# Tag Info

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 ...

### 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 ...

### 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 ...
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### 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 ...
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 ...
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### 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 ...

### 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 ...
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### 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 ...
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### 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 $A$ ...
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$(AA^T)^{-1}Av=(A^{T})^{-1} A^{-1}Av=(A^T)^{-1}v$ $(A^T)^{-1}v=(v^TA^{-1})^{T}$