Idempotent, Nilpotent, And Involutory Matrices

Three Identities That Control Matrix Powers

Repeatedly multiplying a matrix by itself can become tedious. For certain square matrices, however, a simple identity determines the entire sequence of powers. An idempotent matrix repeats after its first power, a nilpotent matrix eventually reaches the zero matrix, and an involutory matrix alternates between two values. Recognizing which identity holds turns many power calculations into short, predictable patterns.

Throughout, matrices are square matrices in \(\mathbb{R}^{n\times n}\), so their powers are defined. We use \(\mathbf{A}^0=\mathbf{I}_n\), and \(\mathbf{A}^k\) for a positive integer \(k\) means the product of \(k\) copies of \(\mathbf{A}\). The symbols \(\mathbf{I}_n\) and \(\mathbf{O}\) denote the identity matrix and zero matrix of the required size.

These are exact identities, not descriptions of how a matrix “usually behaves.” To identify an idempotent or an involutory matrix, calculate its square and compare it with the matrix or identity, respectively. To establish nilpotence, show that at least one positive power is zero. A nonzero square by itself does not rule out nilpotence: the power that becomes zero might be higher.

Idempotent Matrices: Every Positive Power Repeats

If \(\mathbf{A}^2=\mathbf{A}\), multiplying by \(\mathbf{A}\) again does not change the result. The same is true for every later positive power. The exponent zero is an exception to this pattern: by convention it gives \(\mathbf{I}_n\), not necessarily \(\mathbf{A}\).

Proof. We prove the positive-power statement by induction on \(m\). For \(m=1\), \(\mathbf{A}^1=\mathbf{A}\). Suppose \(\mathbf{A}^m=\mathbf{A}\) for some positive integer \(m\). By the definition of powers and associativity of matrix multiplication, \(\mathbf{A}^{m+1}=\mathbf{A}^m\mathbf{A}=\mathbf{A}\mathbf{A}=\mathbf{A}^2=\mathbf{A}\). Thus the statement holds for \(m+1\), and induction proves it for every positive \(m\). The zero-power statement follows from the convention \(\mathbf{A}^0=\mathbf{I}_n\). \(\square\)

Nilpotent Matrices: A Power Eventually Vanishes

For a nilpotent matrix, there is at least one positive exponent \(k\) for which \(\mathbf{A}^k=\mathbf{O}\). The least such \(k\) is the nilpotency index. Once a power is zero, every later power is zero too: if \(m\geq k\), then \(\mathbf{A}^m=\mathbf{A}^k\mathbf{A}^{m-k}=\mathbf{O}\). For \(m=k\), this uses \(\mathbf{A}^0=\mathbf{I}_n\), and \(\mathbf{O}\mathbf{I}_n=\mathbf{O}\).

Proof. Let \(m\geq k\). The power law for powers of the same matrix gives \(\mathbf{A}^m=\mathbf{A}^k\mathbf{A}^{m-k}\). This includes \(m=k\), since then the second factor is \(\mathbf{A}^0=\mathbf{I}_n\). Substituting \(\mathbf{A}^k=\mathbf{O}\) yields \(\mathbf{A}^m=\mathbf{O}\mathbf{A}^{m-k}=\mathbf{O}\), because multiplying a zero matrix by any conformable matrix gives a zero matrix. \(\square\)

Involutory Matrices: Powers Depend on Parity

An involutory matrix satisfies \(\mathbf{A}^2=\mathbf{I}_n\). Thus every pair of consecutive factors \(\mathbf{A}\mathbf{A}\) can be replaced by \(\mathbf{I}_n\). For an even exponent all factors can be paired; for an odd exponent, one factor remains. Consequently, the powers alternate between the identity matrix and \(\mathbf{A}\).

Proof. Let \(m\) be even, so \(m=2q\) for a nonnegative integer \(q\). Grouping the factors in pairs gives \(\mathbf{A}^{2q}=(\mathbf{A}^2)^q=\mathbf{I}_n^q=\mathbf{I}_n\); when \(q=0\), this agrees with \(\mathbf{A}^0=\mathbf{I}_n\). If \(m\) is positive and odd, write \(m=2q+1\) for a nonnegative integer \(q\). Then \(\mathbf{A}^{2q+1}=(\mathbf{A}^2)^q\mathbf{A}=\mathbf{I}_n^q\mathbf{A}=\mathbf{A}\). These two cases cover every nonnegative exponent. \(\square\)

How the Three Conditions Relate

The defining equations should not be conflated: idempotence compares the square with \(\mathbf{A}\), nilpotence compares some positive power with \(\mathbf{O}\), and involution compares the square with \(\mathbf{I}_n\). A matrix may satisfy more than one condition, but the overlap between idempotence and nilpotence is very limited.

Proof. Suppose \(\mathbf{A}\) is idempotent and nilpotent. Idempotence implies \(\mathbf{A}^m=\mathbf{A}\) for every positive integer \(m\). Since \(\mathbf{A}\) is nilpotent, there is a positive integer \(k\) with \(\mathbf{A}^k=\mathbf{O}\). The positive-power identity then gives \(\mathbf{A}=\mathbf{A}^k=\mathbf{O}\). Conversely, if \(\mathbf{A}=\mathbf{O}\), then \(\mathbf{A}^2=\mathbf{O}=\mathbf{A}\), so it is idempotent, and \(\mathbf{A}^1=\mathbf{O}\), so it is nilpotent. \(\square\)

For practical calculations, keep the defining tests and resulting patterns separate:

Common Mistake: Stopping at a Nonzero Square

A common error is to calculate \(\mathbf{A}^2\ne\mathbf{O}\) and conclude that \(\mathbf{A}\) is not nilpotent. That conclusion is not justified: nilpotence requires some positive power to be zero, not necessarily the square. In the nilpotent example above, \(\mathbf{N}^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}\ne\mathbf{O}\), but \(\mathbf{N}^3=\mathbf{O}\). The correction is to continue calculating powers or to provide another valid argument; a nonzero square alone settles neither nilpotence nor non-nilpotence.

Conversely, to show that a matrix is nilpotent, it is enough to display one positive power that equals \(\mathbf{O}\). The index requires one more check: verify that every smaller positive power is nonzero. To show idempotence or involution, calculate the square directly and compare every entry with the claimed result.