Infinite loop spaces by John Frank Adams

By John Frank Adams

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Proof: Let A be left invertible. Then there is an n x n matrix B such that BA = I<

Then H is cogredient to a diagonal matrix (ai \ I I' \ o 2 and that the proposition holds for all (n — 1) x (n — 1) hermitian matrices. If H = 0, our proposition holds automatically. If H =fi 0, we can assume that fin ^ 0. In fact, if An = 0, but there is a nonzero diagonal element ha with i > 1, then interchanging the first and the z-th 38 Chapter 1.

Since S is nonalterate, there is a nonzero diagonal element of 5, say s u ^ 0. Interchanging the first row and the i-th row of S and the first column and the i-th column of S simultaneously, we obtain a symmetric matrix which is cogredient to S and whose element at (1, 1) position is nonzero. Therefore we can assume that s u ^ 0. Write 5 = | ( 311 u V 522 / where U u — and5*22 S22= =(St*)2 Then S is cogredient = (512, (Si2,"• •• •, S, l5i n )n )and r ( l1 5SuU n w \ \ /( 5n Sll V I' jy'u \ ) '« { - 1 uu \ \ (( 11 sjs^u !

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