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1 vote

Proof of Minkowski determinant inequality

We may relax the assumption slightly. Suppose $A\in M_n(\mathbb C)$ is positive definite and $B\in M_n(\mathbb C)$ is positive semidefinite. Let $x_1,x_2,\ldots,x_n$ be the eigenvalues of the positive ...
0 votes

Priority of subscript and superscript operations

To quote different commenters from the deleted duplicate, as the comments are not merged: In general it is context dependent. If, as you say, context here suggests the first then it is the first for ...
-1 votes

How many singular $3\times3$ matrices exist using only the numbers $1$ and $0$?

To count the number of invertible matrices, we can choose each column step by step. For the first column, we need a column that is not zero, which can be achieved in $2^3 - 1 = 7$ ways. The second ...
Kevin's user avatar
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0 votes

Least non-zero singular value for a singular matrix

The proof is straightforward from writing $\|Ax\|^2$ in terms of the SVD $A = UDV^T$. We have $Ax = \sum_{j = 1}^{n}\sigma_{j}(x, v_j)u_j$, so $\|Ax\|^2 = \sum_{j = 1}^{n}\sigma_{j}^2|(x, v_j)|^2 = \...
Kakashi's user avatar
  • 2,054
0 votes

Least non-zero singular value for a singular matrix

Yes, that's true. One way to see this is to think of the nullspace as being spanned by the right singular vectors whose singular value is $0$. Since the right singular vectors are orthogonal, the ...
Qiaochu Yuan's user avatar
2 votes

Finding value of unknown coefficient such that a linear system has a number of solutions

Here is a bit of an unconventional approach, but I think it works nicely here: Take the first two equations and transpose them: $2x+y=1-3z$ $(1)$ $x+2y=2-2z$ $(2)$ Now, solve these for $x$ and $y$ in ...
Red Five's user avatar
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1 vote

Proof of Minkowski determinant inequality

Because $A$ and $B$ are symmetric, so is $A+B$ and by the spectral Theorem it is diagonalizable. Let $\{e_i\}$ be the orthonormal basis for $A+B$. In this case, the determinant is the product of the ...
Kadmos's user avatar
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2 votes

Similar matrices with entries in $p$-adic numbers

Yes, and in fact more generally if $R$ is a unique factorization domain and $F = Frac(R)$, then any matrix over $F$ whose characteristic polynomial is in $R$ is similar over $F$ to a matrix over $R$. ...
Joshua Mundinger's user avatar
0 votes

How to use Givens rotation for complex matrix?

The idea is to tridiagonalize a matrix in order to apply a Cuppen's algorithm to compute the eigenvalues and eigenvectors of the original matrix. That's why I do $G'AG$ to maintain the similarity of ...
superneiluj's user avatar
0 votes

coupled eigenvalue problem with two eigenvalues

Let $\mathbf{v} = \begin{pmatrix}\mathbf{v}_1 \\ \mathbf{v}_2\end{pmatrix}$, and let $I$ be the indentity matrix. $\boxed{\text{If } \omega_1 \ne 0 \text{ and } \omega_2 \ne 0}$: Then the only ...
Daniel P's user avatar
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0 votes

How to use Givens rotation for complex matrix?

The formula for s should include a negative sign: $$s = -\frac{\overline{A_{13}}}{\text{norm}}$$` To see the result of the given rotation, you need to compute $$ G^H A, \text{ not } G' A G. $$
Rajat Sharma's user avatar
0 votes

How to use Givens rotation for complex matrix?

The formula for $s$ is $-b/r$. You have forgotten the minus sign. Also, if you want to see the result of the Givens rotation, you need $G*A$, not $G'AG$. Lastly, Matlab has built in functionality for ...
Matteo's user avatar
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0 votes

How to solve abstract matrix equations involving transposes and inverses?

These are pretty much the main things to keep in mind: Matrix addition works exactly the same as you'd expect. You can add or subtract matrices from both sides of an equation. Matrix multiplication ...
Wyatt Kuehster's user avatar
1 vote
Accepted

Linear transformations and non-eigenvectors

If $T = \lambda$ is a scalar multiple of the identity then every vector in $X$ is an eigenvector of $T$, so the answer is no in this case. The answer is yes in every other case and a "generic&...
Qiaochu Yuan's user avatar
0 votes

Limit of upper triangular matrix is zero

It is equivalent to prove each column $A^k e_j$ of $A^k$ goes to $0$. Set \begin{eqnarray*} c & = & \sup_{1 \leq i \leq n} | a_{i i} |\\ & < & 1. \end{eqnarray*} For $k \in \...
Kakashi's user avatar
  • 2,054
2 votes
Accepted

Finding the root of a $2 \times 2$ matrix

What Cayley-Hamilton tells you about this $2\times2$ matrix $A$ is that it satisfies some quadratic equation, which can be written $$A^2=pA+qI.\tag1\label{eq1}$$ It is then natural to look for some $a,...
Anne Bauval's user avatar
  • 38.7k
1 vote

Finding the root of a $2 \times 2$ matrix

There are a few possible explanations. You may think it's just a strategy to find a solution (not all of them), and if this doesn't work, you may have to try other methods. However, we do know if $A$ ...
Just a user's user avatar
  • 17.3k
0 votes

Confusion regarding standard generator matrix

Yes, your reasoning is correct. Your matrix is in reduced row echelon form. So there is no generator matrix of the form $(I_k | A)$, since those are in reduced row echelon form, too, and the reduced ...
azimut's user avatar
  • 23k
1 vote

Lower bound of Frobenius inner product

Since the Frobenius inner product is just the sum of the column vectors' dot products (or row vectors'), then we could find a lower bound if we found something that relates 'features' of the matrices ...
Surge's user avatar
  • 867
1 vote
Accepted

A sufficient condition for P-matrix

$$ \begin{pmatrix} 1 & 0 & 2 & 0\\ 0 & 1 & 0 & 0\\ 2 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} $$ This is a block matrix with $2\times 2$ blocks with ...
Exodd's user avatar
  • 11.2k
1 vote

What is the square root of a square matrix squared?

You already can't conclude this for ordinary real or complex numbers; the best you could possibly ask for is that $A = \pm x$ (by which I mean $A = \pm x$ times the identity matrix). But actually you ...
Qiaochu Yuan's user avatar
1 vote

If $\mathrm{rank}\left(AB\right)=\mathrm{rank}\left(B\right)$, would $\mathrm{rank}\left(AB^2\right)=\mathrm{rank}\left(B^2\right)$?

Your intuition was good (up to the lapsus "interval" for "injective"). Let us first prove the following key lemma: $$\operatorname{rank}(AB)=\operatorname{rank}(B)\iff A\text{ is ...
Anne Bauval's user avatar
  • 38.7k
3 votes

If $\mathrm{rank}\left(AB\right)=\mathrm{rank}\left(B\right)$, would $\mathrm{rank}\left(AB^2\right)=\mathrm{rank}\left(B^2\right)$?

I hope it is clear to you that $\mathrm{Im}(T) = \{T(v): v \in V\}$ is a subspace of $V$, where $T: V \to V$ is a linear transformation from a vector space to the same vector space. And if $A$ is a ...
Afntu's user avatar
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-1 votes

Square matrices proved by the negative method

For statements of this kind, it is enough (and usually easiest) to provide a counterexample. Thankfully, we don't have to look far to find one because we can just borrow from the fact that this ...
ConMan's user avatar
  • 25.2k
0 votes

Derivatives of multivariate Gaussian

$ \def\c{\cdot} \def\BR#1{\Big[#1\Big]} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\frob#1{\left\| #1 \right\|_F} \def\q{\quad} \def\qq{\qquad} \def\...
greg's user avatar
  • 36.7k
1 vote
Accepted

Matrix multiplication interactions with scalar multiplication

Does Matrix Multiplication "Distribute" over Scalar Multiplication ? NO Is it true that $T^{-1}(|T| ⋅ I) = T^{-1}|T| ⋅ T^{-1}I$ ? NO Consider this : Matrix Multiplication is "...
Prem's user avatar
  • 11.4k
7 votes

Rank of a matrix over integers

There is just not a unique generalization of the rank to the case of non-fields. It depends on what you want to do. An abstract way to define the rank of a linear map $T : V \to W$ between vector ...
Qiaochu Yuan's user avatar
2 votes

Is it possible to efficiently create a matrix M in which the elements are the sum of all possible "path-products" of matrix A?

If I've understood you correctly, we have $$M = A + A^2 + A^3 + \dots = \frac{A}{I - A}.$$ (In general the notation $\frac{X}{Y}$ for two matrices is ambiguous since it could refer either to $XY^{-1}$ ...
Qiaochu Yuan's user avatar
3 votes

Rank of a matrix over integers

If you have a linear dependence relationship with rational coefficients, you can always convert it to a linear dependence relationship with integer coefficients -- just multiply through by the least ...
Robert Shore's user avatar
  • 24.1k
2 votes

Rank of a matrix over integers

Any finitely generated free module over a commutative ring, in this case the integers $\mathbb Z$, has a well-defined cardinality of any basis, which is called its invariant basis number. Thus, it ...
Alex Pawelko's user avatar
1 vote

Number of solutions to underdetermined system of equations in modular arithmetic vs real or complex valued equations

First, if you have a solution to a linear system defined mod $n$ then any vector that is (coordinate-wise) congruent to that solution is also a solution. So if a solution exists, you will always have ...
anankElpis's user avatar
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1 vote
Accepted

Derivative of $x^T \cdot A \cdot x$ with diagonal $A$

It doesn't matter if $A$ is a diagonal matrix or not, because under small perturbation, the off diagonal elements may still be nonzero. This is like the function $f(x)=2x$ has nonzero derivative $2$ ...
Just a user's user avatar
  • 17.3k
2 votes

Derivative of $x^T \cdot A \cdot x$ with diagonal $A$

If $a^T X b = \sum_{ij} x_{ij} a_i b_j $, then the partial derivative wrt. $x_{ij}$ of this is $a_i b_j$, and collecting these into a matrix, we get $ab^T$. In case of a diagonal matrix, the only ...
Eman Yalpsid's user avatar
  • 3,104
2 votes

Given $A_i, B_i \in \mathbb{R}^{k \times d}$, minimize $\sum_{i} \lVert U A_i V^T - B_i \rVert_F^2 $ over orthogonal $U, V$.

$ \def\BR#1{\Big[#1\Big]} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\frob#1{\left\| #1 \right\|_F} \def\q{\quad} \def\qq{\qquad} \def\qif{\q\iff\q} \...
greg's user avatar
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1 vote
Accepted

For a real symmetric matrix, is the product of two factors of its rank decomposition (right times left) also symmetric?

Neither $PQ$ nor $QP$ is necessarily symmetric. As a counterexample, consider $$ A = \pmatrix{1&0&0\\0&0&0\\0&0&0}, \quad P = Q = \pmatrix{ 1&0&0\\ 0&1&1\\0&...
Ben Grossmann's user avatar
1 vote

Derivative of $trace(|A|)$ with respect to $A$

$ \def\h{\odot} \def\BR#1{\Big[#1\Big]} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\sign#1{\op{sign}\LR{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\frob#1{\left\| #1 \right\|_F} \def\q{\...
greg's user avatar
  • 36.7k
0 votes

solving Matrices questions involving some equations

Suppose$$X = \begin{bmatrix} x_{1} & x_{2} \\ x_{3} & x_{4} \\ \end{bmatrix}, C = \begin{bmatrix} -\lambda & b & a & 0 \\ b & -\lambda & 0 & a \\ a & 0 & ...
Saiba Midori's user avatar
0 votes

Finding the derivative of trace $AXBXC^T$ with Respect to $X$

By invariance of the trace under rotation of the factors and cancelling of any basis change matrices inside, in Einstein notation in any basis $$d (A_{ij} X_{jk}B_{kl}X_{lm}C_{im})= A_{ij} dX_{jk}B_{...
Roland F's user avatar
  • 2,956
0 votes
Accepted

Finding the derivative of trace $AXBXC^T$ with Respect to $X$

You might want to employ the index notation and make use of the definition of the trace and partial derivative. Using Einstein summation convention - implied summation over repeated indices - you may ...
Egor Larionov's user avatar
1 vote

For real matrix $M$ and complex vector $v$, is $M(\operatorname{Re}(v))=\operatorname{Re}(M(v))$?

Just write $v=x+iy$ with $x,y$ real vectors. $Mv=M(x+iy)=Mx+M(iy)=\underbrace{Mx}_{\in\mathbb R}+i\,\underbrace{My}_{\in\mathbb R}\quad$ by linearity. But then $\Re(Mv)=Mx=M\Re(v)$ and same for ...
zwim's user avatar
  • 28.8k
1 vote
Accepted

How to treat juxtaposition between vector and column and how to handle them?

If $s,t \in F^n$ are column vectors; i.e., we can write $$ s = \begin{pmatrix} s_1 \\ \vdots \\ s_n \end{pmatrix} \text{ and } t = \begin{pmatrix} t_1 \\ \vdots \\ t_n \end{pmatrix},$$ then they are ...
Dheeran Wiggins's user avatar
0 votes

Inverse of an invertible triangular matrix (either upper or lower) is triangular of the same kind

Here's an alternative approach using flags: Definition (Upper Triangular w.r.t. Flag). Let $T \in \mathscr{L}(V)$, and let $\mathfrak{F}$ be a complete flag in $V$; i.e., $\mathfrak{F}$ is a ...
Dheeran Wiggins's user avatar
0 votes
Accepted

Tensor chain rule

$ \def\p{\partial} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\gradLR#1#2{\LR{\grad{#1}{#2}...
greg's user avatar
  • 36.7k
1 vote
Accepted

A matrix inequality $X - \frac{1}{u^\top Xu} Xuu^\top X \succeq Y - \frac{1}{u^\top Yu} Yuu^\top Y$

Approach 1. Since $X\succeq Y\succ0$, we have \begin{align} &A:=\pmatrix{u^\top\\ I}X\pmatrix{u&I} =\pmatrix{u^\top Xu&u^\top X\\ Xu&X}\\ \succeq\ &B:=\pmatrix{u^\top\\ I}Y\pmatrix{...
user1551's user avatar
  • 141k
2 votes

Sensitive eigenvectors to small perturbations in the matrix?

This has nothing to do with perturbations. Your approximation matrix is conjugated with regards to the original, so you would expect your eigenvector to be close to the conjugate of the original $\...
Matthew Spam's user avatar
1 vote

Sensitive eigenvectors to small perturbations in the matrix?

This is hardly surprising, as the new matrix is not close to the old one. In fact, the relative error $\|\rho_{num}-\rho\|/\|\rho\|=0.6176$ between the two matrices is very large. What happens here is ...
user1551's user avatar
  • 141k
0 votes
Accepted

For real matrix $M$ and complex vector $v$, is $M(\operatorname{Re}(v))=\operatorname{Re}(M(v))$?

It suffices to write it out. For every $1\leq i\leq n$, we have: $$(Mv)_i=\sum_{j=1}^n M_{ij}v_j.$$ Taking real parts: $$\mathrm{Re}(Mv)_i=\mathrm{Re}\left(\sum_{j=1}^n M_{ij}v_j\right)=\sum_{j=1}^n\...
Zuy's user avatar
  • 4,733
1 vote
Accepted

If a diagonally-dominant matrix is similar to a positive-definite matrix, is the positive-definite matrix also diagonally-dominant?

$$ A=\pmatrix{{11+3\sqrt{13}\over2}&0\cr0&{11-3\sqrt{13}\over2}\cr} \ \ \ \ \ \ \ B=\pmatrix{10&3\cr3&1\cr} $$
Gerry Myerson's user avatar
6 votes
Accepted

Is there a matrix with rational entries similar to a given matrix?

In the second case, it's equivalent to find a matrix $A$ with rational entries such that $A^2=2I$ (because $x^2-2$ is irreducible over $\mathbb Q$, it must be the minimal polynomial of $A$, hence $A$ ...
Just a user's user avatar
  • 17.3k
2 votes
Accepted

An elementary matrix multiplication.

$(AA^T)^{-1}Av=(A^{T})^{-1} A^{-1}Av=(A^T)^{-1}v$ $(A^T)^{-1}v=(v^TA^{-1})^{T}$
Bowei Tang's user avatar
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