Some Special Types of Matrices

Idempotent Matrix:

A square matrix \(\mathbf{A}\) is called idempotent if, when multiplied by itself, it yields the same matrix. In other words, \(\mathbf{A}\) satisfies the equation:

\[ \mathbf{A}^2 = \mathbf{A} \]

This means that applying the matrix operation multiple times has the same effect as applying it once.

An example of an idempotent matrix is:

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

To verify that it is idempotent, we compute \(\mathbf{A}^2\):

\[ \mathbf{A}^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \mathbf{A} \]

Since \(\mathbf{A}^2 = \mathbf{A}\), it is idempotent.

Properties of Idempotent Matrices:

If \(\mathbf{A}\) is idempotent, then \(\mathbf{A}^n = \mathbf{A}\) for all \(n \geq 2\):

This property follows directly from the definition of an idempotent matrix, which states that \(\mathbf{A}^2 = \mathbf{A}\). Now, if we compute \(\mathbf{A}^n\) for any \(n \geq 2\), we have:

\[ \mathbf{A}^n = \mathbf{A} \cdot \mathbf{A}^{n-1} \]

But since \(\mathbf{A}^2 = \mathbf{A}\), we can replace any higher power of \(\mathbf{A}\) by \(\mathbf{A}\) itself. Therefore, for any \(n \geq 2\):

\[ \mathbf{A}^n = \mathbf{A} \]

This means applying the matrix \(\mathbf{A}\) multiple times has the same effect as applying it once, which is a core characteristic of idempotent matrices.
If \(\mathbf{A}\mathbf{B} = \mathbf{A}\) and \(\mathbf{B}\mathbf{A} = \mathbf{B}\), then \(\mathbf{A}\) and \(\mathbf{B}\) are idempotent:

To prove this, we start by showing that \(\mathbf{A}\) is idempotent:

\[ \mathbf{A}^2 = \mathbf{A}\mathbf{A} = \mathbf{A}\mathbf{B}\mathbf{A} = \mathbf{A}(\mathbf{B}\mathbf{A}) = \mathbf{A}\mathbf{B} = \mathbf{A} \]

Hence, \(\mathbf{A}\) is idempotent. Similarly, we show that \(\mathbf{B}\) is idempotent:

\[ \mathbf{B}^2 = \mathbf{B}\mathbf{B} = \mathbf{B}\mathbf{A}\mathbf{B} = \mathbf{B}(\mathbf{A}\mathbf{B}) = \mathbf{B}\mathbf{A} = \mathbf{B} \]

Therefore, \(\mathbf{B}\) is also idempotent.
1. If \(\mathbf{A}\) is idempotent, then \((\mathbf{I} + \mathbf{A})^2 = \mathbf{I} + 3\mathbf{A}\):
  
  Let’s compute \((\mathbf{I} + \mathbf{A})^2\):
  
  \[ (\mathbf{I} + \mathbf{A})^2 = (\mathbf{I} + \mathbf{A})(\mathbf{I} + \mathbf{A}) \]
  
  Expanding the right-hand side:
  
  \[ (\mathbf{I} + \mathbf{A})(\mathbf{I} + \mathbf{A}) = \mathbf{I}^2 + \mathbf{I}\mathbf{A} + \mathbf{A}\mathbf{I} + \mathbf{A}^2 \]
  
  Since \(\mathbf{A}^2 = \mathbf{A}\), this simplifies to:
  
  \[ \mathbf{I} + \mathbf{A} + \mathbf{A} + \mathbf{A} = \mathbf{I} + 3\mathbf{A} \]
  
  Therefore, we have:
  
  \[ (\mathbf{I} + \mathbf{A})^2 = \mathbf{I} + 2\mathbf{A} \]
  
  This shows that when \(\mathbf{A}\) is idempotent, adding the identity matrix to \(\mathbf{A}\) and squaring the result gives \(\mathbf{I} + 2\mathbf{A}\).
  
  \((\mathbf{I} + \mathbf{A})^3 = \mathbf{I} + 7\mathbf{A}\) for an idempotent matrix \(\mathbf{A}\).
2. Similarly, \((\mathbf{I} + \mathbf{A})^3 = \mathbf{I} + 7\mathbf{A}\)
  
  To prove, we start by expanding \((\mathbf{I} + \mathbf{A})^3\). Using the binomial expansion for matrices, we have:
  
  \[ (\mathbf{I} + \mathbf{A})^3 = \mathbf{I}^3 + 3\mathbf{I}^2\mathbf{A} + 3\mathbf{I}\mathbf{A}^2 + \mathbf{A}^3 \]
  
  we get,
  
  \[ (\mathbf{I} + \mathbf{A})^3 = \mathbf{I} + 3\mathbf{A} + 3\mathbf{A} + \mathbf{A} \]
  
  Simplifying the right-hand side:
  
  \[ (\mathbf{I} + \mathbf{A})^3 = \mathbf{I} + 7\mathbf{A} \]
  
  Thus, we have shown that:
  
  \[ (\mathbf{I} + \mathbf{A})^3 = \mathbf{I} + 7\mathbf{A} \]
3. \((\mathbf{I} + \mathbf{A})^n = \mathbf{I} + (2^n - 1)\mathbf{A}\) for an idempotent matrix \(\mathbf{A}\).
  
  To understand this, consider the example of \((\mathbf{I} + \mathbf{A})^4\):
  
  \[ (\mathbf{I} + \mathbf{A})^4 = \mathbf{I}^4 + 4\mathbf{I}^3\mathbf{A} + 6\mathbf{I}^2\mathbf{A}^2 + 4\mathbf{I}\mathbf{A}^3 + \mathbf{A}^4 \]
  
  Since \(\mathbf{A}\) is idempotent (\(\mathbf{A}^n = \mathbf{A}\) for \(n \geq 2\)), this simplifies to:
  
  \[ (\mathbf{I} + \mathbf{A})^4 = \mathbf{I} + 4\mathbf{A} + 6\mathbf{A} + 4\mathbf{A} + \mathbf{A} = \mathbf{I} + 15\mathbf{A} \]
  
  We notice a pattern: for \(n = 2\), \((\mathbf{I} + \mathbf{A})^2 = \mathbf{I} + 3\mathbf{A}\); for \(n = 3\), \((\mathbf{I} + \mathbf{A})^3 = \mathbf{I} + 7\mathbf{A}\). The coefficients \(3\), \(7\), and \(15\) follow the form \(2^n - 1\). Therefore, for any \(n\), we expect:
  
  \[ (\mathbf{I} + \mathbf{A})^n = \mathbf{I} + (2^n - 1)\mathbf{A} \]
  
  Proof:
  
  To prove this, use the binomial theorem to expand \((\mathbf{I} + \mathbf{A})^n\):
  
  \[ (\mathbf{I} + \mathbf{A})^n = \binom{n}{0} \mathbf{I} + \binom{n}{1} \mathbf{A} + \binom{n}{2} \mathbf{A}^2 + \dots + \binom{n}{n} \mathbf{A}^n \]
  
  Since \(\mathbf{A}^k = \mathbf{A}\) for all \(k \geq 2\) (because \(\mathbf{A}\) is idempotent), this simplifies to:
  
  \[ (\mathbf{I} + \mathbf{A})^n = \mathbf{I} + \mathbf{A} \left( \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} \right) \]
  
  We know from the binomial expansion that the sum of the binomial coefficients is:
  
  \[ \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n - 1 \]
  
  Thus, we conclude:
  
  \[ (\mathbf{I} + \mathbf{A})^n = \mathbf{I} + (2^n - 1)\mathbf{A} \]

Involutory Matrix

An involutory matrix is a square matrix \(\mathbf{A}\) such that \(\mathbf{A}^2 = \mathbf{I}\), where \(\mathbf{I}\) is the identity matrix. This means that the matrix is its own inverse, i.e., \(\mathbf{A}^{-1} = \mathbf{A}\).

An example of an involutory matrix is:

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

To verify that it is involutory, we compute \(\mathbf{A}^2\):

\[ \mathbf{A}^2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{I} \]

Since \(\mathbf{A}^2 = \mathbf{I}\), it is an involutory matrix.

Nilpotent Matrix

A nilpotent matrix is a square matrix \(\mathbf{A}\) such that \(\mathbf{A}^k = \mathbf{O}\) for some positive integer \(k\), where \(\mathbf{O}\) is the zero matrix. The smallest such \(k\) is called the index of nilpotency.

Consider the matrix:

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

We compute \(\mathbf{A}^2\):

\[ \mathbf{A}^2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \mathbf{O} \]

Since \(\mathbf{A}^2 = \mathbf{O}\), \(\mathbf{A}\) is a nilpotent matrix with an index of nilpotency \(k = 2\).

Another example:

\[ \mathbf{S} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \]

Let's compute its powers:

\[ \mathbf{S}^2 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

\[ \mathbf{S}^3 = \mathbf{S}^2 \cdot \mathbf{S} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \mathbf{O} \]

Thus, \(\mathbf{S}^3 = \mathbf{O}\), showing that this matrix is nilpotent with an index of nilpotency \(k = 3\).

Periodic Matrix

A periodic matrix is a square matrix \(\mathbf{A}\) that satisfies the equation:

\[ \mathbf{A}^{k+1} = \mathbf{A} \]

for some positive integer \(k\). The smallest such \(k\) is called the period of the matrix. This means that after raising \(\mathbf{A}\) to the power of \(k+1\), the matrix returns to itself.

Consider the matrix:

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

Let's compute powers of \(\mathbf{A}\):

\[ \mathbf{A}^2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{I} \]

\[ \mathbf{A}^3 = \mathbf{A}^2 \cdot \mathbf{A} = \mathbf{I} \cdot \mathbf{A} = \mathbf{A} \]

Since \(\mathbf{A}^3 = \mathbf{A}\), \(\mathbf{A}\) is a periodic matrix with period \(k = 2\), satisfying \(\mathbf{A}^{k+1} = \mathbf{A}\).

Orthogonal Matrices

A square matrix \( A \) is called orthogonal if it satisfies the condition:

\[ A A^T = I \]

where \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix.

Determinant of an Orthogonal Matrix

Taking the determinant on both sides of \( A A^T = I \), we get:

\[ |A A^T| = |I| \]

Using the property \( |A B| = |A||B| \) for the product of matrices, we obtain:

\[ |A| |A^T| = 1 \]

Since \( |A^T| = |A| \), it follows that:

\[ |A|^2 = 1 \implies |A| = \pm 1 \]

Thus, the determinant of an orthogonal matrix is either \( 1 \) or \( -1 \).

Therefore, orthogonal matrices are always non-singular.

If \( A A^T = I \), then multiplying both sides by \( A^{-1} \) from the left yields:

\[ A^{-1} A A^T = A^{-1} I \implies \boxed{A^T = A^{-1}} \]

Multiplying \( A^T = A^{-1} \) on the right by \( A \), we have: [ \boxed{A^T A = I} ]

Properties of Orthogonal Matrices

If \( A \) is an orthogonal matrix, then the following properties hold:

\[ A A^\top = I \quad \text{and} \quad A^\top A = I \]

Additionally, the inverse of an orthogonal matrix is equal to its transpose:

\[ A^{-1} = A^\top \]
If \( A \) is orthogonal, then the determinant of \( A \) is either \( +1 \) or \( -1 \):

\[ |A| = \pm 1 \]
If \( A \) and \( B \) are orthogonal matrices, then \( A B \) is orthogonal.