Discussion
Let \(A\) and \(B\) be \(n \times n\) matrices with real entries.
Consider the following statements:
P: If \(A\) is symmetric then \(rank(A)\) = Number of nonzero eigenvalues (counting multiplicity) of \(A\).
Q: If \(AB = \mathbf{0}\) then \(rank(A) + rank(B) \le n\).
Then
Answer: A
Both P and Q are TRUE.
P: a real symmetric matrix is orthogonally diagonalizable, \(A = QDQ^{T}\), so its rank equals the rank of the diagonal matrix D of eigenvalues, which is the number of nonzero eigenvalues counted with multiplicity. (For a non-symmetric matrix this can fail: a nonzero nilpotent matrix has rank at least 1 and only zero eigenvalues.)
Q: if \(AB = \mathbf{0}\), every column of B lies in the null space of A, so \(rank(B) \le n - rank(A)\) by the rank-nullity theorem, which gives \(rank(A) + rank(B) \le n\). This is also the case of Sylvester's inequality \(rank(AB) \ge rank(A) + rank(B) - n\).
Since both statements hold, the options calling either one false are wrong.
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