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 Incoming links: Similarity invariance of trace

Past exam of the mathematical methods I course of the University of Oxford / 2014 / 12 / iv / Solution 2026-10-25

Apply the cyclic property of the trace and

B B^{- 1} = I

:

tr (B^{- 1} A B) = tr (A B B^{- 1}) = tr A .

This proves similarity invariance of trace without requiring

A

or

B

to be symmetric.

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 10 / a / Solution 2026-10-20

Use the Einstein summation convention, with repeated component indices summed. The following three expressions are listed separately below. For cyclic invariance of the trace,

tr (A BC) = a_{ij} b_{jk} c_{ki} = c_{ki} a_{ij} b_{jk} = tr (C A B),

where the middle equality commutes scalar entries, not the matrices themselves. If

A

is invertible, this gives

tr (A B A^{- 1}) = tr (B A^{- 1} A) = tr B .

This is similarity invariance of trace.

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