Let $A, B \in M_n$ be positive definite and $A \circ B = \left[ {{a_{ij}}{b_{ij}}} \right]$.

Why does $A \circ {B^{ - 1}} + {A^{ - 1}} \circ B \ge 2{I_{n \times n}}$ ?


First observe that for any two elements $X,Y \in M_{n}$, we have $X \circ Y$ is a one of the principal minor of $X \otimes Y$. And it is also straightforward to observe that any principal minor of positive definite matrix is also positive definite.

So to prove $A\circ B^{-1} + A^{-1} \circ B \geq 2 I_n$, It is sufficient to prove $$A\otimes B^{-1} + A^{-1} \otimes B \geq 2 I_{n^2}.$$

Now $A,B$ are given positive definite. So both the element $A\otimes B^{-1}$ and its inverse $(A \otimes B^{-1})^{-1} = A^{-1}\otimes B$ is postive definte.

Now using spectral theorem we will prove that for any positive definite matrix $T \in M_k$, one will have $$T+T^{-1} \geq 2I_k.$$

Since $T$ is positive definite, there exist a orthonormal eigen basis with respect to which $T$ will be of the diagonal form and $T +T^{-1}$ will look like \begin{align} \begin{pmatrix} t_1 + \frac{1}{t_1} &0 &\cdots &0\\ 0 & t_2 + \frac{1}{t_2} &\cdots &0\\ 0& 0 &\cdots &0\\ 0& 0& \cdots & t_k + \frac{1}{t_k} \end{pmatrix} \end{align} where $t_i$'s are positive eigen value of $T$. And since $x+\frac{1}{x} \geq 2$ for all $x>0$, we will have $$T+T^{-1} \geq 2I_k.$$

  • $\begingroup$ This is awesome!!. I didn't understand the sufficiency part for the kronecker product. Can you expand a bit on that? Also, if you can take the trouble, can you try helping with my approach as well? $\endgroup$ – dineshdileep Dec 1 '15 at 5:37
  • $\begingroup$ @dinesh one matrix repn for $A \otimes B$ will be of the form $(a_{ij} B)$. Now take the principal minor corresponding to row $1,n+2, 2n+3,...,(n-1)n+ n$ and associated column then you will get the schur product matrix $\endgroup$ – Timon Dec 1 '15 at 6:52
  • $\begingroup$ Then consider $A \otimes B^{-1} + A^{-1} \otimes B - I$ and see the corresponding principal minor. $\endgroup$ – Timon Dec 1 '15 at 9:17
  • $\begingroup$ So is the proof like, if $A\geq B$ , then the corresponding principle minors also obey the same identity? $\endgroup$ – dineshdileep Dec 1 '15 at 9:32
  • $\begingroup$ Yes it is like A-B \geq 0 imply corresponding principal minor \ geq 0 and hence principal minor of A \geq principal minor of B $\endgroup$ – Timon Dec 1 '15 at 9:36

This answer was completed with the help of user timon who has already provided an excellent answer which is quite general. This is a different approach.

For $x>0$, you have that $x+\frac{1}{x}\geq 2$. Now, Let $A=\sum_{i}\alpha_i x_i x_i^H$ be its Eigen-decomposition so that $\alpha_i>0$ are the eigenvalues. Similarly $B=\sum_{j}\beta_i y_i y_i^H$ so that $\beta_i>0$ are the eigenvalues. Then, try to show that

$$C=A\circ B^{-1}+A^{-1}\circ B=\sum_{i,j}\left(\frac{\alpha_i}{\beta_j}+\frac{\beta_j}{\alpha_i}\right)(x_i\circ y_j)(x_i\circ y_j)^H$$

Now, note that co-efficients of all terms are greater than two. Define the vectors $r_{ij}=x_i\circ y_j$ for all $i,j$ and also constants $\gamma_{ij}=\frac{\alpha_i}{\beta_j}+\frac{\beta_j}{\alpha_i}$. Note that $\gamma_{ij}\geq 2$

Thus, we have \begin{align} C&=\sum_{i,j}\left(\frac{\alpha_i}{\beta_j}+\frac{\beta_j}{\alpha_i}\right)(x_i\circ y_j)(x_i\circ y_j)^H \\ &\geq 2\sum_{i,j}(x_i\circ y_j)(x_i\circ y_j)^H \\ &=2\left(\left(\sum_{i}x_ix_i^H\right)\circ \left(\sum_{j}y_jy_j^H\right)\right) \\ &= 2I \end{align} This proves the needed inequality. Steps above follow from following facts $$\sum_{i}x_ix_i^H=I$$ This follows since $x_i$'s are a set of orthonormal vectors by definition. Similar property holds for $y_i$'s. Also we have $$(x_i\circ y_j)(x_i\circ y_j)^H = \left(x_ix_i^H\right)\circ \left(y_jy_j^H\right)$$ Try to prove this yourself.

  • $\begingroup$ Could you please tell what does $r_{ij}$ represent? What is $x_I \circ y_j$? $\endgroup$ – Timon Dec 1 '15 at 18:46
  • $\begingroup$ $x_i \circ y_j$ represents the hadamard product (or schur product) of vectors $x_i$ and $y_j$. I defined the vector $r_{ij}=x_i\circ y_j$. The idea is that $(x_ix_i^H)\circ (y_jy_j^H)=(x_i\circ y_j)(x_i\circ y_j)^H$ $\endgroup$ – dineshdileep Dec 2 '15 at 4:43
  • $\begingroup$ Isn't this true that as $x_i$s are orthonormal and $y_j$s are also orthonormal, then the collections $r_{ij}$ also orthonormal? $\endgroup$ – Timon Dec 2 '15 at 5:46
  • $\begingroup$ No, there are $NC_2$ vectors $r_{ij}$. Moreover, hadamard product of two orthogonal matrices can't be hadamard, right? $\endgroup$ – dineshdileep Dec 2 '15 at 6:18
  • $\begingroup$ what I understood that if $x x^H$ denote the linear map $e_k \to \langle e_k, x \rangle x$ , then $(x\circ y)(x \circ y)^H$ denote the linear map on the span of $\{e_k \otimes e_k| k=1,2,..,n\}$ defined by $e_k \otimes e_k \to \langle e_k, x \rangle .\langle e_k,y \rangle$. Now it can easily be seen that $\sum _{ij}|r_{ij}^H u|^2 = ||u||^2 =1$ for all $u \in span \{e_k \otimes e_k| k=1,2,..,n\}$. To know detail send me your email id. $\endgroup$ – Timon Dec 3 '15 at 18:05

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