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Hello.
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So here we consider a to be the unitary matrix where a is equal to 3, negative i, i3.
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And we have the t -s -a, so maps c -square to c -square, by multiplication, by the hermation matrix a.
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So the characteristic equation of a is going to be, we take the determinant of a -minus lambda i, set equal to zero.
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So we have 3 -limda -i, negative -i, and then 3 - minus -lamda -lamda, set -equal to 0.
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And we're then going to get a quadratic, taking the determinant, and the quadratic there is going to be lambda squared minus 6th lambda plus 8 is equal to 0.
00:51
So then the roots then of the equation of lambda is equal to 4 and 2.
00:57
So therefore, the eigenvalues of a are going to be 4 and 2.
01:01
And then we get the eigenvector...