Adjoint Of Adjoint Of A Matrix

The adjoint of a matrix, also known as the adjugate, plays a central role in linear algebra because it connects the concepts of determinant, inverse, and cofactor expansion. When we apply the adjoint operation twice—taking the adjoint of the adjoint—we obtain a matrix that exhibits a striking pattern tied to the original matrix’s size and determinant. Understanding this double‑adjoint relationship not only deepens our grasp of matrix theory but also provides useful shortcuts in computations involving inverses and characteristic polynomials And it works..

Definition of the Adjoint (Adjugate)

For an (n \times n) matrix (A = [a_{ij}]) over a field (commonly (\mathbb{R}) or (\mathbb{C})), the adjoint of (A), denoted (\operatorname{adj}(A)) or sometimes (A^{\ast}), is the transpose of its cofactor matrix.

1. Compute the cofactor (C_{ij}=(-1)^{i+j}\det(M_{ij})), where (M_{ij}) is the ((n-1)\times (n-1)) submatrix obtained by deleting row (i) and column (j).
2. Form the cofactor matrix (C = [C_{ij}]).
3. Take the transpose: (\operatorname{adj}(A)=C^{\mathsf T}).

A fundamental identity links the adjoint to the determinant and the inverse (when it exists):

[ A \cdot \operatorname{adj}(A) = \operatorname{adj}(A) \cdot A = \det(A), I_n . \tag{1} ]

If (\det(A)\neq 0), we can rewrite (1) as (A^{-1}= \frac{1}{\det(A)}\operatorname{adj}(A)) Easy to understand, harder to ignore..

Properties of the Adjoint

Before examining the adjoint of the adjoint, recall several useful properties that hold for any square matrix (A):

• Linearity in each entry: (\operatorname{adj}(A)) is a polynomial of degree (n-1) in the entries of (A).
• Transpose: (\operatorname{adj}(A^{\mathsf T}) = (\operatorname{adj}(A))^{\mathsf T}).
• Scalar multiplication: For a scalar (c), (\operatorname{adj}(cA)=c^{,n-1}\operatorname{adj}(A)).
• Product rule: (\operatorname{adj}(AB)=\operatorname{adj}(B)\operatorname{adj}(A)) (note the reversed order).
• Power: (\operatorname{adj}(A^k)=\bigl(\operatorname{adj}(A)\bigr)^k) when (A) is invertible.
• Determinant: (\det\bigl(\operatorname{adj}(A)\bigr)=\bigl(\det(A)\bigr)^{,n-1}).

These facts will be instrumental when we iterate the adjoint operation.

Adjoint of the Adjoint: Derivation

Let (B = \operatorname{adj}(A)). We want an explicit expression for (\operatorname{adj}(B)=\operatorname{adj}\bigl(\operatorname{adj}(A)\bigr)). Start from identity (1) applied to (B):

[ B \cdot \operatorname{adj}(B) = \det(B), I_n . \tag{2} ]

Using the determinant property of the adjoint, (\det(B)=\det\bigl(\operatorname{adj}(A)\bigr)=\bigl(\det(A)\bigr)^{,n-1}). Substituting (B=\operatorname{adj}(A)) into (2) gives

[ \operatorname{adj}(A) \cdot \operatorname{adj}\bigl(\operatorname{adj}(A)\bigr) = \bigl(\det(A)\bigr)^{,n-1} I_n . \tag{3} ]

Now multiply (3) on the left by (A). Using (1) again, (A\operatorname{adj}(A)=\det(A)I_n):

[ A\bigl[\operatorname{adj}(A) \cdot \operatorname{adj}\bigl(\operatorname{adj}(A)\bigr)\bigr] = \bigl(A\operatorname{adj}(A)\bigr) \cdot \operatorname{adj}\bigl(\operatorname{adj}(A)\bigr) = \det(A)I_n \cdot \operatorname{adj}\bigl(\operatorname{adj}(A)\bigr) = \det(A),\operatorname{adj}\bigl(\operatorname{adj}(A)\bigr). ]

But the left‑hand side can also be regrouped as ((A\operatorname{adj}(A)) \cdot \operatorname{adj}\bigl(\operatorname{adj}(A)\bigr)) which we already simplified. Equating this to the right‑hand side of (3) after left‑multiplication by (A) yields

[ \det(A),\operatorname{adj}\bigl(\operatorname{adj}(A)\bigr) = A\bigl[\bigl(\det(A)\bigr)^{,n-1} I_n\bigr] = \bigl(\det(A)\bigr)^{,n} A . ]

Assuming (\det(A)\neq 0) (the invertible case), we can divide both sides by (\det(A)) to obtain

[ \boxed{\operatorname{adj}\bigl(\operatorname{adj}(A)\bigr)=\bigl(\det(A)\bigr)^{,n-2} A } . \tag{4} ]

Equation (4) is the core result: the adjoint of the adjoint of an invertible (n\times n) matrix equals the original matrix scaled by ((\det A)^{,n-2}).

When (\det(A)=0) (singular case), the formula must be interpreted carefully because division by zero is not allowed. We treat singular matrices separately in the next section And that's really what it comes down to. Nothing fancy..

Special Cases

1. Invertible Matrices ((\det A \neq 0))

For any invertible matrix, (4) holds directly. Notable consequences:

• If (n=2), then (\operatorname{adj}(\operatorname{adj}(A)) = (\det A)^{0} A = A). Hence for (2\times2) matrices the double adjoint returns the original matrix exactly.
• If (n=3), (\operatorname{adj}(\operatorname{adj}(A)) = (\det A)^{1} A = (\det A) A). The double adjoint is simply the original matrix multiplied by its determinant.
• As (n) grows, the scaling factor ((\det A)^{,n-2}) can become very large or very small, reflecting the high‑degree polynomial nature of the adjoint operation.

2. Singular Matrices ((\det A = 0))

When (\det A = 0), identity (1) tells us that (A\operatorname{adj}(A)=0). Because of this, the rank of (\operatorname{adj}(A)) is at most 1 (in fact, it is either zero or has rank 1 unless (A) is the zero matrix). Applying the adjoint a