Question

suppose v is a complex vector space. suppose t 2 l.v / is such that 5 and 6 are eigenvalues of t and that t has no other eigenvalues. prove that .t 5i /n1.t 6i /n1 d 0, where n d dim v.

Answer 1

Since the (i,j)-entry of A is equal to 0 whenever i ≠ j, it follows that A is a diagonal matrix with all diagonal entries equal to 0. Therefore, .t 5i /n1.t 6i /n1 = 0

To prove that $.t 5i /n1.t 6i /n1 = 0$, we will show that the vectors .t 5i / and .t 6i / are linearly dependent.

Since 5 and 6 are the only eigenvalues of t, it follows that the characteristic polynomial of t is given by

$p.t / d .t /5 .t /6$

Let .t 5i / and .t 6i / be the corresponding eigenvectors for the eigenvalues 5 and 6, respectively. Since the eigenvectors .t 5i / and .t 6i / are corresponding to different eigenvalues, they are linearly independent.

However, since the characteristic polynomial p.t / is a polynomial of degree 2, it follows that the eigenvectors .t 5i / and .t 6i / must span the entire vector space V. Therefore, any vector in V can be expressed as a linear combination of .t 5i / and .t 6i /. In particular, we can express the zero vector 0 as a linear combination of .t 5i / and .t 6i /, which means that .t 5i / and .t 6i / are linearly dependent.

Thus, we have shown that .t 5i /n1.t 6i /n1 = 0, as desired.

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