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11

Here is an analytic proof (which I think I have learnt somewhere else, although not in this form). We first show an algebraic fact: Theorem 1. Let $\mathbb{K}$ be a commutative ring and let $n\in\mathbb{N}$. Let $p_{1},p_{2},\ldots,p_{n-1}$ be $n-1$ elements of $\mathbb{K}$. For any $\left( i,j\right) \in\left\{ 1,2,\ldots,n\right\} ^{2}$ satisfying ...

7

This answer is a suggestion as to how you may prove the result, but does not contain a proof. Instead, I'll show how a similar problem has been answered, and offer pointers to claims that would show your result: Consider the matrix: $$A_{ij} = \frac{1}{1 + |a_i - a_j|^2}$$ Ultimately we will find that $\det(A)$ is always non-negative. The reason is that ...

6

First assume that $B$ is invertible. We then have $$A = B^{-1}AB$$ and $$\det(A+BC) = \det(B^{-1}(A+BC)B) = \det(B^{-1}AB + CB) = \det(A + CB)$$ Since invertible matrices are dense and determinant is continuous, equality extends to non-invertible $B$'s by continuity.

6

Since the most of related results seem to be published in articles devoted to the perturbation of determinant, let me rewrite your bound $$\big\vert \, \det \left(\,A\,\right) - \det \left(\,B\,\right)\,\big\rvert \le n^{n+1} \left\|\,A-B\,\right\|_\infty \max \big\{ \left\|\,A\,\right\|^{n-1}_\infty, \,\left\|\,B\,\right\|^{n-1}_\infty ... 5 Based on the dimensions of A and G (as defined in the paper), it appears that \det(A,G) is the determinant of the matrix you get by augmenting A by G. The identity you mention would be written as$$\left|\det(M)\right| = \sqrt{\det (M^\top M)}$$in modern notation. 5 We could use eigenvalues: If \lambda is an eigenvalue of M, then \mu = \lambda + x is an eigenvalue of A = M + xI. Equivalently, if \mu is an eigenvalue of A, then \mu - x is an eigenvalue of M. Note that A is given by$$ A = \pmatrix{ 0 & a_2 & a_3 & \cdots & a_n\\ a_1+x & 0 & 0 &\cdots & 0\\ ...

5

Let $A$ and $D$ be the identity matrix. Does $B=C$ imply $EB=CE$?

4

The matrix $M_n$ is a circulant matrix where $c_0=c_1=c_{n-1}=0$ and other $c_k=1$. Eigenvalues and eigenvectors of circulant matrices can be readily calculated.

4

Adding to a row another row multiplied by any number doesn't change the determinant: \begin{align} \begin{bmatrix} x-3 & x-4 & x-a \\ x-2 & x-3 & x-b \\ x-1 & x-2 & x-c \end{bmatrix} &\to \begin{bmatrix} -1 & -1 & b-a \\ x-2 & x-3 & x-b \\ x-1 & x-2 & x-c \end{bmatrix} &&R_1\gets R_1-R_2 ...

3

$(1,-2,1)$ is a left eigenvector associated with a zero eigenvalue, so the determinant is trivially zero since that matrix is not invertible.

3

$A \ =\begin{pmatrix} 7 & 1 & 3 & -2\\ -2 & 1 & -12 & -1 \\ 1 & 16 & -4 & a \\ 2 & 4 & 2 & 2 \\ \end{pmatrix}\cong \begin{pmatrix} 13 & 1 & 5 & -5\\ -5 & 1 ... 3 Hint:$\det(M) = \det(M^T)$for any square matrix$M$. Solution: (after comment exchange below) Let$M = cI_n - A$. Note that$M^T = cI_n^T - A^T = cI_n - A^T$. Then, from the hint, you get$\det(cI_n - A) = \det(cI_n - A^T)$3 If$J$is the$n\times n$matrix with all entries 1, then$M_n=J-I-C_n$, where$C_n$is the adjacency matrix of the cycle. The all-ones vector is an eigenvector for$C_n$with eigenvalue 2. Since$C_n$is symmetric, if$\lambda\ne2$is an eigenvalue for$C_n$with eigenvector$z$, then$z$must be orthogonal to the all-ones vector. Hence$Jz=0$and therefore ... 2 The first question is not true as the following counterexample shows: $$A = B = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} \\ C = D = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\\ E = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$ We get $$AB = 0 = CD$$ but $$AEB = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\\ CED = ... 2 I don't think this works by sandwiching.....$$A=\begin{pmatrix} 1 & 3 \\ 0 & 1 \\ \end{pmatrix}, B=\begin{pmatrix} 1 & 2 \\ 0 & 1 \\ \end{pmatrix}, C=\begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix}, D=\begin{pmatrix} 1 & 4 \\ 0 & 1 ... 2 The "sandwich" statement doesn't generally hold. For example, take $$A = B = I = \pmatrix{1&0\\0&1}, \quad C = D = \pmatrix{0&1\\1&0}$$ Verify that$AB = CD$. However, if we take $$E = \pmatrix{1&0\\0&0}$$ we find that$AEB \neq CED$2 Yes of course, for example the matrix $$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\in\operatorname{SL}_2(\Bbb R)$$ but$A\not\in\mathcal O_2(\Bbb R)$since$AA^T\ne I_2$. 2 So you have a map from$\mathbb R^4 \to \mathbb R^4$;$x \mapsto L^t x$, which maps the set you are interested in to the unit sphere. Since multiplication by a matrix$M$causes sets to be mapped to new sets whose volume is$\det(M)$times as large as its original volume, you deduce that the volume of your set is$\frac{\pi^2}2 \div \text{det}(L^t) = ...

2

The property $\det(E^iE^j) = \det(E^i)\det(E^j)$ doesn't just hold for two elementary matrices $E^i,E^j$, it holds for any elementary matrix $E$ and any matrix $A$: $$\det(EA) = \det(E)\det(A)$$ Proof: There are three types of elementary matrices: row-scaling row-swapping row-adding If I have some matrix $A$ with determinant $\det(A)$, and I ...

2

Call the determinant $D_n(a_1,a_2,\ldots,a_n;x)$. Perform the following operations: row $n$ replaced by row $n$ minus row $n-1$; and then column $n-1$ replaced by column $n-1$ plus column $n$. This gives an equal determinant, in which the last row is all zero except for $-x$ at the end. Expanding along the last row, ...

2

For (2), note that $$cI_n-A^T = (cI_n)^T-A^T = (cI_n-A)^T,$$ and then that $\det A = \det A^T$ for any $A$.

2

It should be $\text{tr}(\text{adj}(A)H)$. The derivative of $\det A$ is easiest to obtain from Laplace's expansion. Denote $A_{ij}$ the cofactor of the element $a_{ij}$. Then $$\det A=a_{ij}A_{ij}+\text{independent on a_{ij} terms}\quad\Rightarrow\quad \frac{\partial\det A}{\partial a_{ij}}=A_{ij}.$$ Then in your notations $D\det_A(H)$ is the linear ...

2

In order for a matrix to not be diagonalizable, you need that the geometric multiplicity is less than the algebraic multiplicity. Meaning.. the number of (linearly independent) eigenvectors is less than the number of roots of the characteristic equation. Writing out the characteristic equation: $$(3-\lambda)((a-\lambda)(-\lambda)-(a-2)(-2)) = ... 2 OK I figured it out myself... someone please check if this is right Instead of proving A has non-negative determinant, we can actually prove it's positive semidefinite. The proof is done by induction on n, and all we need is to prove \det(A)\ge0. First note that permuting a_1,\cdots,a_n, or adding a constant to all of them doesn't change the ... 2 No, not even for diagonal matrices. Consider for instance$$C = \left[\begin{array}{cc}\sqrt{2}-1 & 0 \\ 0 & \sqrt{2}-1\end{array}\right]B = \left[\begin{array}{cc}1 & 0\\0 & \epsilon\end{array}\right]A = I.$$Then$$\det(A+B) = 2+2\epsilon \geq 2 = \det(A+C)$$but$$\det(B) = \epsilon < 3-2\sqrt{2} = \det(C)$$for ... 2 You have reduced the proof to this: If E is an n\times n matrix in reduced row echelon form, and \det(E) \neq0, then E is the identity matrix. To show this, I think you can work directly from the definition of RREF. The leading coefficient in the first row must be in the first column. Otherwise e_{11} = 0 and \det(E) would be zero. So ... 2 In other words, you multiply each row and column of the square matrix A by the same \omega_q. Therefore the determinant changes by \prod \omega_q^2 = 1. 2 It is not linear, or more precisely it is linear only for matrices of size 1. For a matrix of size n\times n, the determinant, as a function of matrix columns, is multilinear. If A=[a_1,a_2,\dots, a_n], where a_i are columns (with n rows), then$$\det(\lambda_1a_1, a_2,\dots, a_n] = \lambda \det(A)$$which is the definition of multilinearity. ... 2 Consider a 2\times 2 matrix$$ A=\left[\matrix{a_{11} & a_{12}\\ a_{21} & a_{22}}\right]. $$Using the column notations$$ A_1=\left[\matrix{a_{11}\\ a_{21}}\right],\quad A_2=\left[\matrix{a_{12}\\ a_{22}}\right] $$we can write$$ A=[A_1\ A_2], \qquad \det A=\det[A_1\ A_2]=f(A_1,A_2)=a_{11}a_{22}- a_{21}a_{12}  that is the determinant is a ...

2

The rank of the matrix $xx'$ is $1$ because all of its columns are scalar multiples of just one column, $x$, so just one column spans the column space. As with columns, in this case so also with rows: they are all scalar multiples of just one row vector, $x'$. If the rank of an $n\times n$ matrix is less than $n$, then its determinant is $0$.

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