (10e)}

Answer. $B^{-1}$

Roughwork.

Let

$\begin{aligned} A_{m\times n}&:m\textrm{-by-}n \\ B_{n\times o}&:n\textrm{-by-}o \\ (C^{\mathrm{T}})_{n\times p}}&:n\textrm{-by-}p \\ (D^{\mathrm{T}})_{q\times n}&:q\textrm{-by-}n \\ C_{p\times n}&:p\textrm{-by-}n \\ D_{n\times q}&:n\textrm{-by-}q \\ (AB)_{m\times o}&:m\textrm{-by-}o \\ ((AB)^{\mathrm{T}})_{o\times m}&:o\textrm{-by-}m \\ ((AC)^{\mathrm{T}})_{m\times p}&:m\textrm{-by-}p \\ (D^{\mathrm{T}}C^{\mathrm{T}})_{q\times p}&:q\textrm{-by-}p \\ ((D^{\mathrm{T}}C^{\mathrm{T}})^{\mathrm{T}})_{p\times q}&:p\textrm{-by-}q \\ \end{aligned}$

such that after simplification the original array of matrices (here transpose in place of inverse) should give an $o$ -by- $n$ matrix. For merely satisfying this requirement one candidate can be $B_{n\times o}$ ‘s transpose, i.e., $(B^{\mathrm{T}})_{o\times n}$ , but not much evidence than coincidence here.

Observe, that the simplified expression, in general, should $\textrm{\scriptsize{NOT}}$ necessarily be a span (/linear combination) of matrices in the list above, for none any pair of matrices have necessarily the same dimension for which matrix addition is possible, apart from square matrices.

Hence, assume $AB$ , $C$ , and $D$ to be three square matrices which are invertible.

Note, that the simplified expression should be an arrangement of matrix multiplication amongst some of $8$ matrices $A$ and $A^{-1}$ , $B$ and $B^{-1}$ , $C$ and $C^{-1}$ , and $D$ and $D^{-1}$ .

Consider, by simplification the matrix product should have strictly less than $r=7$ factors, i.e., at most $6$ ; and in adding an extra identity matrix $I$ to the preceding (*such as to avoid terms $AA^{\mathrm{-1}}=A^{\mathrm{-1}}A=I$ to go nearby) we have $n=9$ matrices at hand; the possible outcomes is a permutation

$\text{}^9P_{6}=9^6=531441$

with replacement, minus

$8\times 6=48$ .

The chance of getting by is

$\displaystyle{\frac{1}{531441-48}=\frac{1}{531393} \approx 0.000188\%$ .

This problem is not to be attempted.

February 20, 2023February 20, 2023

202302200931 Review Problem 6.5.7

Let $z\in\mathbb{C}$ . Recall that $z=x+iy$ for some $x,y\in\mathbb{R}$ , and we can form the complex conjugate of $z$ by taking $\overline{z}=x-iy$ . The function $c:\mathbb{R}^2\to\mathbb{R}^2$ which sends $(x,y)\mapsto (x,-y)$ agrees with complex conjugation.

(a) Show that $c$ is a linear map over $\mathbb{R}$ (i.e., scalars in $\mathbb{R}$ ).
(b) Show that $\overline{z}$ is not linear over $\mathbb{C}$ .

_{Extracted from D. Cherney, et al. (2013). Linear Algebra.}

Roughwork.

A function $L:V\to W$ is linear if $V$ and $W$ are vector spaces and

$L(ru+sv)=rL(u)+sL(v)$

for all $u,v\in V$ and $r,s\in\mathbb{R}$ .

(a)

$\begin{aligned} c(az_1+bz_2) & = c(a(x_1+iy_1)+b(x_2+iy_2)) \\ & = c((ax_1+bx_2)+i(ay_1+by_2)) \\ & = (ax_1+bx_2)+i(-(ay_1+by_2)) \\ & = (ax_1+bx_2)-i(ay_1+by_2) \\ & = a(x_1-iy_1) + b(x_2-iy_2) \\ & = a(x_1+i(-y_1)) + b(x_2+i(-y_2)) \\ & = a(c(x_1+iy_1)) + b(c(x_2+iy_2)) \\ & = ac(z_1) + bc(z_2) \\ \end{aligned}$

(b)

Not to be attempted.

September 25, 2020June 5, 2024

202009251201 Homework 1 (Q1)

Solve the system $\mathbf{A}\mathbf{x} = \mathbf{0}$ for each matrix $\mathbf{A}$ below.

(a)

$\mathbf{A}= \begin{bmatrix} 1 & 3 & 5 & 7 \\ 3 & 5 & 7 & 9 \\ 5 & 7 & 9 & 1 \\ \end{bmatrix}$

(b)

$\mathbf{A}= \begin{bmatrix} 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & -2 & 0 \\ \end{bmatrix}$

(c)

$\mathbf{A}= \begin{bmatrix} 4 & 6 & 0 & 1 & -9 \\ 1 & 2 & -4 & 5 & 7 \\ 2 & 3 & 6 & 4 & 2 \\ 1 & 0 & 3 & 2 & -5 \\ \end{bmatrix}$

(d)

$\mathbf{A}= \begin{bmatrix} 1 & 2 & 3 & 1 & 1 \\ 1 & 4 & 0 & 1 & 2 \\ 0 & 2 & -3 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$

Recall

Observe that $\mathbf{A}$ ‘s in (a) and (b) are 3-by-4 matrices, and in (c) and (d) are 4-by-5 matrices. Multiplication of two matrices $\mathbf{A}$ and $\mathbf{B}$ results in a matrix product $\mathbf{A}\mathbf{B}$ . By convention, should an $m$ -by- $n$ matrix $\mathbf{A}$ be written on the left, $\mathbf{A}$ is meant the multiplicand, and should an $n$ -by- $p$ matrix $\mathbf{B}$ be written on the right, $\mathbf{B}$ is meant the multiplier. The matrix product $\mathbf{AB}$ will become an $m$ -by- $p$ matrix.

Solution.

(a)

Let a $4$ -by- $1$ column vector $\mathbf{x}$ be

$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$ .

Then,

$\begin{aligned} \mathbf{A}\mathbf{x} & = \mathbf{0} \\ \begin{bmatrix} 1 & 3 & 5 & 7 \\ 3 & 5 & 7 & 9 \\ 5 & 7 & 9 & 1 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{bmatrix} & = \mathbf{0} \\ \begin{bmatrix} (1)(x_1)+(3)(x_2)+(5)(x_3)+(7)(x_4) \\ (3)(x_1)+(5)(x_2)+(7)(x_3)+(9)(x_4) \\ (5)(x_1)+(7)(x_2)+(9)(x_3)+(1)(x_4) \\ \end{bmatrix} & = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \end{aligned}$

Here we have four unknowns but three equations.

$\begin{Bmatrix} 15x_1+ 45x_2+75x_3+105x_4 & = 0 \\ 15x_1+25x_2+35x_3+45x_4 & = 0 \\ 15x_1+21x_2+27x_3+3x_4 & = 0 \\ \end{Bmatrix}$

It is feasible to carry through all the calculations, but do let us not work in so awkward a manner.

Performing row operations of matrices and using shorthand notations (e.g., $R_1$ stands for Row 1; $5R_2$ stands for 5 times each entry in Row 2; $R_1-R_3$ means every entry in Row 3 is to be subtracted from the corresponding entry in Row 1.)

(1) $R_3-5R_1$

$\begin{bmatrix} 1 & 3 & 5 & 7 \\ 3 & 5 & 7 & 9 \\ 0 & -8 & -16 & -34 \\ \end{bmatrix}$

(2) $R_2 - 3R_1$

$\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 12 \\ 0 & -8 & -16 & -34 \\ \end{bmatrix}$

(3) $R_3 - 2R_2$

$\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 12 \\ 0 & 0 & 0 & -58 \\ \end{bmatrix}$

(4) $R_3 \div (-58)$

$\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 12 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

(5) $R_2-12R_3$

$\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

(6) $R_1-7R_3$

$\begin{bmatrix} 1 & 3 & 5 & 0 \\ 0 & -4 & -8 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

(7) $R_2 \div (-4)$

$\begin{bmatrix} 1 & 3 & 5 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

(8) $R_1-3R_2$

$\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

The matrix above is called the reduced row-echelon form of $\mathbf{A}$ .

Now that the simplification has come handy:

$\begin{aligned} x_1 -x_3 &=0 \\ x_2 +2x_3 & = 0 \\ x_4 & = 0 \\ \end{aligned}$

and the answer is $\mathbf{x}=\begin{bmatrix} x \\ -2x \\ x \\ 0 \end{bmatrix}$ .

Questions (b), (c), and (d) are left the readers.