Let A be any n × n-matrix, A = (aij). Then the determinant can be computed by the development after a column or row, e.g.,

detA = n∑

aij(−1)i+j detA(i,j),

where A(i,j) is the (i, j)-th complement, i.e., the matrix obtained by omitting the i-th row and the j-th column from the matrix A. This means that if we define a matrix B = (bjk) as

bjk = (−1)k+j detA(k,j),

then we have traced the inverse of the matrix A, provided it exists. At least we can write

aijbjk = δik detA,

or in matrix form, AB = (detA) I . (4.28)

p(ξ) = det (ξ I−A) = ξn + an−1ξn−1 + · · ·+ a0 (4.29)

is the characteristic polynomial for the matrix A, then the matrix p (A) satisﬁes

p (A) = An + an−1An−1 + · · ·+ a0 I = O. (4.30)

Proof: We apply (4.28) to the matrix λ I−A to get

p(λ) I = (λ I−A)B(λ), (4.31)

where B(λ) = (bij(λ)) is a matrix of polynomials in λ, defined as the (j, i)-th complement of the matrix λ I−A. Hence we can write

B(λ) = λn−1Bn−1 + · · ·+ λB1 +B0

as a polynomial in λ with coefficients which are matrices independent of λ. For any k ≥ 1 we can write

Ak − λk I = (A− λ I) (Ak−1 + λAk−2 + · · ·+ λk−1 I) .