1. Let a be a nonzero matrix of order n, a * be the adjoint matrix of a, and a * = at. It is proved that | a | ≠ 0

1. Let a be a nonzero matrix of order n, a * be the adjoint matrix of a, and a * = at. It is proved that | a | ≠ 0

There is a formula
r(A*)=
n. When R (a) = n
1, when R (a) = n
0, when R (a) = n
Here, a * = at, so r (a *) = R (at) = R (a)
Obviously, it is the first case in the formula, so a is full rank, | a | ≠ 0

Let a be an invertible matrix of order n, and B be any n * m matrix. How to prove it
If a is transformed into identity matrix I by a series of elementary row transformations, then B is transformed into a ^ - 1B by the same series of elementary row transformations

Elementary row transformation is equivalent to multiplying a series of Elementary Matrices on the left side of the matrix
The product of elementary matrix is invertible matrix
P(A,B)=(E,X)
PA=E
PB=X
P = a ^ - 1, x = a ^ - 1B

Linear algebra problem: let a be a real symmetric matrix of order n and n be an odd number. If a ^ n = I, prove that a = I

The real symmetric matrix A is orthogonal and similar to the diagonal matrix. The diagonal elements are the eigenvalues of A
That is, there is an orthogonal matrix P such that p'ap = D = diag (D1, D2,..., DN), where Di is the eigenvalue of a (because a is symmetric, the eigenvalues are all real numbers)
A ^ n = I, and using P'p = I
D ^ n = (p'ap) ^ n = P '* a ^ n * P = P' * P = I
It is proved that (DI) ^ n = 1 holds for any I
Because Di is a real number and N is an odd number, di = 1 holds for any I
So p'ap = I
So a = P * p '= I
It's over