By Dragoslav S. Mitrinovic, J. Pecaric, A.M Fink

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Then (M T x)T y = xT (My) = 0 for every y ∈ K m . 2 Matrices as Linear Maps 25 is ker M T . 5, we ﬁnd also that the orthogonal of ker M T is R(M). Exchanging the roles of M and MT leads to the following. 6 The orthogonal of R(M) is ker M T and that of ker M is R(M T ). The following consequence is sometimes called the Fredholm principle. 1 Let M ∈ Mn×m (K) and b ∈ K n . In order that the linear equation Mx = b be solvable, it is necessary and sufﬁcient that zT b = 0 for every z ∈ ker(M T ). 6, we obtain the following identities for a matrix M ∈ Mn×m (K): n = dim ker M T + rk M T , m = dim ker M + rk M, n = dim ker M T + rk M, m = dim ker M + rk M T .

Xnn ] is an integral domain with a unit, in which we may apply the previous results. Let us deﬁne the matrix X = (xi j )1≤i, j≤n , which belongs to Mn (A[x11 , . . , xnn ]). Its determinant is a polynomial, which we denote DetA . We observe that DetA does not really depend on the ring A, in the sense that it is the image of DetZ through the canonical ring homomorphism Z → A. For this reason, we simply write Det. Given M ∈ Mn (A), the determinant of M is obtained by substituting the entries mi j to the indeterminates xi j in Det.

Actually, it must be understood in the following way. Let us consider the expression XIn −M as a matrix with entries in K[X]. When one substitutes a matrix N for the indeterminate X in XIn − M, one obtains a matrix of Mn (A), where A is the subring of Mn (K) spanned by In and N (denoted above by K(N)). The ring A is commutative (but is not an integral domain in general), because it is the set of the q(N) for q ∈ K[X]. Therefore, ⎞ ⎛ N − m11 In ⎟ ⎜ .. ⎟ ⎜ . −mi j In ⎟. ⎜ PM (N) = ⎜ ⎟ . ⎠ ⎝ N − mnn In The Cayley–Hamilton theorem expresses that the determinant (which is an element of Mn (K), rather than a scalar) of this matrix is zero.