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:
This means that applying the matrix operation multiple times has the same effect as applying it once.
An example of an idempotent matrix is:
To verify that it is idempotent, we compute \(\mathbf{A}^2\):
Since \(\mathbf{A}^2 = \mathbf{A}\), it is idempotent.
Properties of Idempotent Matrices:
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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.
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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.
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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 \(\mathbf{A}\).
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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:
\[ (\mathbf{I} + \mathbf{A})^3 = \mathbf{I} + 7\mathbf{A} \] -
\((\mathbf{I} + \mathbf{A})^n = \mathbf{I} + (2^n - 1)\mathbf{A}\) 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:
\[ (\mathbf{I} + \mathbf{A})^n = \mathbf{I} + (2^n - 1)\mathbf{A} \]
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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:
To verify that it is involutory, we compute \(\mathbf{A}^2\):
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:
We compute \(\mathbf{A}^2\):
Since \(\mathbf{A}^2 = \mathbf{O}\), \(\mathbf{A}\) is a nilpotent matrix with an index of nilpotency \(k = 2\).
Another example:
Let's compute its powers:
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:
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:
Let's compute powers of \(\mathbf{A}\):
Since \(\mathbf{A}^3 = \mathbf{A}\), \(\mathbf{A}\) is a periodic matrix with period \(k = 2\), satisfying \(\mathbf{A}^{k+1} = \mathbf{A}\).