Page 94 - Textos de Matemática Vol. 39

84 CHARLES R. JOHNSON AND YULIN ZHANG
case, we note that the repeated eigenvalue of A can not be among α1, α2, ..., αk because of the rank conditions on A; in the former case, A could also have a repeated eigenvalue that might lie among α1, α2, ..., αk.
We note also that the conditions (iii) are precisely equivalent to the state- ments that the 3, 2 entry of L and the 2, 3 entry of U be nonzero. Suppose
⎡⎣ 1 0 0 ⎤⎦ ⎡⎣ 1 1 y 3 ⎤⎦ ⎡⎣ 1 1 y 3 ⎤⎦ B=x110 0α1=x1α+x1 1+x1y3 .
x3 1 1 0 0 γ x3 α+x3 1+x3y3+γ
Then, condition (i) and (ii) are automatically satisfied and we choose
γ = det A , so that det B = αγ = det A, one of the conditions for similarity.
Nowα
1 1 B[1, 2] = x1 α + x1 ,
so that det(B[1, 2] − αiI) = −αix1 + α − αi(1 + α − αi).
Since αi ̸= 0 and this expression is linear in x1, we choose x1 ̸= 0 to
be any one of the infinitely many values that makes it nonzero. There will be another (finite) restriction placed upon x1 later, but, otherwise, we henceforth imagine it to be data, as are α and γ. At this point all required conditions on B are met, except that it must be a similarity of A.
To ensure that B be a similarity of A, we first choose x3,y3 so as to achieve the characteristic polynomial of A. This may require another restriction upon x1, which may be made without loss of generality. Then we consider the nonderogatory case, which entails one final, and feasible, restriction upon x1. First suppose that Σ1 = traceA and Σ2 are the first and second elementary symmetric functions of the eigenvalues of A. Since det B = det A, if we arrange Σ1 astraceBandΣ2 asthesumofthe2-by-2minorsofB,thenBandAwill have the same spectra. Thus, by calculation from B we have
and
Σ1 =2+α+γ+x1 +x3y3
Σ2 =α+γ+1+x1 +x1γ+αx3y3 +αγ−x3 −αx1y3
From these we first obtain
x3y3 =Σ1 −(2+α+γ+x1),
which is nonzero by one more restriction upon x1. We may then write y3 = Σ1 −(2+α+γ+x1),
x3
assuming x3 ̸= 0, which will be justified later. Now, upon substitution into the
desired expression for Σ2, we have
Σ2 =(1+α)(1+γ)+x1(1+γ)+α(Σ1−(2+α+γ+x1))−x3−αx1(Σ1−(2+α+γ+x1)) x3