If the 3-dimensional column vectors α and β satisfy α 'β = 2, how to find the nonzero eigenvalues of the matrix β α'?

If the inner product of the vector is 2, we can see that the trace of matrix β α transpose is 2, that is, the sum of eigenvalues is 2. Because the matrix rank obtained by vector multiplication is 1, there is only one non-zero eigenvalue, so the non-zero eigenvalue is 2

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