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Exercise 12 from SEC. 82 of Finite-Dimensional Vector Spaces - 2nd Edition by Paul R. Halmos.

If $A$ and $B$ are linear transformations on a finite-dimensional inner product space, and if $\textbf{0} \le A \leq B$, then $\det A \leq \det B$.

Notation: $0 \le A$ means that $A$ is positive.

If $\det B = 0$, choose $x \in \ker B$, then $0 \le ((B-A)x,x) = (Bx, x) - (Ax, x) = -(Ax, x) \le 0$. Therefore $\ker B \subseteq \ker A$ and $0 = det A \le det B = 0$.

How to prove for the case when $B$ is invertible? Intuitively, I feel that since both $A$ and $B$ are positive their proper values are also positive and $B$'s proper values should be greater the $A$'s and same for determinants.

AMathStudent
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Andreo
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  • Why did you remove the "self-adjoint"? How do you define the $\leq$ relation on general linear transformations? – darij grinberg Feb 09 '20 at 21:11
  • I wanted to write a message to you but did not find how to do that. In the book Finite Dimensional Vector Spaces the notation $0 \le A$ is equivalent to saying that $A$ is positive. And $A \le B$ means that $0 \le B - A$ is positive. Both $A$ and $B$ are self-adjoint because they are positive. Do you think it should be stated explicitly that $A$ and $B$ are self-adjoint? – Andreo Feb 09 '20 at 22:15
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    Yes, it should be stated explicitly, partly because there are several relations denoted by $\leq$ (for example, entrywise inequality) for general (non-symmetric) matrices, and partly because it gives a good keyword for searching. (math.stackexchange doesn't have private messages.) – darij grinberg Feb 09 '20 at 22:21
  • I've added "self-adjoint" back. – Andreo Feb 09 '20 at 22:38

2 Answers2

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Hint: Reduce your inequality to a simpler case as follows: $$ 0 \preceq A \preceq B \iff 0 \preceq B^{-1/2}AB^{-1/2} \preceq B^{-1/2}BB^{-1/2} $$

Ben Grossmann
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    You could also prove that the eigenvalues (proper values) of $A$ are smaller than those of $B$, but I don't see a quick way to prove this result. The quickest approach I can think of is to use the Courant-Fischer theorem. – Ben Grossmann Jan 20 '20 at 00:42
  • I wonder if there is some sort of dependency between eigen spaces of $A$ and of $B$? Say, they have to coincide or be a subspaces of each other? – Andreo Jan 20 '20 at 00:51
  • I have reduced the inequality to the form that you suggested, but can't see how to prove inequality for the determinants ... – Andreo Jan 20 '20 at 01:26
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    a bridge for the first comment. if $B$ were diagonal: $\text{diagonal of A} \preceq \text{eigenvalues of A}$ and $0 \leq \text{diagonal of A} \leq B$ where here $\preceq$ denotes majorized and $\leq $ is a point-wise inequality. In any case this indicates the Hadamard Determinant Inequality plays a useful role in this problem (and is one way to finish off the hint). – user8675309 Jan 20 '20 at 02:31
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    @Andreo We can simplify the inequality further to get $B^{-1/2}AB^{-1/2} \preceq I$. Note that a real-symmetric/Hermitian matrix $M$ will satisfy $M \preceq I$ if and only if all of its proper values satisfy $\lambda \leq 1$. – Ben Grossmann Jan 20 '20 at 07:21
  • @Omnomnomnom, thank you for the explanation. My mistake was: I built inequality with $B^{1/2}$ instead of $B^{-1/2}$. – Andreo Jan 21 '20 at 01:12
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A verbose ("for dummies") version of Omnomnomnom's and Andreo's proof is as follows.

We have two cases: (a) $B$ is non-invertible, and (b) $B$ is invertible.

Case (a): since $B$ is not invertible, det $B = 0$. Also, the null-space of $B$ has some non-zero vector $x$. It follows from the positivity of $(B-A)$ that $\big\langle(B-A)x, x \big\rangle \geq 0$

$\implies \langle Bx-Ax, x\rangle \geq 0 \implies \langle Bx, x\rangle - \langle Ax, x\rangle \geq 0\\ \implies 0 - \langle Ax, x\rangle \geq 0 \implies \langle Ax, x\rangle = 0$ [$\because \langle Ax, x\rangle \geq 0$ due to the positivity of $A$] $\implies \langle \sqrt A \sqrt Ax, x\rangle = 0$ [$\because$ positivity of $A$ implies that it has a positive square-root] $\implies \langle \sqrt Ax, {\sqrt A}^*x \rangle = 0$ $\implies \langle \sqrt Ax, \sqrt Ax \rangle = 0$ [$\because \sqrt A$ is positive, it is self-adjoint too] $\implies \left \Vert \sqrt Ax \right \Vert^2 = 0 \implies \sqrt Ax = 0$ $\implies \sqrt A \sqrt Ax = 0 \implies Ax = 0$

$\implies$ (non-zero) $x$ is in the null-space of $A \implies A$ is not invertible $\implies$ det $A = 0$ $\implies$ det $A =$ det $B$ (since both det are $0$).

Case (b): in finite dimensions, the invertibility of $B \geq \textbf{0}$ implies that $B > \textbf{0}$ (positive-definite). Thus, $B$ has a positive-definite square root, say $B^\frac{1}{2}$ which is also invertible. If $B^{-\frac{1}{2}}$ is the inverse of $B^\frac{1}{2}$, then $B^{-\frac{1}{2}}$ is also positive-definite.

Now, because $B-A$ is given as positive, $C^*(B-A)C$ is positive for an arbitrary $C$, i.e, $C^*(B-A)C \geq \textbf{0}$. If we let $C = B^{-\frac{1}{2}}$, then we have $C^* = {B^{-\frac{1}{2}}}^* = B^{-\frac{1}{2}}$. Substituting $C$ and $C^*$ in the inequality on hand yields $\textbf{1}- B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \geq \textbf{0}$, i.e., $B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \leq \textbf{1}$. We observe that $B^{-\frac{1}{2}}AB^{-\frac{1}{2}}$ is positive transformation, and thus unitarily diagonalizable. This fact, together with the inequality $B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \leq \textbf{1}$, reveals that each Eigenvalue (diagonal entry in the diagonal form) of $B^{-\frac{1}{2}}AB^{-\frac{1}{2}}$ is $\leq 1$. This implies that det $B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \leq 1$. Further, because det $CD$ = det $C \cdot$ det $D$ holds for invertible transformations $C$ and $D$, it follows that det $B^{-\frac{1}{2}}\cdot$ det $A \cdot$ det $B^{-\frac{1}{2}} \leq 1$ where det $B^{-\frac{1}{2}} \neq 0$. This fact, together with the finding that det $B^{-\frac{1}{2}}\cdot$ det $B \cdot$ det $B^{-\frac{1}{2}} =$ det $B^{-\frac{1}{2}}BB^{-\frac{1}{2}} =$ det $B^{-\frac{1}{2}} B^\frac{1}{2} B^\frac{1}{2} B^{-\frac{1}{2}} = $ det $\textbf{1} = 1$, informs that det $A \leq$ det $B$.

AMathStudent
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  • Nice work! Just saw this and wanted to give some (hopefully welcome) feedback. In case (a), you prove that $A \preceq B \implies \ker(A) \subset \ker(B)$. For some alternative proofs, see my post here. The proofs are written for matrices, but can of course be adapted to general operators over inner product spaces. Your proof for case (b) can be shortened. In particular, after you arrive at the fact $B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \leq \textbf{1}$, you can simply state that $\textbf{1} - B^{-\frac{1}{2}}AB^{-\frac{1}{2}}$ is positive and... – Ben Grossmann Jul 30 '20 at 08:53
  • ...that the eigenvalues of $\mathbf 1 - M$ are equal to $1 - \lambda$ for the eigenvalues of $M$ (for any operator $M$). So, we conclude that since both $B^{-1/2}AB^{-1/2}$ and $\mathbf 1 - B^{-1/2}AB^{-1/2}$ are positive, the eigenvalues of $B^{-1/2}AB^{-1/2}$ lie on the interval $[0,1]$. Actually, now that I've written this out, I'm not sure that it's any shorter... – Ben Grossmann Jul 30 '20 at 08:56
  • As for the determinant part of the computation: as long as we're being verbose, it might be helpful to explicitly state that $$ \det(B^{-1/2}AB^{-1/2}) = \det(A)/\det(B). $$ Once you have done this, we have already used the fact that $B^{-1/2} B B^{-1/2} = \mathbf 1$, so for your last sentence, we can simply say that $$ \det (B^{-1/2}BB^{-1/2}) = \det \mathbf 1 = 1, $$ and of course $\det(A)/\det(B) \leq 1 \implies \det(A) \leq \det(B)$. – Ben Grossmann Jul 30 '20 at 09:02
  • Appreciate your acknowledgment, Ben Grossmann! I am a beginner in math, and a positive gesture does build interest and confidence. Thanks. By the way, in your first comment, I guess you wanted to say that $A \leq B \implies$ ker $B \subset$ ker $A$. – AMathStudent Jul 31 '20 at 01:57
  • Your third comment certainly offers a clearer articulation of the proof than the vague phrasing I have used (viz. "this fact together with that ... shows that ..."). Thanks again. – AMathStudent Jul 31 '20 at 02:13