Skip to content

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:

  1. 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.

  2. 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 \(\mathbf{A}^{k+1} = \mathbf{A}\).