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Inequality for a Hermitian matrix

Let \mathbb{O} \neq A \in \mathcal{M}_n(\mathbb{C}) be a Hermitian matrix. Prove that

    \[{\rm rank}(A) \geq \frac{\left({\rm tr}(A) \right)^2}{{\rm tr}\left(A^2 \right)}\]

Solution

It is known that if A is Hermitian then A is diagonisable and of course each of its eigenvalues is real. It is also known that if a_1, \dots, a_m \in \mathbb{R} then

    \[\left ( a_1+\cdots+a_m \right )^2 \leq m \left ( a_1^2 +\cdots +a_m^2 \right )\]

Having said the above we have that if \lambda_1, \dots, \lambda_m are the distinct eigenvalues of the matrix A then {\rm rank}(A)=m as well as

    \[{\rm tr}(A) = \sum_{i=1}^{m} \lambda_i \quad \text{and} \quad {\rm tr} \left ( A^2 \right ) = \sum_{i=1}^{m} \lambda_i^2\]

Making use of the inequality above we get the result.

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