If $0 \leq A \leq B$, can I conclude that $(A \otimes I) \leq (B \otimes I)$?

Question

Let $A \in \mathbb{R}^{n \times n}, B\in \mathbb{R}^{n \times n}$ are two positive semi-definite matrices and $A \leq B$ means $B - A$ is positive semi-definite. Also let $I \in \mathbb{R}^{m \times m}$ be the identity matrix. And $\otimes$ denotes the Kronecker product of matrix.

If $0 \leq A \leq B$, can I conclude that $(A \otimes I) \leq (B \otimes I)$?

Answer 1

Let $A \geq 0$. It is enough to show that $$ A \otimes 1_{\mathbb{R}^{m \times m}} \geq 0. $$ The following two approaches are based on https://math.stackexchange.com/q/1316606, where it has been shown that, for $B \in \mathbb{R}^{n \times n}$, we can identify $B \otimes 1_{\mathbb{R}^{m \times m}}$ with the $m \times m$ diagonal matrix with $B$ on the diagonal, which will be denoted by $\tilde{B}$.

1. Let $v \in (\mathbb{R}^{n})^{m}$. Then we have $$ \langle \tilde{A} v, v\rangle = \langle \begin{pmatrix} Av_{1}\\ \vdots \\ Av_{m}\end{pmatrix}, v\rangle = \sum_{i=1}^{m} \langle A v_{i}, v_{i}\rangle \geq 0, $$ since $A \geq 0$. Hence, $\tilde{A} \geq 0$ and therefore also $A \otimes 1_{\mathbb{R}^{m \times m}} \geq 0$.

2. We also can use the fact that $$ \det(B \otimes 1_{\mathbb{R}^{m \times m}}) = \det(\tilde{B}) = \det(B)^{m} $$ for all $B \in \mathbb{R}^{n \times n}$. With $$ \det(B \otimes 1_{\mathbb{R}^{m \times m}} - \lambda (1_{\mathbb{R}^{n \times n}} \otimes 1_{\mathbb{R}^{m \times m}})) = \det(\tilde{B} - \lambda 1_{(\mathbb{R}^{n \times n})^{m \times m}}) = \det(B-\lambda 1_{\mathbb{R}^{n \times n}})^{m} $$ for all $B \in \mathbb{R}^{n \times n}$ and $\lambda \in \mathbb{C}$ we see that $A \otimes 1_{\mathbb{R}^{m \times m}}$ and $A$ have the same eigenvalues. Hence, $A \otimes 1_{\mathbb{R}^{m \times m}}$ is positive. (Here we used the fact that a matrix is positive semidefinit if and only if all eigenvalues are non-negative.)

Answer 2

While both of Levi's arguments are correct another way to see this is via the square root: because $A\leq B$ we know that $\sqrt{B-A}\geq 0$ exists so $$ 0\leq (\sqrt{B-A}\otimes {\bf 1})^*(\sqrt{B-A}\otimes {\bf 1})=(\sqrt{B-A})^2\otimes {\bf 1}^2=B-A\otimes{\bf 1}\,, $$ or, equivalently, $A\otimes{\bf 1}\leq B\otimes{\bf 1}$.

As a side note observe that one does not need $A\geq 0$, only $A\leq B$.