The definition of diagonalization states that, An n × n matrix A is diagonalizable if re exists an n × n invertible matrix P and an n × n diagonal matrix D such that,
P^-1 AP = D
A = PDP^-1
Aⁿ = (PDP^-1)ⁿ = PDⁿP^-1

Hadamard matrix is named after a famous French mamatician, Jacques Hadamard. It is defined as ?a square matrix whose entries are only 1 or -1 and whose column (or row) vectors orthogonal?.

For a matrix A, A^k+1 = A, where k is a positive integer is known as periodic matrix.
Whereas, if A² = A, i.e. k=1, n it is known as idempotent matrix and if A^k = 0, n it is known as Nilpotent matrix.

To prove that A = A^T, let us consider an example,
\(A = \begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix}\)
?A ? ?I ? = 0
\( \begin{vmatrix} 1-? & 0 \\ 2 & 3-? \end{vmatrix} = 0\)
(1-?) (3-?) = 0
3 ? ? – 3? +?² = 0
?² – 4? + 3 = 0
(? – 3) (? – 1) = 0
? = 1, 3
Consider \(A^T = \begin{bmatrix}1 & 2\\ 0 & 3 \end{bmatrix}\)
?A ? ?I ? = 0
\(\begin{bmatrix}1-? & 2 \\ 0 & 3-? \end{bmatrix} = 0\)
(1-?) (3-?) = 0, which is similar to result obtained for A, hence eigen values are same.

The general form of a skew symmetric matrix is given by,
\(\begin{bmatrix}0 & w1 & w2 \\ w1 & 0 & w3 \\ w2 & w3 & 0\end{bmatrix} \)

Therefore, from given matrix,
x = 6,
x+y = 3 → 6+y=3 → y = -3

To prove above statement, let us consider an example,
A = \(\begin{bmatrix} 0 & -4 & 1 \\ 4 & 0 & -3 \\-1 & 3 & 0\end{bmatrix} \)

Therefore, A + A = \(\begin{bmatrix} 0 & -4 & 1 \\ 4 & 0 & -3 \\ -1 & 3 & 0\end{bmatrix} \) + \(\begin{bmatrix}0&-4 & 1 \\ 4 & 0&-3 \\-1 & 3 & 0\end{bmatrix} \) = \(\begin{bmatrix} 0 & -8 & 2 \\ 8 & 0 & -9\\ -2 & 6 & 0\end{bmatrix} \) which is also a skew-symmetric matrix.

There are generally two types of integrals,
I. Definite Integrals: These are furr classified as,

• Proper Definite Integrals
• Improper Definite Integrals

II. Indefinite Integrals

We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
Given that, λ=3
[A-λI]=\(\begin{bmatrix}3&10&5\\-2&-3&-4\\3&5&7\end{bmatrix}-(3)\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)
[A-λI]=\(\begin{bmatrix}3-3&10&5\\-2&-3-3&-4\\3&5&7-3\end{bmatrix}\)
[A-λI]=\(\begin{bmatrix}0&10&5\\-2&-6&-4\\3&5&4\end{bmatrix}\)
Since, [A-λI]X=0
\(\begin{bmatrix}0&10&5\\-2&-6&-4\\3&5&4\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}\)
Using Row transformation

Procedure to find invertible matrix is as follows,
Step 1: Find eigen values of given matrix.
A = \(\begin{bmatrix}
2 & 0 & 0 \\
1 & 2 & 1\\
-1 & 0 & 1
\end{bmatrix} \)
?A ? ?I ? = 0
\(\begin{vmatrix}
2-? & 0 & 0\\
1 & 2-? & 1\\
-1 & 0 & 1-?
\end{vmatrix} \) = 0 ????????? (i)
(2-?) ((2- ?) (1- ?)) = 0
(2- ?)² (1-?) = 0
? = 2, 2, 1
Step 2: Compute eigen vectors
Consider ? = 2,
(A – ?I) \(\vec{X} = \vec{0} \)
\(\begin{bmatrix}
0 & 0 & 0\\
1 & 0 & 1 \\
-1 & 0 & -1
\end{bmatrix}
\vec{X} = \vec{0} \quad ^{\underrightarrow{Reducing \,furr, \,we get,\,}}
\begin{bmatrix}0 & 0 & 0 \\
1 & 0 & 1 \\
0 & 0 & 0\end{bmatrix}
\vec{X}=\vec{0}
\)
x2 is free variable, hence, x2 = s
Let x1 = -t, x3 = t, since x1+x3 = 0
\(\vec{X} = s\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \) + t\(\begin{bmatrix}
-1\\
0\\
1
\end{bmatrix} \)
\(\vec{X_1} = \begin{bmatrix}
0\\
1\\
0
\end{bmatrix} \) \(\vec{X_2} = \begin{bmatrix}
-1\\
0\\
1\end{bmatrix} \)
Consider ? = 1
(A – ?I)\(\vec{X} = \vec{0} \)
\(\begin{bmatrix}
1 & 0 & 0\\
1 & 1 & 1 \\
-1 & 0 & 0
\end{bmatrix}
\vec{X} = \vec{0} \quad ^{\underrightarrow{Reducing \,furr, \,we get,\,}}
\begin{bmatrix}1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 0\end{bmatrix}
\vec{X}=\vec{0}
\)
x1 = 0, Let x2 = -s and x3 = s since x2+x3=0
\(\vec{X}= s\begin{bmatrix}
0\\
-1\\
1
\end{bmatrix} \)
\(\vec{X_3} = \begin{bmatrix}
0\\
-1\\
1
\end{bmatrix} \)
Step 3: Formation of invertible matrix.
\(P = [\vec{X_1} \vec{X_2} \vec{X_3}]\)
P = \(\begin{bmatrix}
0 & -1 & 0 \\
1 & 0 & -1\\
0 & 1 & 1
\end{bmatrix} \)