53 votes
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Is this proof of Perron's theorem correct, and if so is it original?

(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 ...
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52 votes
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Is there an explicit formula for the hessian of "Determinant"?

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}{\...
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  • 1,222
35 votes
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Consequences of eigenvector-eigenvalue formula found by studying neutrinos

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 ...
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31 votes

Proof for the derivative of the determinant of a matrix

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} \...
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29 votes
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A curious determinantal inequality

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 ...
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  • 91k
27 votes

Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ matrices?

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 ...
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21 votes
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Does the matrix exponential preserve the positive-semi-definite ordering?

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 ...
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  • 27.8k
20 votes
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Inverse of special upper triangular matrix

Let $A$ be the nilpotent matrix $$\begin{pmatrix}0 & 1 & 1 & \cdots & 1 \\ & 0 & 1 & \cdots & 1 \\ & & \cdots & \cdots & \cdots \\ & & & 0 &...
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  • 3,120
18 votes
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Determinant of a $k \times k$ block matrix

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 $$ \...
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18 votes

When the sum of positive definite matrices converges, does the sum of the norm of the associate matrices converges?

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.
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  • 88.5k
18 votes

Is this proof of Perron's theorem correct, and if so is it original?

As someone who lives and dies by the Perron-Frobenius theorem (PFT) and who is familiar with a good deal of the literature on nonnegative matrices (and generalizations of nonnegative matrices), your ...
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18 votes

Matrix trace & norm

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 ...
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  • 4,184
17 votes

Matrix trace & norm

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}^...
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  • 27.8k
17 votes

Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ matrices?

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 $N$ ...
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  • 2,653
16 votes
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On the positivity of matrices

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 ...
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16 votes

Operator norm of square root of matrix vs original

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 $$ ...
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16 votes
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Counting eigenvalues without diagonalizing a matrix

Here is an efficient method. First of all, I must quote that diagonalizing $M$ is not a method, because there is no explicit way to carry this out. It amounts to calculating the roots of a polynomial !...
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  • 47.5k
15 votes

How much redundancy resides in an $n \times n$ orthogonal matrix?

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 ...
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15 votes
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is it possible to have two non-isomorphic non-regular graphs with the same adjacent spectrum and the same laplacian spectrum?

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....
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  • 10.8k
15 votes
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Trace of non-commutable matrices

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 ...
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  • 27.8k
15 votes
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The abc-conjecture as an inequality for inner-products?

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 ...
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  • 47.5k
15 votes
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Closed form solution for $XAX^{T}=B$

$B^{-1/2}XAX^TB^{-1/2}=I$, so $B^{-1/2}XA^{1/2}=Q$ must be orthogonal. On the other hand, for any orthogonal $Q$, it is simple to verify that $X = B^{1/2}QA^{-1/2}$ solves the equation, so this is a ...
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14 votes
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Finding the closest matrix to $\text{SO}_n$ with a given determinant

(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 ...
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14 votes
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Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ matrices?

Elementary proof. The linear space $E$ spanned by $SO_n$ is the orthogonal of those matrices $M$ such that $\langle M,Q\rangle:={\rm Tr}(MQ)=0$ for every $Q\in SO_n$. Let $M=SR$ be a polar ...
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  • 47.5k
13 votes

Determinant of a $k \times k$ block matrix

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-...
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13 votes
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Packing obtuse vectors in $\mathbb{R}^d$

You can prove it by induction. The case $d=1$ is obvious. Assume it's true for dimension $d \geq 1$, and consider any set of $d+3$ vectors in $\mathbb{R}^{d+1}$. Fix one of these, say $\vec{v}_0$, ...
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  • 14.8k
13 votes

Packing obtuse vectors in $\mathbb{R}^d$

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 ...
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13 votes
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Almost commuting unitary matrices

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 ...
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13 votes
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The space of positive definite orthogonal matrices

You may find the Cayley transform to be useful here: As is well-known and easy to prove, every orthogonal $n$-by-$n$ matrix $R$ that does not have $-1$ as an eigenvalue can be written uniquely in the ...
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13 votes
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Are Diagonally dominant Tridiagonal matrices diagonalizable?

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 ...
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