math class note

Math Class Notes - MATH 2109

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Page 1

Properties of eigen value —

1. The sum of the eigen values is equal to the trace of the matrix.
2. The product of the eigen values is the determinant of the matrix.
3. If the eigen value of $A$ are ${\lambda }_1,{\lambda }_2,{\lambda }_3,\dots$ then the eigen values of $A^{-1}$ are $\frac{1}{{\lambda }_1},\frac{1}{{\lambda }_2},\frac{1}{{\lambda }_3},\dots$
4. The eigen value of $kA$ are $k{\lambda }_1,k{\lambda }_2,k{\lambda }_3,\dots$
5. $A^n$ matrix’s eigen values ${\lambda }_1^n,{\lambda }_2^n,{\lambda }_3^n,\dots$

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Page 2

$|\lambda I-A|=0\text{— characteristic eqn.}$

$$A=\left(\begin{array}{cc}1& 2\\ 3& 2\end{array}\right)$$ $$\left|\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)-\left(\begin{array}{cc}1& 2\\ 3& 2\end{array}\right)\right|=0$$ $$\Rightarrow \left|\begin{array}{cc}\lambda -1& -2\\ -3& \lambda -2\end{array}\right|=0$$ $$\Rightarrow \lambda =\text{?}\text{— characteristic root or eigen value}$$

(Note: There is a blank or scribble in the box after $\lambda =$ in the handwritten notes)

$|\lambda I-A|⟶\text{expand}=\text{characteristic polynomial}$

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* 4 cases depending on $\lambda$ —

1. $0<\lambda <1$

          λx     x
   o----->----->
   

   $\lambda x<x$

2. $\lambda >1$

          x      λx
   o----->----->
   

   $\lambda x>x$

3. $-1<\lambda <0$

      λx     o     x
   <-----o----->
   

   $\lambda x<x$

4. $\lambda <-1$

         λx          o     x
   <----------o----->
   

   $\lambda x>x$

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Page 3

$$\{\begin{array}{l}(2-3)x+2y+3z=0\\ x+2y+3z=0\\ 2x+y+(2+4)z=0\end{array}$$

(Note: The system above is crossed out in the original notes)

--- LE done

06.08.26 (Arif Sir)

Eigen Value and Eigen Vector

$$Ax=\lambda x$$

Annotations from the diagram:

• $A$: square matrix (annotated as: “corresponding eigen matrix”)
• $x$ (in $Ax$): vector
• $\lambda$: any scalar value (annotated as: “eigen value”)
• $x$ (in $\lambda x$): corresponding eigen vector
• The term $\lambda x$ is annotated with: “dimension doesn’t change”

Formal def. from book —

$$Ax=\lambda x$$ $$\Rightarrow \lambda x-Ax=0$$ $$\Rightarrow (\lambda I-A)x=0\text{— homogeneous system of L}$$

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Page 4

$$\sim \left[\begin{array}{ccccc}1& -1& -1& -1& 0\\ 0& 4& 0& 2& 0\\ 0& 0& 2& -1& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right]$$

Back to the eqn:

$$x_1-x_2-x_3-x_4=0$$ $$4x_2+2x_4=0$$ $$2x_3-x_4=0$$

Let, $x_4=a$ where $[a=\text{free variable}]$

$$x_2=-\frac{a}{2}$$ $$x_3=\frac{a}{2}$$ $$x_1=a$$ $$=\left(a,-\frac{a}{2},\frac{a}{2},a\right)⟶\text{general soln.}$$ $$[Ax=0]⟶\text{homogeneous}$$ $$[Ax=b]⟶\text{non-homogeneous}$$

(i) $|A|\ne 0$, unique soln. (ii) $|A|=0$, infinitely many soln.

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Page 5

System of LE

          system of LE
         /            \
  homogeneous      non-homogeneous
    system             system

$$[Ax=0]⟶\text{always consistent}.$$ $$\{\begin{array}{l}{a}_{11}{x}_1+a_{12}{x}_2+\cdots +a_{1n}{x}_n=0\\ ⋮\\ a_{n1}{x}_1+a_{n2}{x}_2+\cdots +a_{nn}{x}_n=0\end{array}$$

homogeneous system can never be inconsistent.

$$Ax=0⟶$$

1. Trivial soln. (zero) $⟶$ always exists, that’s why consistent
2. Non-trivial soln. (non-zero)

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Q.

$$\{\begin{array}{l}{x}_1-x_2-x_3-x_4=0\\ x_1+3x_2-x_3+x_4=0\\ 3x_1-7x_2-x_3-6x_4=0\\ 2x_1+2x_2-2x_3=0\\ 6x_1-2x_2-4x_3-5x_4=0\end{array}$$

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Page 6

Back to the eqn:

$$\{\begin{array}{l}{x}_1+2x_2-x_3=1\\ -8x_2+5x_3=-1\end{array}$$

$1,-8⟶$ leading entry or pivot entry, $⟶$ their columns $⟶$ pivot column.

$$\text{No. of variable}-\text{Number of eqn. after converting into echelon}=\text{free variable}$$

Here, free variable $=3-2=1$ Here, there are two eqn. and three variables, so, there are $3-2=1$ free variable.

Let, $x_3=k$ (free v.)

$$\Rightarrow -8x_2+5k=-1$$ $$\therefore x_2=\frac{1+5k}{8}$$ $$x_1+2\left(\frac{1+5k}{8}\right)-k=1$$ $$\Rightarrow x_1+\frac{1+5k}{4}-k=1$$ $$\Rightarrow x_1=\frac{3-k}{4}$$

$\}$ General solution.

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Page 7

$$\ast \left(\begin{array}{cccccc}1& 2& 3& 4& 1& 6\\ 0& 1& 2& 0& 2& 3\\ 0& 0& 0& 1& 3& 5\end{array}\right)$$

Free variables $=2$

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04.08.26 (Arif Sir)

Q. Determine the value of $\lambda$ and $\mu$ for which the system

$$\{\begin{array}{l}x+y+z=6\\ x+2y+3z=10\\ x+2y+\lambda z=\mu \end{array}$$

has (i) unique soln. (ii) no soln. (iii) infinitely many soln.

$\to$ Augmented matrix

$$\left[\begin{array}{cccc}1& 1& 1& 6\\ 1& 2& 3& 10\\ 1& 2& \lambda & \mu \end{array}\right]\underset{R_3^{\prime }\to R_3-R_2}{\overset{R_2^{\prime }\to R_2-R_1}{\to }}\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& 2& 4\\ 0& 0& \lambda -3& \mu -10\end{array}\right]$$

No solution: when, $\lambda =3$, $\mu \ne 10$

Infinitely many: $\lambda =3$, $\mu =10$

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Page 8

# Gauss - Jordan Elem. Elimination.

8.

$$\{\begin{array}{l}{x}_2-4x_3=8\\ 2x_1-3x_2+2x_3=-1\\ 5x_1-8x_2+7x_3=1\end{array}$$

(Note: The first equation $x_2-4x_3=8$ appears to have been written twice, with the first occurrence crossed out)

$$\Rightarrow \left[\begin{array}{cccc}0& 1& -4& 8\\ 2& -3& 2& -1\\ 5& -8& 7& 1\end{array}\right]⟶\text{Inconsistent.}$$

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*

$$\{\begin{array}{l}{x}_1+2x_2-x_3=1\\ 3x_1-2x_2+2x_3=2\\ 7x_1-2x_2+3x_3=5\end{array}$$

(Note: In the first equation, a duplicate $x_3$ is written above the $-x_3$ term in the notes)

$$\underset{R_2^{\prime }\to R_2-3R_1}{\overset{R_3^{\prime }\to R_3-7R_1}{\to }}\left[\begin{array}{cccc}1& 2& -1& 1\\ 0& -8& 5& -1\\ 0& -16& 10& -2\end{array}\right]\stackrel{R_3^{\prime \prime }\to R_3^{\prime }-2R_2^{\prime }}{\to }\left[\begin{array}{cccc}1& 2& -1& 1\\ 0& -8& 5& -1\\ 0& 0& 0& 0\end{array}\right]⟶\text{consistent}$$

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Page 9

Let $c=[A|b]$

If $\text{Rank}(A)=\text{Rank}(c)$ $\to$ The matrix is consistent

If $\text{Rank}(A)\ne \text{Rank}(c)$ $\to$ The matrix is inconsistent

$$\left[\begin{array}{cccc}1& 1& -1& 7\\ 2& -1& 2& 5\\ 3& 1& -1& 3\end{array}\right]\underset{R_3^{\prime }\to R_3-3R_1}{\overset{R_2^{\prime }\to R_2-2R_1}{\to }}\left[\begin{array}{cccc}1& 1& -1& 7\\ 0& -3& 4& -9\\ 0& -2& 2& -18\end{array}\right]$$ $$\stackrel{R_3^{\prime \prime }\to R_3^{\prime }-\frac{2}{3}{R}_2^{\prime }}{\to }\left[\begin{array}{cccc}1& 1& -1& 7\\ 0& -3& 4& -9\\ 0& 0& -2/3& -12\end{array}\right]\text{— now - echelon form}$$

Here, in now-echelon form, $\text{Rank}(A)=\text{Rank}(c)$ $\therefore$ Solution exists.

Back substitution:

$$x+y-z=7$$ $$-3y+4z=-9$$ $$-\frac{2}{3}z=-12$$ $$\therefore z=18$$

(Note: The z is written with a crossbar $\text{ƶ}$ in the original notes. At the bottom right, “Gaussian Elimination” is written vertically).

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Page 10

Inverse of a matrix by elementary transformation

Algorithm: $[A|I]\sim [I|A^{-1}][A:I]$

$$\left[\begin{array}{cccccc}1& 2& 3& 1& 0& 0\\ 2& 5& 3& 0& 1& 0\\ 1& 0& 8& 0& 0& 1\end{array}\right]\underset{R_3^{\prime }\to R_3-R_1}{\overset{R_2^{\prime }\to R_2-2R_1}{\to }}\left[\begin{array}{cccccc}1& 2& 3& 1& 0& 0\\ 0& 1& -3& -2& 1& 0\\ 0& -2& 5& -1& 0& 1\end{array}\right]$$ $$\stackrel{R_3^{\prime \prime }\to R_3^{\prime }+2R_2^{\prime }}{\to }\left[\begin{array}{cccccc}1& 2& 3& 1& 0& 0\\ 0& 1& -3& -2& 1& 0\\ 0& 0& -1& -5& 0& 1\end{array}\right]$$

(Note: In the final matrix, row 3, column 5 is written as 0 in the notes, though mathematically it should be 2. There are also faint erased outlines of matrices at the bottom of this page).

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Page 11

Arif Sir (03.08.26)

$$\{\begin{array}{l}{a}_{11}{x}_1+a_{12}{x}_2+\cdots +a_{1n}{x}_n=b_1\\ a_{21}{x}_1+a_{22}{x}_2+\cdots +a_{2n}{x}_n=b_2\\ ⋮\\ a_{m1}{x}_1+a_{m2}{x}_2+\cdots +a_{mn}{x}_n=b_m\end{array}$$

$\begin{array}{rl}& \text{coefficient matrix }A\\ & \text{solution vector }x\\ & \text{RHS constant vector }b\end{array}\}Ax=b$

$$A=\left(\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ a_{21}& a_{22}& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{m1}& a_{m2}& \dots & a_{mn}\end{array}\right)$$

• System of L.E.
  • consistent
    • unique soln.
    • infinitely many soln.
  • inconsistent
    • no solution

$$Ax=b$$ $$\{\begin{array}{l}x+y-z=7\\ 2x-y+2z=5\\ 3x+y-z=3\end{array}$$

• $[A|b]\to$ Augmented matrix

$$\left[\begin{array}{cccc}1& 1& -1& 7\\ 2& -1& 2& 5\\ 3& 1& -1& 3\end{array}\right]$$

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Page 12

Q. Find the rank of,

$$\left[\begin{array}{cccc}3& 1& 4& 3\\ 2& 0& 1& 2\\ 1& 2& 3& 5\\ 0& 1& 2& 4\end{array}\right]$$

(Note: There is a curly brace pointing from this matrix to: “interchange as 3rd $\leftrightarrow$ 1 (R1), better for calc.“)

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28.07.26 class-05

Q. Convert into normal form —

$$\left[\begin{array}{cccc}1& 2& -1& 3\\ 4& 1& 2& 1\\ 3& -1& 1& 2\\ 1& 2& 0& 1\end{array}\right]\underset{R_4^{\prime }\to R_4-R_1}{\overset{R_2^{\prime }\to R_2-4R_1,R_3^{\prime }\to R_3-3R_1}{\to }}\left[\begin{array}{cccc}1& 2& -1& 3\\ 0& -7& 6& -11\\ 0& -7& 4& -7\\ 0& 0& 1& -2\end{array}\right]$$ $$\stackrel{R_3^{\prime }\to R_3-R_2}{\to }\left[\begin{array}{cccc}1& 2& -1& 3\\ 0& -7& 6& -11\\ 0& 0& -2& 4\\ 0& 0& 0& 0\end{array}\right]\underset{C_3^{\prime }\to C_3+C_1,C_4^{\prime }\to C_4-3C_1}{\overset{C_2^{\prime }\to C_2-2C_1}{\to }}\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& -7& 6& -11\\ 0& 0& -2& 4\\ 0& 0& 0& 0\end{array}\right]$$ $$\underset{C_4^{\prime }\to C_4-11/7C_2}{\overset{C_3^{\prime }\to C_3+6/7C_2}{\to }}\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& -7& 0& 0\\ 0& 0& -2& 4\\ 0& 0& 0& 0\end{array}\right]$$

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Page 13

Rank:

• The no. of independent rows of a matrix is called rank of the matrix.
• $\to$ zero row always dependent.
• $\to$ Any other row except zero row in row-echelon form, is an independent row.
• $\to$ All rows in identity matrix — independent
• $\to$ In matrix, no. of ind. row = ind. column no.

Q. Find the rank of,

$$\left[\begin{array}{cccc}2& 1& 3& 2\\ 3& 0& 1& 4\\ 5& 1& 2& 3\\ -11& -4& -5& 0\end{array}\right]\underset{R_3^{\prime }\to R_3-5R_1,R_4^{\prime }\to R_4+11R_1}{\overset{R_2^{\prime }\to R_2-3R_1}{\to }}\left[\begin{array}{cccc}1& 1& 3& 2\\ 0& -3& -8& -2\\ 0& -4& -13& -7\\ 0& 7& 28& 22\end{array}\right]$$ $$\underset{R_4^{\prime }\to R_4^{\prime }+7/3R_2^{\prime }}{\overset{R_3^{\prime }\to R_3^{\prime }-4/3R_2^{\prime }}{\to }}\left[\begin{array}{cccc}1& 1& & \\ 0& -3& & \\ 0& 0& & \\ 0& 0& & \end{array}\right]$$

(Note: The first row of the first matrix is written as $2132$ in the notes but acts as $1132$ in operations. The last matrix is left incomplete in the original notes).

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Page 14

Row-Echelon form: A matrix is said to be in row echelon form if it follows the following properties —

1. All non-zero rows are above any rows of all zero.
2. Each leading entry of a row is in a column to the right of the leading entry of the row above it.
3. All entries in a column below leading entry is zero.

Examples:

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

(Note: The circled entries above are annotated as “leading entry”. There is a Bengali note below them: [leading এর নিচেরটা zero]).

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Page 15

3nd cycle 27.06.26 class-04

Row-echelon form —

$$\left[\begin{array}{ccccc}\stackrel{◯}{2}& 3& 4& 5& 6\\ 0& \stackrel{◯}{1}& 2& 3& 1\\ 0& 0& 0& \stackrel{◯}{2}& 3\\ 0& 0& 0& 0& \stackrel{◯}{1}\\ 0& 0& 0& 0& 0\end{array}\right]$$

(Note: The circled entries represent leading entries).

Reduced row-echelon form — $\to$ Follows all row-echelon rules, Moreover,

1. The leading entry must be 1.
2. The leading entry is the only non-zero entry of th is it’s column.

Normal form — $\to$ 4 forms:

1. $I_n$
2. $\left[\begin{array}{cc}{I}_n& 0\end{array}\right]⟶\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\end{array}\right]$
3. $\left[\begin{array}{c}{I}_n\\ 0\end{array}\right]⟶\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]$
4. $\left[\begin{array}{cc}{I}_n& 0\\ 0& 0\end{array}\right]⟶\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right]$

[In = Identity matrix]

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Page 16

2nd cycle 20.07.26 class-03

Elementary transformation — (Elementary row operations)

Any of the following operations is called transformation:

1. Interchanging any two rows
2. Multiply any rows by a non-zero constant
3. Addition of constant multiplication of elements of any rows with another row. i.e. $R_i+k{R}_j$

$\to$ All these matrices are equivalent, not equal.

$$\left[\begin{array}{ccc}3& 1& 2\\ 5& 2& 0\\ 1& 2& 4\end{array}\right]\stackrel{R_1\leftrightarrow R_3}{\to }\left[\begin{array}{ccc}1& 2& 4\\ 5& 2& 0\\ 3& 1& 2\end{array}\right]$$

(Note: Annotated with “interchange” and “eqv. sign” below the $\sim$ equivalent arrow).

$$\stackrel{R_2\to \frac{R_2}{5}}{\to }\left[\begin{array}{ccc}3& 1& 2\\ 1& 2/5& 0\\ 1& 2& 4\end{array}\right]$$ $$\stackrel{R_2^{\prime }\to R_2-R_1}{\to }\left[\begin{array}{ccc}1& 2& 4\\ 0& -8/5& -4\\ 3& 1& 2\end{array}\right]$$

(Note: This operation is applied to the row-interchanged matrix with $R_2$ scaled by $1/5$).

$\begin{array}{rl}& R_2^{\prime }\to R_1+R_3\\ & \text{won’t exist}\end{array}$ (Note: In the notes, the arrow in $R_2^{\prime }\to R_1+R_3$ is crossed out, showing this is not a valid elementary row operation).

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Page 17

$R_2^{\prime }\to R_2\pm \dots$ $R_2^{\prime }\to R_1-R_2(\text{X})$ $R_2^{\prime }\to -R_2+R_1[R_2\text{ আগে দিতে হবে}]$

$$\begin{array}{r}{R}_2^{\prime }\to R_2-R_1\\ R_1^{\prime }\to R_1-R_2\end{array}\}\begin{array}{rl}& \text{can’t be done}\\ & \text{zero হয়ে যাবে।}\end{array}$$

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# If a matrix $B$ is obtained from matrix $A$ by one or more elementary transformations, then $B$ is said to be equivalent to $A$. i.e. $A\sim B$ $[\text{If only one operation is done}\to \text{called elementary matrix to }A]$

Rank of a matrix:

The number of independent rows of a matrix is called rank of a matrix.

$$\left[\begin{array}{ccc}1& 2& 3\\ 5& 1& 6\\ 2& 4& 6\end{array}\right]\left[\begin{array}{ccc}5& 2& 6\\ 13& 13& 24\\ 1& 3& 4\end{array}\right]\to 2r_1+3r_2$$

(Note: In the second matrix, Row 2 is $2R_1+3R_3$. The note on the right is written as $2r_1+3r_2$ where $r_2$ refers to Row 3 or is an arithmetic typo).

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Page 18

$$A_{12}{B}_{22}=\left[\begin{array}{c}4\\ 5\end{array}\right]\left[\begin{array}{cc}7& 8\end{array}\right]=\left[\begin{array}{cc}28& 32\\ 35& 40\end{array}\right]$$ $$A_{11}{B}_{12}+A_{12}{B}_{22}=\left[\begin{array}{cc}39& 55\\ 45& 58\end{array}\right]$$ $$A_{21}{B}_{11}=\left[\begin{array}{ccc}6& 2& 3\\ 1& 2& 3\end{array}\right]\left[\begin{array}{c}3\\ 5\\ 1\end{array}\right]=\left[\begin{array}{c}18+6+9\\ 3+10+3\end{array}\right]=\left[\begin{array}{c}33\\ 16\end{array}\right]$$

(Note: Mathematically, $A_{21}{B}_{11}=\left[\begin{array}{c}31\\ 16\end{array}\right]$, but the notes have $18+6+9=33$ in the first row. The second element in the final vector is written as $16$, but acts as $32$ in subsequent additions).

$$A_{22}{B}_{21}=\left[\begin{array}{c}4\\ 4\end{array}\right]\left[\begin{array}{c}4\end{array}\right]=\left[\begin{array}{c}16\\ 16\end{array}\right]$$ $$A_{21}{B}_{11}+A_{22}{B}_{21}=\left[\begin{array}{c}49\\ 48\end{array}\right]$$

(Note: Mathematically, $\left[\begin{array}{c}33\\ 32\end{array}\right]+\left[\begin{array}{c}16\\ 16\end{array}\right]=\left[\begin{array}{c}49\\ 48\end{array}\right]$).

$$A_{21}{B}_{12}=\left[\begin{array}{ccc}6& 2& 3\\ 1& 2& 3\end{array}\right]\left[\begin{array}{cc}1& 2\\ 2& 6\\ 2& 3\end{array}\right]=\left[\begin{array}{cc}6+4+6& 12+12+9\\ 1+4+6& 2+12+9\end{array}\right]=\left[\begin{array}{cc}16& 33\\ 11& 23\end{array}\right]$$ $$A_{22}{B}_{22}=\left[\begin{array}{c}4\\ 4\end{array}\right]\left[\begin{array}{cc}7& 8\end{array}\right]=\left[\begin{array}{cc}28& 32\\ 28& 32\end{array}\right]$$ $$A_{21}{B}_{12}+A_{22}{B}_{22}=\left[\begin{array}{cc}44& 65\\ 39& 55\end{array}\right]$$

$\therefore$ Product of multiplication,

$$AB=\left[\begin{array}{ccc}32& 39& 55\\ 29& 45& 58\\ 49& 44& 65\\ 48& 39& 55\end{array}\right]$$

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Page 19

$$\left[\begin{array}{ccc}0& 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{array}\right]\left[\begin{array}{ccc}0& \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{array}\right]=I$$ $$\Rightarrow \left[\begin{array}{ccc}4{\beta }^2+{\gamma }^2& 2{\beta }^2-{\gamma }^2& -2{\beta }^2+{\gamma }^2\\ 2{\beta }^2-{\gamma }^2& {\alpha }^2+{\beta }^2+{\gamma }^2& {\alpha }^2-{\beta }^2-{\gamma }^2\\ -2{\beta }^2+{\gamma }^2& {\alpha }^2-{\beta }^2-{\gamma }^2& {\alpha }^2+{\beta }^2+{\gamma }^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$ $$2{\beta }^2-{\gamma }^2=0\Rightarrow 2{\beta }^2={\gamma }^2$$ $${\alpha }^2-{\beta }^2-{\gamma }^2=0$$ $$\Rightarrow {\alpha }^2-{\beta }^2-2{\beta }^2=0\Rightarrow {\alpha }^2=3{\beta }^2$$ $${\alpha }^2+{\beta }^2+{\gamma }^2=1\Rightarrow 3{\beta }^2+{\beta }^2+2{\beta }^2=1\Rightarrow \beta =\frac{1}{\sqrt{6}}$$ $${\gamma }^2=2\cdot \frac{1}{6}=\frac{1}{3}\Rightarrow \gamma =\frac{1}{\sqrt{3}}$$ $${\alpha }^2=3{\beta }^2=3\cdot \frac{1}{6}=\frac{1}{2}\Rightarrow \alpha =\frac{1}{\sqrt{2}}$$

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Page 20

* Partitioning Multiplication:

$$A={\left[\begin{array}{cccc}1& 2& 3& 4\\ 0& 1& 4& 5\\ 6& 2& 3& 4\\ 1& 2& 3& 4\end{array}\right]}_{4\times 4}B={\left[\begin{array}{ccc}3& 1& 2\\ 5& 2& 6\\ 1& 2& 3\\ 4& 7& 8\end{array}\right]}_{4\times 3}$$ $$AB=\left[\begin{array}{cc}{A}_{11}{B}_{11}+A_{12}{B}_{21}& A_{11}{B}_{12}+A_{12}{B}_{22}\\ A_{21}{B}_{11}+A_{22}{B}_{21}& A_{21}{B}_{12}+A_{22}{B}_{22}\end{array}\right]$$ $$A_{11}{B}_{11}=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 4\end{array}\right]\left[\begin{array}{c}3\\ 5\\ 1\end{array}\right]=\left[\begin{array}{c}3+10+3\\ 0+5+4\end{array}\right]=\left[\begin{array}{c}16\\ 9\end{array}\right]$$ $$A_{12}{B}_{21}=\left[\begin{array}{c}4\\ 5\end{array}\right]\left[\begin{array}{c}4\end{array}\right]=\left[\begin{array}{c}16\\ 20\end{array}\right]$$ $$A_{11}{B}_{11}+A_{12}{B}_{21}=\left[\begin{array}{c}32\\ 29\end{array}\right]$$ $$A_{11}{B}_{12}=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 4\end{array}\right]\left[\begin{array}{cc}1& 2\\ 2& 6\\ 2& 3\end{array}\right]=\left[\begin{array}{cc}1+4+6& 2+12+9\\ 0+2+8& 0+6+12\end{array}\right]=\left[\begin{array}{cc}11& 23\\ 10& 18\end{array}\right]$$

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Page 21

1st matrix column par. = 2nd matrix row partitioning partitioning

[HW] Prove that,

Square matrix = Hermitian + Skew-hermitian

Here, $A=\frac{1}{2}(A+A^{\ast })+\frac{1}{2}(A-A^{\ast })$ (Note: Beneath the first term is written $B$, and beneath the second term is written $C$).

$$B=\frac{1}{2}(A+A^{\ast })$$ $$\Rightarrow B^{\ast }=\frac{1}{2}(A+A^{\ast }{)}^{\ast }=\frac{1}{2}(A^{\ast }+A)=B$$ $$C=\frac{1}{2}(A-A^{\ast })$$ $$\Rightarrow C^{\ast }=\frac{1}{2}(A-A^{\ast }{)}^{\ast }=\frac{1}{2}(A^{\ast }-A)=-C$$

Q. Determine $\alpha ,\beta ,\gamma$ when,

$$\left[\begin{array}{ccc}0& 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{array}\right]\text{is orthogonal.}$$

We know, orthogonal, $A{A}^T=I$

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Page 22

* Determine $\alpha ,\beta ,\gamma$ when,

$$\left[\begin{array}{ccc}0& 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{array}\right]\text{is orthogonal}$$ $$[][]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$

(Note: The matrices on the LHS of the equation are drawn as empty brackets).

$$\Rightarrow \left[\begin{array}{ccc}4{\beta }^2+{\gamma }^2& 2{\beta }^2-{\gamma }^2& -2{\beta }^2+{\gamma }^2\\ 2{\beta }^2-{\gamma }^2& {\alpha }^2+{\beta }^2+{\gamma }^2& {\alpha }^2-{\beta }^2-{\gamma }^2\\ -2{\beta }^2+{\gamma }^2& {\alpha }^2-{\beta }^2-{\gamma }^2& {\alpha }^2+{\beta }^2+{\gamma }^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$

Partitioning:

$$A=\left[\begin{array}{cccccc}3& 0& 1& 5& 9& -2\\ -5& 2& 4& 0& -3& 1\\ -8& -6& 3& 1& 7& -4\end{array}\right]$$

(Note: Annotated with “column partitioning” pointing to the vertical partition lines, “row partitioning” pointing to the right side of the matrix, and “block matrix” pointing to the bottom-left).

Matrix Multiplication by partitioning:

$$A={\left[\begin{array}{cccc}1& 2& 3& 4\\ 0& 1& 4& 5\\ 6& 2& 3& 4\\ 1& 2& 3& 4\end{array}\right]}_{4\times 4}B={\left[\begin{array}{ccc}3& 1& 2\\ 5& 2& 6\\ 1& 2& 3\\ 4& 7& 8\end{array}\right]}_{4\times 3}$$

(Note: Annotated with submatrix labels $A_{11},A_{12},A_{21},A_{22},B_{11},B_{12},B_{21},B_{22}$ respectively).

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Method of contradiction:

If possible, $A=P+Q$ where $P\to \text{symmetric}$ and $Q\to \text{skew-symmetric}$ --- (1)

$$P^T=P{Q}^T=-Q$$

Then,

$$A^T=(P+Q{)}^T$$ $$=P^T+Q^T$$ $$=P-Q\text{— (ii)}$$

(1) + (ii),

$$A+A^T=2P$$ $$\Rightarrow P=\frac{1}{2}(A+A^T)$$

(1) - (ii),

$$Q=\frac{1}{2}(A-A^T)$$

Hermitian Matrix: If $A=[a_{ij}]$ is a square-matrix over the complex field and $A^{\ast }=(\overline{A}{)}^T$ (conjugate transpose) i.e. $a_{ij}=\overline{a_{ji}}$ for all $i,j$ then $A$ is called Hermitian matrix.

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

(Note: The diagonal elements $2,3,4$ are circled/slashed and annotated with: “diagonal always real num”).

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Skew-Hermitian:

$$A^{\ast }=(\overline{A}{)}^T=-A$$ $$\left[\begin{array}{ccc}0& 1+i& 3+2i\\ -1+i& 0& i\\ -3+2i& i& 0\end{array}\right]$$

(Note: Annotated with: ”$\to$ diagonal can be either zero or pure imaginary (real part zero)”).

HW * Square matrix = (Hermitian + Skew-hermitian) $\to$ prove.

$$A=\frac{1}{2}(A+A^{\ast })+\frac{1}{2}(A-A^{\ast })$$

More types of matrix — (Def.)

1. Singular : $|A|=0$
2. Orthogonal : $A{A}^T=I$
3. Idempotent : $A^2=A$
4. Nilpotent : $A^n=0$ but $A^{n-1}\ne 0$, nilpotent of order $n$.
5. Periodic : $A^{m+1}=A\to A$ is said to be periodic with period $m$.
6. Involutory : $A^2=I$
7. Unitary : $A{A}^{\ast }=I=A^{\ast }A$

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ii. Upper and lower triangular matrix:

A square matrix whose elements $a_{ij}=0$ for $i>j$ is called an upper triangular matrix and whose elements $a_{ij}=0$ for $i<j$ is called a lower triangular matrix.

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14.07.26 class-02

Every square matrix can be uniquely expressed as the sum of a symmetric and skew-symmetric matrices.

$$\text{square matrix}=\text{(symmetric)}+\text{(skew-symmetric)}$$

when $A$ be a square matrix,

$$A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)$$

(Note: Beneath the first term is written $B$, and beneath the second term is written $C$).

$$B^T={\left(\frac{1}{2}(A+A^T)\right)}^T=\frac{1}{2}(A^T+(A^T{)}^T)$$ $$=\frac{1}{2}(A^T+A)$$ $$=B$$ $$C^T={\left(\frac{1}{2}(A-A^T)\right)}^T$$ $$=\frac{1}{2}(A^T-(A^T{)}^T)$$ $$=\frac{1}{2}(A^T-A)$$ $$=-C$$

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Definition: —

A matrix is a rectangular array of numbers (real or complex) enclosed by a pair of brackets. The numbers in the array are called the entries or the elements of the matrix.

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Types of Matrix (Def. and examples) — (Note: “formal def.” is written in green next to the title).

1. Row Matrix : 1 row
2. Column matrix : 1 column (Note: There is a vertical arrow $\downarrow$ drawn below this line).
3. Rectangular : Unequal numbers of row and column.
4. Square matrix : Equal number of row and column.
5. Scalar matrix : A diagonal whose elements are same.
6. Diagonal matrix : A square matrix whose elements $a_{ij}=0$ when $i\ne j$ is called a diagonal matrix.

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7. Identity matrix
8. Null matrix
9. Symmetric matrix : A matrix is equal to its transpose. i.e. $[A=A^T]$ A square matrix such that $a_{ij}=a_{ji}$ for $1\le i,j\le n$ is said to be symmetric matrix.

$$\left[\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right]$$

(Note: “Symmetric” is written below the matrix with a wavy underline).

10. Skew-symmetric matrix : Same as before — such that $[a_{ij}=-a_{ji}][A=-A^T]$ (diagonal always zero)

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

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1st cycle 13.07.2026 class-01

Topics

1. Vector analysis
2. Fourier series, integral, transform (Arif sir) (Note: The word “trans” was written before “series” but is crossed out in the notes).
3. Laplace transformation
4. Matrices (Arif sir) (Note: The number is written with a prefix that looks like an “8” or “B” or section symbol, i.e., “84.” or “§4.“)

Books

1. Linear Algebra with applications — Howard Anton
2. Linear Algebra with applications — David C. Lay (Note: A curly bracket connects these two books, pointing to “matrix”).

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H.W. # Importance of Matrix (Applications, usage) —

• $\to$ RGB coloring, photo editing
• $\to$ signal processing
• $\to$ —