Majorization inequality involving positive semidefinite matrices and orthogonal projections

Question

For a symmetric $n \times n$ matrix $A$, we will write $\lambda(A)$ for the vector of its eigenvalues in descending order. In other words, $\lambda_1(A) \ge \lambda_2(A) \dots \ge \lambda_n(A)$ are the eigenvalues of $A$. For two vectors $v, w \in \mathbb{R}^n$, we will write $v \succeq w$ ($v$ majorizes $w$), if $\sum_{i=1}^k v_i \ge \sum_{i=1}^k w_i$ for all $k=1 \dots n-1$, and $\sum_{i=1}^n v_i = \sum_{i=1}^n w_i$. We can state the problem now:

Let $A$ be a $n \times n$ positive semidefinite matrix, and let $P$ be any orthogonal projection (also a $n \times n$ matrix). Is it true that $$ \lambda(A) \preceq \lambda(PAP) + \lambda((I-P)A(I-P))?$$

My thoughts on the problem:

• the majorization condition for $k=1$ follows from the characterization of largest eigenvalue as $\lambda_1(A) = \max_{||v||=1} \langle Av, v \rangle $ and Cauchy-Schwarz,
• the majorization equality condition for $k=n$ follows quickly since $tr(A) = tr(AP) + tr(A(I-P)) = tr(PAP) + tr((I-P)A(I-P))$,
• numerically it seems to be true.

Answer 1

The answer is yes.

We can extend your approach for the largest eigenvalue using the fact that $QR$ and $RQ$ have the same eigenvalues (up to algebraic multiplicity) for square $Q,R$ and the Ky-Fan maximization principle (it's hard to find a free direct source, but this should do for our purposes).

In particular, we have \begin{align} \sum_{j=1}^k\lambda_j(A) &= \max_{X^TX = I_k} \operatorname{tr}(X^TAX) \\ & = \max_{X^TX = I_k} \operatorname{tr}(X^TA^{1/2}(P + [I-P])A^{1/2}X) \\ & \leq \max_{X^TX = I_k} \operatorname{tr}(X^TA^{1/2}PA^{1/2}X) + \max_{X^TX = I_k} \operatorname{tr}(X^TA^{1/2}[I-P]A^{1/2}X) \\ & = \sum_{j=1}^k\lambda_j(A^{1/2}PA^{1/2}) + \sum_{j=1}^k\lambda_j(A^{1/2}[I-P]A^{1/2}) \\ & = \sum_{j=1}^k\lambda_j(AP) + \sum_{j=1}^k\lambda_j(A[I-P]) \\ & = \sum_{j=1}^k\lambda_j(AP^2) + \sum_{j=1}^k\lambda_j(A[I-P]^2) \\ & = \sum_{j=1}^k\lambda_j(PAP) + \sum_{j=1}^k\lambda_j([I-P]A[I-P]). \end{align}

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Note: For more about the Ky Fan inequality and the associated norm, see this post and this post. It is also discussed further in "Matrix Analysis" by Rajendra Bhatia.

This result can be proved even quicker if we simply apply Lidskii's inequality, which is typically proved with an approach similar to that taken here.