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25 votes
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Proof that block matrix has determinant $1$

Write the SVD of $A$, say $A=PDQ^T$ with $D$ diagonal with non-negative entries and $P\in O(n),Q\in O(m)$. Then $\sqrt{I_n + AA^T} = P\sqrt{1+D^2}P^T$ and $\sqrt{I_m+ A^TA} = Q\sqrt{1+D^2}Q^T$. This ...
jlewk's user avatar
  • 1,364
22 votes

Proof that block matrix has determinant $1$

We have $Af(A^TA)=f(AA^T)A$ for any reasonable function $f$, including $f(x)=\sqrt{1+x}$. This suffices to check for $f(x)=x^k$ when it is obvious, then approximate your function by a polynomial.
Fedor Petrov's user avatar
15 votes
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When does the determinant distribute over addition?

let me assume $A$ is invertible, then you ask when $$\det(1+X)=1+\det X,\;\;X=A^{-1}B $$ so if $X$ has eigenvalues $x_i$, $i=1,2,\ldots n$, you would need $$\prod_{i}(1+x_i)=1+\prod_i x_i$$ basically ...
Carlo Beenakker's user avatar
14 votes
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Off-diagonalize a matrix

This is a so-called chiral symmetry. The restriction on the symmetry of the spectrum of $M$ is the only restriction you need, you can then bring $M$ to the desired off-diagonal form by a unitary ...
Carlo Beenakker's user avatar
11 votes

Conjugated subgroups in $\mathsf{GL}(m+n,\mathbb{Z})$ implies conjugated subgroups in $\mathsf{GL}(n,\mathbb{Z})$?

$\def\ZZ{\mathbb{Z}}\def\GL{\text{GL}}$We can make partial progress using: Warfield, R. B. jun., Cancellation of modules and groups and stable range of endomorphism rings, Pac. J. Math. 91, 457-485 (...
David E Speyer's user avatar
6 votes
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Block matrices and their determinants

Flip the order of the Kronecker products to get $M'=A_n(I_n,I_n)+B_n\otimes T_n$, where $T_n=A_n(1,0)$. Note that $\det M=\det M'$. Since all blocks are polynomial in $A$, they commute, and therefore ...
MTyson's user avatar
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6 votes
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One question on block-circulant matrices

The formula for the specific case is $$\det K=\det(A+B+C+D)\det(A-B+C-D)\det(A+iB-C-iD)\det(A-iB-C+iD).$$ More generally, for a block-circulant matrix with $n$ square blocks $A_0,\ldots,A_{n-1}$, the ...
Denis Serre's user avatar
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4 votes
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Is there a formula for the determinant of a block matrix of this kind?

Just a sketch of an idea that seems to work: You can get rid of the corrections $C$ using the matrix determinant lemma (or, better, replace them with $B$, which makes the matrix block circulant). ...
Federico Poloni's user avatar
4 votes
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Spectrum of this block matrix

If $\lambda_\max$ is the greatest eigenvalue of $T$, the least eigenvalue of $A$ is between $-\lambda_\max$ and $\max(b_1, b_n) - \lambda_\max$.
Robert Israel's user avatar
4 votes
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A closed-form expression for the inverse of a block-matrix

Say that $$B^{-1}=:\begin{pmatrix} b & X^T \\ Y & M \end{pmatrix}.$$ Then using Schur's complement formula (thanks to Nathaniel), $b=(x-{\bf1}^TA^{-1}{\bf1})^{-1}$ and $M=(A-x^{-1}{\bf11}^T)^{-...
Denis Serre's user avatar
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4 votes

Relation between the eigenvalues of a block matrix and the eigenvalues of its diagonal blocks

To see what you might expect for a relation, consider the case of a $2\times 2$ matrix $M=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$, with eigenvalues $\lambda_\pm=\tfrac{1}{2}(a+d)\pm\sqrt{4bc+(a-...
Carlo Beenakker's user avatar
4 votes
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Eigenvalues of a specific matrix

For the signed circulant matrix $$U:=\left[\begin{matrix} & 1 & & & \\ & & \ddots & & \\ & & & 1 &\\ -1 & & & & \end{matrix}\...
Narutaka OZAWA's user avatar
4 votes
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Solving a recursion for polynomials defined by a matrix product

Your polynomial is precisely $$ \sum_{k_1+2k_2+\cdots+nk_n=n}\binom{k_1+\cdots+k_n}{k_1,\ldots,k_n}X_1^{k_1}\cdots X_n^{k_n}. $$ The proof is straightforward by induction: you have $$ p_n(X)=\sum_{i=...
Vladimir Dotsenko's user avatar
3 votes
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The normalizer of block diagonal matrices

By request, from my comment: Your guess is correct. Because there are clearly elements in the normaliser arbitrarily permuting same-sized blocks, if you've got an element of the normaliser, you may ...
LSpice's user avatar
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3 votes
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Conditions to solve linear system with matrix blocks

You find various conditions in Section 3 of the classical review paper by Benzi-Golub-Liesen on this kind of problems, which are known as saddle-point problems.
Federico Poloni's user avatar
3 votes

When does $\det \begin{pmatrix} A & X \\ X^T & A \end{pmatrix} = (\det A)^2 + (\det X)^2$?

$$B=\left[\matrix{1&1&1&1\\1&1&-1&-1\\-1&-1&-1&1\\-1&1&-1&1}\right]$$ and probably many other solutions. I'm also voting to close because you didn't ...
Brendan McKay's user avatar
2 votes
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Solve linear system with bordered positive definite matrix

By combining the useful comments of Rodrigo and Todd, the methodology to solve this system is shown here below. One caveat is that the method is probably not very efficient, since you need to use the ...
fusiled's user avatar
  • 39
2 votes
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Inverse of a larger matrix where the inverse of the submatrix is known

You know $\begin{pmatrix} A & 0 \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} & 0 \\ 0 & 1 \end{pmatrix}$, and from there you can make two successive rank-$1$ modifications, ...
Zach Teitler's user avatar
  • 6,207
2 votes

Non-singular matrix with restricted entries

[EXPANDED] PART 1 (also done by Peter) If $x,y$ are coprime and have opposite sign, there is a singular symmetric matrix with 1 on the diagonal and only $x$ and $y$ off the diagonal. Say $x<0,y>...
Brendan McKay's user avatar
2 votes

Non-singular matrix with restricted entries

Disclaimer: this is only a partial answer. If $S = \{x, y\}$ (considered as variables), the determinant must be a polynomial with integer coefficients and constant coefficient 1. Therefore by Gauss's ...
Peter Taylor's user avatar
  • 6,766
2 votes

Non-singular matrix with restricted entries

This is perhaps a minor observation (Edit: thanks to Peter Taylor’s comment below). If $S$ has every admissible symmetric matrix being non-singular, then $-1\notin S$. This is due to the singular $2\...
Jack L.'s user avatar
  • 1,433
2 votes

Eigenvalues and eigenvectors of k-blocks matrix

$X$ is simply a tensor product $C\otimes D$ where $C$ is the matrix with all diagonal entries $a$ and non-diagonal entries $b$ and where each entry in $D$ is $1$. If $R$, $S$ are diagonalizable, then ...
Joseph Van Name's user avatar
1 vote

Eigenvalues in unit disk for a 2×2 block matrix

Clearly all eigenvalues apart from the eigenvalue $\lambda(\epsilon)$ with $\lambda(0) = 1$ stay in the open unit disk for $\epsilon$ sufficiently small. To see what happens to the last eigenvalue, ...
Federico Poloni's user avatar
1 vote

Eigenvalues in unit disk for a 2×2 block matrix

Part 1 - Determinant of $X$ As a partial result, it is possible to show that $\lvert\det X\rvert < 1$ for sufficiently small $\epsilon$. In fact, because $B$ and $C$ commute, then thanks to a known ...
MathMax's user avatar
  • 203
1 vote
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Eigenvalues of a block matrix with zero diagonal blocks

If you decompose $M=\begin{pmatrix} X_{q\times q}&Y_{q\times k_3}\\ (Y_{q\times k_3})^{\rm T}&0_{k_3\times k_3}\end{pmatrix}$ into four block matrices, with $q=k_1+k_2$, then the determinant ...
Carlo Beenakker's user avatar
1 vote

Sufficient conditions for invertibility of a block tridiagonal matrix

The following is a list of answers I know for some specific cases. However, they are not strong enough for my uses. Simple conditions A sufficient, but weak condition is that $C_i = 0$ for each $i$. ...
kaba's user avatar
  • 387
1 vote
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If the direct sum of $L$ and $M$ has a pseudoinverse, then do $L$ and $M$ have pseudoinverses?

I propose an answer to my own question: The following observation is based on Section 3.1 of "Theory of generalized inverses over commutative rings" by K.P.S. Bhaskara Rao. The matrix $ALA$ ...
wlad's user avatar
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