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Let $A$ be an invertible $n\times n$ matrix. Without loss of generality, we may assume that the spectrum of $A$ does not meet the half-line $(-\infty,0]$: in fact, it is possible to find a half-line w …

Let $B=A+D$. With $B_1,\dots,B_n$ the columns of $B$, $d_1,\dots,d_n$ the diagonal $D$ $$ \det A=(B_1-d_1e_1)\wedge\dots\wedge (B_n-d_ne_n) $$ so that $\det A$ is an explicit polynomial in $d$, whose …

Take $A=-I$. Then both $\Vert A\Vert_1,\Vert A^{-1}\Vert_1 $ are bounded although $I+A=0$ is not invertible.

May I attract your attention on the following formula, where $B$ is a $n\times n$ non-singular symmetric real-valued matrix $$ \int_{\mathbb R^n}e^{-2i\pi x\cdot \xi} e^{i\pi Bx\cdot x}dx=e^{\frac{i\p …

Some definitions and comments on the logarithm of a nonsingular symmetric matrix. The set $\mathbb C\backslash\mathbb R_{-}$ is star-shaped with respect to 1, so that we can define the principal deter …

You can be completely explicit in this matter. For $T_j$ in a commutative algebra $$ T_1T_2\dots T_k=\frac{1}{2^k k!}\sum_{\epsilon_j=\pm 1} \epsilon_1\dots\epsilon_k(\epsilon_1T_1 +\dots+\varepsilon_ …

Complexification. Let $A$ be a real $n\times n$ matrix such that $A^T=-A$. Let us define $ B=iA, i=\sqrt{-1}. $ We have $$ B^*=\overline{B}^T=-iA^T=iA=B, $$ so that $B$ is self-adjoint on $\mathbb C^n …

Just a remark. Following your assumptions, your matrix is hyperbolic in the sense that you know that all eigenvalues are real-valued. I understand that it depends on some (real) parameters, then from …

Well, you have $$ [\frac{1}{idx},x]=1/i $$ although $\frac{1}{idx}$ is far from the Identity.

With $z=x+iy$, we use the Fourier transformation in $(x,y)$ to see that $H$ is unitarily equivalent to $$ \frac12\begin{pmatrix}2m&\xi-i\eta\\ \xi+i\eta&-2m\end{pmatrix}, \text{whose eigenvalues are } …

No. Take the unitary $ A=\begin{pmatrix} 0&1\\ -1&0 \end{pmatrix} $ which satisfies your assumption. The matrix $B=\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$ has norm 1.

Let $A$ be an invertible $n\times n$ matrix which is antisymmetric: $A^T=-A$, e.g. the symplectic matrix $$ \begin{pmatrix} 0&1 \\\\ -1&0 \end{pmatrix}. $$ The equation $AX+XA^T=I$ cannot have a matri …

The answer by Bazin (https://mathoverflow.net/users/21907/bazin), Faa di Bruno's formula for vector valued functions, URL (version: 2012-09-04): https://mathoverflow.net/q/106339 is providing a formul …