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Let a and B be m * N and N * m matrices respectively. It is proved that AB and Ba have the same nonzero eigenvalues
If a is a nonzero eigenvalue of AB, then there is a nonzero vector x such that ABX = ax * *
But BX is not equal to zero, otherwise if BX = 0 has AX = 0, it contradicts a nonzero and X nonzero
Note: BX = y
By multiplying B by * * left, we can see that bay = ay. Since y is a non-zero vector, a is also the eigenvalue of Ba
Similarly, the nonzero eigenvalue of Ba is also the eigenvalue of ab
That is to prove the conclusion
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