Let A be n x n matrix which is both Hermitian and unitary, thena)A2 = Ib)A is realc)the eigenvalues of A are 0,1, -1d)the characteristic and minimal polynomial of A are the same.Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Explanation:

Hermitian Matrix:
A matrix is said to be Hermitian if it is equal to its conjugate transpose. In other words, A = A* (where * denotes the conjugate transpose of the matrix).

Unitary Matrix:
A matrix is said to be unitary if its inverse is equal to its conjugate transpose. In other words, A^-1 = A* (where * denotes the conjugate transpose of the matrix).

Now, let's prove the given statement:

a) A^2 = I
To prove this, we can use the fact that A is both Hermitian and unitary.
We know that A = A* and A^-1 = A*, so we can write:
A^2 = AA = A*A = I
Therefore, A^2 = I.

b) A is real
This statement is not necessarily true. A Hermitian matrix is not necessarily real. It can have complex entries as well.

c) The eigenvalues of A are 0, 1, -1
This statement is not necessarily true. Hermitian matrices have real eigenvalues, but unitary matrices can have complex eigenvalues as well. So, we cannot say anything about the eigenvalues of A without additional information.

d) The characteristic and minimal polynomial of A are the same
This statement is not necessarily true either. The characteristic polynomial of A is given by det(A - λI), where λ is the eigenvalue and I is the identity matrix. The minimal polynomial of A is the smallest polynomial that satisfies p(A) = 0 (where p is a polynomial). The minimal polynomial divides the characteristic polynomial, but they need not be the same. So, we cannot say anything about the characteristic and minimal polynomial of A without additional information.

Therefore, the correct answer is option 'A': A^2 = I.

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