Given that a is a real symmetric matrix of order 3, satisfying a ^ 4 + 2A ^ 3 + A ^ 2 + 2A = 0 and rank r (a) = 2, find all eigenvalues of matrix A and rank r (a + e) I can find out the eigenvalue of matrix A is 0 or - 2, but the answer is that because the real symmetric matrix must be similar diagonalized and the rank r (a) = R (similar diagonalization sign) = 2, so the eigenvalues of a are 0, - 2, - 2

Given that a is a real symmetric matrix of order 3, satisfying a ^ 4 + 2A ^ 3 + A ^ 2 + 2A = 0 and rank r (a) = 2, find all eigenvalues of matrix A and rank r (a + e) I can find out the eigenvalue of matrix A is 0 or - 2, but the answer is that because the real symmetric matrix must be similar diagonalized and the rank r (a) = R (similar diagonalization sign) = 2, so the eigenvalues of a are 0, - 2, - 2

Because a can be similar to diagonalization
So a is similar to diagonal matrix B, and the elements on the main diagonal of B are the eigenvalues of A
The rank of similar matrix is the same
So the rank of diagonal matrix B is also 2
So the number of nonzero eigenvalues of a is 2
So the eigenvalues are 0, - 2, - 2
Conclusion: the rank of diagonalizable matrix is equal to the number of nonzero eigenvalues of matrix

It is known that a is a real positive definite symmetric matrix of order n and B is an anti real symmetric matrix of order n. proof: det (a + b) > 0

A is a real positive definite symmetric matrix of order n, = = > A = PP ^ t (with P invertible) B is an anti real symmetric matrix of order n = = > p ^ {- 1} BP ^ {- 1} ^ t is an anti real symmetric matrix of order n, = = > p ^ {- 1} BP ^ {- 1} ^ t is a complex number whose real part is 0, = = = > det (a + b) = | a | e + P ^ {- 1} BP ^ {- 1} ^ t | = > 0 method 2: counter proof | a + B

If a satisfies a ^ 2-2a-4e = 0, it is proved that both a + E and a-3e are invertible and inverse to each other. If a satisfies a ^ 2 + 2A + 3E = 0, it is proved that a is invertible and a ^ (- 1)
(1) If a satisfies a ^ 2-2a-4e = 0, it is proved that a + E and a-3e are both invertible and inverse matrices;
(2) If a satisfies a ^ 2 + 2A + 3E = 0, we prove that a is an invertible matrix and find a ^ (- 1)

(1) From (a + e) (a-3e) = A & # 178; - 2a-3e = (A & # 178; - 2a-4e) + e = 0 + e = e
A + E and a-3e are both invertible and inverse matrices
(2) From a ^ 2 + 2A + 3E = 0, there is
A(A+2E) =-3E
That is a · - (a + 2e) / 3 = E
So a is reversible and a inverse = - (a + 2e) / 3