It is proved that there exists an n-square matrix (over an appropriate field) with prescribed eigenvalues and n prescribed entries except in the following cases: (i) the n prescribed entries are the principal entries, (ii) the n prescribed entries are on a row or column and the nonprincipal ones are zero. In [1] the following theorem has been proved. THEOREM. Let Al, * * *, An and a,, * * *, an_ 1 be prescribed complex numbers and let (it, jt), t = 1, * * *, n 1, be prescribed different positions in an n-square matrix. Then there exists an n-square matrix with eigenvalues Al, * , An and with the prescribed entries at in the prescribed positions (it, jt), t = 1, , n 1. The number n 1 of prescribed entries cannot in general be increased. The authors show that the number of prescribed entries cannot be increased with two examples: (i) the prescribed entries are the n diagonal entries, (ii) the prescribed entries are on a row or column and are all equal to zero. In the present paper we shall show that, with a small modification for the second example, these two cases are the only ones in which the number of prescribed entries cannot be increased to n. Moreover, our result is valid for a more general field and our proof provides an effective method for constructing the matrix. First we state the problem we shall treat. PROBLEM. Let (F be afield and Al ... * *, a,, ... , a E (. Let (i1 , j1), ..., (in jn) be distinct positions in an n-square matrix. Construct an n-square matrix over (D with eigenvalues Al ..., An and at in the position (it, jt), t = 1, * *, n. Concerning the positions (i 1 5 jl) *... , (in jin) one and only one of the following cases is possible: (i) The positions (it,jt), t = 1, * * , n, are the principal positions. (ii) The positions (it, jt), t = 1, ... , n,form a nonprincipal diagonal. (iii) The positions (it, jt), t = 1, ... , n, are on a row or column. (iv) None of the above cases. In case (i), it is well known that the problem has a solution if and only if a1 + ... + an = Al + ... + An. In case (ii), there always exists a solution (in certain cases we have to impose a restriction on (F) as has been shown in [2]. In case (iii), if all the at corresponding to nonprincipal positions are zero, it is obvious that the problem has a solution if and only if the at corresponding to the principal position is equal to one of the prescribed eigenvalues. Now we prove that in case (iii), under the additional condition that not all the at corresponding to nonprincipal positions are zero, our problem has a solution. We assume that the prescribed positions are on a row (if they are on a column, the problem can be reduced to this case by considering the transposed matrix). Performing a suitable permutation on the rows and then the same permutation on the columns, we see that, without loss of generality, we may assume that the * Received by the editors March 21, 1972, and in revised form August 3, 1972. t Coimbra, Portugal. This research was supported by the C. Gulbenkian Foundation. 414 This content downloaded from 157.55.39.152 on Sat, 26 Nov 2016 04:22:31 UTC All use subject to http://about.jstor.org/terms MATRICES WITH PRESCRIBED ENTRIES 415 prescribed positions are on the first row, i.e., they are (1, 1), (1, 2), * , (1, n) and that the corresponding entries are (after reordering) a1, a2, * * , an. Suppose that a1 is equal to one of the Ai (e.g., Al). In this case, [A a2 .. an A2 is a solution of the problem. Suppose now that a1 # Ai, i 1, * -, n. First we consider the case an # 0. To solve our problem we need the L-operator [2]: if E, G, F are matrices of type m x m, (n m) x (n m), m x (n m) respectively, by L(FG)E (i : j; i, j< n) we denote the matrix obtained from [E F _O G_ by adding the ith row to its jth row and subtracting its jth column from its ith column. Now let Al1 a2 ... an 1 C = A . D =[in], _ ~~~~An1_J