51

The formula you're looking for can be obtained by differentiating Jacobi's formula
$$ \frac{\mathrm{d}}{\mathrm{d}t} \det A(t) = \det A(t) \cdot \operatorname{tr}\left( A^{-1} \frac{\mathrm{d}A}{\mathrm{d}t} \right) $$
with respect to a second parameter, say $s$:
\begin{multline}
\frac{\partial^2}{\partial s \partial t} \det A(s,t) =
\det A(s,t) \cdot \bigg[
...

34

(1) Correctness: I read all arguments in detail and couldn't find anything wrong with them. Of course, this doesn't mean too much...
(2) Orginality: I think in a topic which has such an extensive historical record as Perron-Frobenius theory does, the question of "originality" or "novelty" of any particular proof is a very delicate one.
There is an enormous ...

34

The OP asks about generalisations and applications of the formula in arXiv:1908.03795.
$\bullet$ Concerning generalisations: I have found an older paper, from 1993, where it seems that the same result as in the 2019 paper has been derived for normal matrices (with possibly complex eigenvalues) --- rather than just for Hermitian matrices: On the eigenvalues ...

answered Nov 18 '19 at 12:57

Carlo Beenakker

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28

Let $C := A^{1/2} (A+B) A^{1/2} + B^{1/2} (A+B) B^{1/2}$; this is a positive semi-definite matrix with the same trace as $(A+B)^2$. We show that the eigenvalues of $C$ are majorised by the eigenvalues of $(A+B)^2$, that is to say that the sum of the top $k$ eigenvalues of $C$ is at most the sum of the top $k$ eigenvalues of $(A+B)^2$ for any $k$. By the ...

27

Another way to obtain the formula is to first consider the derivative of the determinant at the identity:
$$
\frac{d}{dt} \det (I + t M) = \operatorname{tr} M.
$$
Next, one has
$$
\begin{split}
\frac{d}{dt} \det A (t)
&=\lim_{h \to 0} \frac{\det \bigl(A (t + h)\bigr) - \det A (t)}{h}\\
&=\det A (t) \lim_{h \to 0} \frac{\det \bigl(A (t)^{-1} A (...

answered Aug 17 '15 at 8:17

Jean Van Schaftingen

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25

For $n>2$, the span of the matrices in $\mathrm{SO}(n)$ is the full space $M_n(\mathbb{R})$ of $n$-by-$n$ matrices with real entries.
One proof is using representation theory: If we let $S\subset M_n(\mathbb{R})$ denote the span of $\mathrm{SO}(n)$, then notice that this span is invariant under the action of $\mathrm{SO}(n)\times\mathrm{SO}(n)$ defined ...

answered Sep 11 '18 at 9:42

Robert Bryant

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21

By "corresponding submatrices" I presume you mean those $2\times2$ minors obtained by deleting $n-2$ colums and $n-2$ rows, where these columns and rows have the same $n-2$ indices. Once you've calculated the determinants of these submatrices you recover the action of $A$ on the exterior square $\Lambda^2 V$.
Now the following paper: "An algorithm for ...

20

To supplement Robert's counterexample, let me mention below some interesting facts about the matrix exponential, along with what may be regarded as the "correct" way of obtaining matrix exponential like operator inequalities.
Assume throughout that $A \ge B$ (in Löwner order).
The map $X \mapsto X^r$ for $0 \le r \le 1$ is operator monotone, i.e., $A^r \...

19

Let $A$ be the nilpotent matrix $$\begin{pmatrix}0 & 1 & 1 & \cdots & 1 \\ & 0 & 1 & \cdots & 1 \\ & & \cdots & \cdots & \cdots \\ & & & 0 & 1 \\ & & & & 0\end{pmatrix},$$ then the matrix $P$ is equal to $1 + \beta A + \alpha A^2$.
This gives the inverse: \begin{eqnarray*}P^{-1} ...

18

Yes. The norm of a positive definite matrix does not exceed its trace, and the sum of traces is finite, since the sum of diagonal elements is finite for each of $n$ places.

18

For hermitian $B$, the inequality is very easy to prove: use that $\|B\| - B$ is positive semidefinite together with the fact that the product of two positive semidefinite matrices has nonnegative trace. This gives $\mathrm{tr}(A(\|B\| - B)) \geq 0$, from which the inequality follows immediately.
For general $B$, we can therefore argue that
$$\mathrm{Re}(\...

17

Expanding my comment into an answer, which offers a more general result.
Theorem (von Neumann). Let $A$ and $B$ be arbitrary $n\times n$ complex matrices. Then, $$|\text{trace}(AB)| \le \sum_{i=1}^n \sigma_i(A)\sigma_i(B),$$ where $\sigma_i(\cdot)$ denotes the $i$-th singular value.
As an immediate corollary we have the OP's result, because for a PSD ...

16

We can just manipulate $C$ in the usual way by row operations: Subtract the last "row" from all the other "rows" (this is really several traditional row operations done at once). This produces
$$
\begin{pmatrix} A- B &0& 0 & \ldots & 0 &B-A \\
0 & A-B &0 &\ldots & 0 & B-A\\
&& \ldots &&&\\
B & B ...

answered Aug 25 '14 at 19:18

Christian Remling

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16

Such matrices are called copositive in the literature.
Moreover, the statement you want to show is known to be false.
While I don't recall a counterexample right away,
an intuition for this is that it's NP-hard to check copositivity, but computationally easy to check the condition in your statement, via semidefinite programming, see e.g. this text by ...

16

The matrix $L_n$ is positive definite.
Proof. The matrix $G_n$ with entries ${\rm gcd}(a,b)$ is positive definite because of $G=D^T\Phi D$ where $\Phi={\rm diag}(\phi(1),\ldots,\phi(n))$ ($\phi$ the Euler's totient function) and $d_{ij}=1$ if $i|j$ and $0$ otherwise. Then the matrix $H_n$ with entries $\frac1{a+b}$ is positive definite because
$$h_{ij}=\...

15

One simple example with a special matrix, which has somehow "a continuum" as eigenvalue...
Consider some function $ f(x) = K + ax + bx^2 + cx^3 + ... $ having a nonzero radius of convergence. Then think of the infinite matrix of the form
$$ \small \begin{bmatrix}
K & . & . & . & \cdots \\\
a & K & . & . & \cdots \\\
...

15

Let $P$ be any $(n-2)\times (n-2)$ orthogonal matrix and $Q$ any
$2\times 2$ orthogonal matrix. Let $M=P\oplus Q$. If we erase the four
entries of $Q$, then there are uncountably many ways to fill them in
to obtain an orthogonal matrix.
Perhaps the "correct" condition on $M$ should be that it is an
orthogonal matrix with no zero entries.

answered Jun 8 '15 at 1:37

Richard Stanley

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15

Yes, Brendan McKay showed that almost all trees have mates that are simultaneously cospectral in both adjacency and Laplacian spectra. And more.
http://users.cecs.anu.edu.au/~bdm/papers/SpectralTrees.pdf
Edit: I wondered briefly what the smallest pair of such trees would be, and a few minutes of Sage told me that there are two on 11 vertices.
And here ...

15

Your conjecture is a special case of the following result which essentially follows from the Lieb-Thirring inequality.
Let $A$ and $B$ be Hermitian matrices. Then, for every positive integer $p$ we have
\begin{equation*}
|\text{tr}(AB)^{2p}| \le \text{tr}A^{2p}B^{2p}
\end{equation*}

15

Let $S$ be the span of $SO(N)$ .
Then it's obvious that if $A \in S$ and $D_1, D_2 \in SO(N)$ then $D_1^{-1} A D_2 \in S$ .
Therefore it's enough to show that $B := diag(1,0,...0) \in S$ .
If $N$ is odd and $> 1$ then we can write $$B = \frac{1}{2} (\left[\begin{matrix}1&0\cr0&I_{N-1}\end{matrix}\right] + \left[\begin{matrix}1&0\cr0&-I_{...

15

This is not true. For example, if
$$
A = \frac{2}{1 + \sqrt{5}}\begin{bmatrix}
1 & 1 \\ 0 & 1\end{bmatrix}
$$
then $\|A\| = 1$. Furthermore, $A$ has exactly $2$ square roots, which are
$$
B_{\pm} = \frac{\pm\sqrt{2}}{\sqrt{1 + \sqrt{5}}}\begin{bmatrix}
1 & 1/2 \\ 0 & 1\end{bmatrix}.
$$
It is straightforward to check that $\|B_{\pm}\| \...

answered Dec 6 '19 at 0:42

Nathaniel Johnston

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13

Edit Now this answers the first question for the operator norm and the normalized Hilbert-Schmidt norm.
The answer depends on the norm you are considering. The answer is no for the operator norm, but is yes for the normalized Hilbert-Schmidt norm (at least if you replace $O(\varepsilon)$ by $o(1)$, see the answers to this question).
Here are some details ...

answered Nov 4 '15 at 8:16

Mikael de la Salle

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13

Counterexample:
$$
\begin{bmatrix}
-1 & 1 & 0 & 0\\
0 & -1 & 1 & 0\\
0 & 0 & -2 & 2\\
0 & 0 & 2 & -2
\end{bmatrix}
$$
is defective: its eigenvalues are $-1,-1, 0, -4$ (it is block triangular, so its eigenvalues are those of the $2\times 2$ blocks on the diagonal), but $A+I$ has rank 3.
Strategy to construct ...

answered Dec 17 '16 at 18:30

Federico Poloni

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13

(Basically) Full answer
For $s \geq 1$ we always take the matrix with diagonals $s^{1/n}$.
For $s < 1$ and we have two possibilities. For $n$ small enough we take (still) the matrix with diagonals $s^{1/n}$. For $n$ large enough we instead solve
$$a(1-a)^{n-1} = s$$
and take the smaller of the (at most two) solutions. The optimizer will be the matrix ...

13

I want to add to the list of places where this identity has been used previously.
For a recent one, see arXiv:1710.02181 “State transfer in strongly regular graphs with an edge perturbation”.
Equation (2) in Section 2 of this reads
\[
\frac{\phi(X\setminus a,t)}{\phi(X,t)} = \sum_r \frac{(E_r)_{a,a}}{t-\theta_r}
\]
If the eigenvalue $\theta_r$ is ...

12

Update: This recent paper on this topic may also be of interest; it's quite short and claims to have a fully constructive approach.
I think the closest to answering your question is the following paper.
"Column subset selection, matrix factorization, and eigenvalue optimization", by J. A. Tropp.
In Proc. 2009 ACM-SIAM Symp. Discrete Algorithms (SODA), pp. ...

12

The following recent paper: "An exact duality theory for semidefinite programming based on sums of squares" by I. Klep, and M. Schweighofer (both are on MO I think) addresses exactly your question: When is there a $\lambda \in \mathbb{R}^m$ such that $\sum_i \lambda_iA_i \succeq 0$.
If you want something simpler, then the following Lemma, cf. L.Lovasz ...

12

For example, try $$A = \pmatrix{1 & 0\cr 0 & 0\cr},\ B = A + t C\ \text{where}\ C = \pmatrix{1 & 1\cr 1 & 1\cr}$$
where $t>0$ is small. Then $A ≼ B$ but
$$ e^B - e^A = t \pmatrix{e & e-1 \cr
e-1 & 1 \cr} + O(t^2)$$
so that $$ \det(e^B - e^A) = (e - (e-1)^2) t^2 + O(t^3) < 0$$

12

Subtracting the last row of blocks from the first $k-1$ rows of blocks, we obtain
$$\begin{bmatrix}A-B & O & O & \dots & O & B-A\\ O & A-B & O & \dots & O & B-A\\ O & O & A-B & \dots & O & B-A\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots &...

answered May 21 '16 at 15:09

Rodrigo de Azevedo

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12

I always give this as an exercise in my linear algebra class in the form:
If $k+1$ vectors in $\mathbb{R}^n$ form obtuse angles with one another, then after removing any one vector from that collection, the remaining vectors are linearly independent.
Proof. Assume that $v_1,\ldots,v_k$ are the remaining vectors, and that they are linearly dependent. Then $...

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