Let a and B be n-dimensional nonzero column vectors, and a = 2e (n-dimensional) - ABT (transpose vector of B) If the square of a = a + 2E, what is the transpose vector of a multiplied by B

Let a and B be n-dimensional nonzero column vectors, and a = 2e (n-dimensional) - ABT (transpose vector of B) If the square of a = a + 2E, what is the transpose vector of a multiplied by B

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α. If β is n-dimensional sequence vector, then the transpose of determinant / α × β + transpose of β × α / = 0
The process of proof

In order to prove that the determinant is equal to 0, we only need to prove that the corresponding matrix is not of full rank. In the proof, we need to use two inequalities about rank, R (a + b) ≤ R (a) + R (b), R (AB) ≤ min {R (a), R (b)}. If α 'denotes the transpose of α, then R (α β' + β α ') ≤ R (α β

Let a be an n-dimensional nonzero real vector, a = e + AAT (transpose of a), n > = 3, then a has several eigenvalues of 1?

Here, let's first draw a conclusion, which is well proved
If x is the eigenvalue of matrix C, then the eigenvalue of E + C is 1 + X
If a ≠ 0, we can know that AA '(a' denotes transpose) will not be 0, and R (AA ')