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Prove symmetric matrix $A$ is congruent to $A ^2$ iff $A$ is PSD

Since $A$ is symmetric matirx, there exists an orthogonal matrix $T$ such that $$T^{t}AT=\mathbb{diag}\{\lambda_1,\cdots,\lambda_n\}$$ where $\lambda_1,\cdots,\lambda_n$ are all eigenvalues of $A$ and ...
fusheng's user avatar
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$\int_{x_i>0} f(x_1+x_2, x_3+x_4)^* f(x_1+x_3, x_2+x_4) \;(x_1x_2x_3x_4)^\alpha \geq0$ for any function $f$ and $\alpha >-1$?

With some constant $c$ which is later to be determined, define for $(i = 1 \cdots 4)$ new variables $y_i = x_i^{1+c}$ and let $\beta = 1 + c$. We require $\beta > 0$ (which will be substantiated ...
Andreas's user avatar
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How to verify positive definitiveness of the given Kinetic term?

The last terms are cubic under scaling, that is, multiplying all $c_k$ and thus also the $\dot c_k$ with the same (constant) scalar. Now consider the whole expression as polynomial in this scalar. ...
Lutz Lehmann's user avatar
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$\int_{x_i>0} f(x_1+x_2, x_3+x_4)^* f(x_1+x_3, x_2+x_4) \geq0$ for any function $f$?

Too long for a comment on the post of @abacaba, hence this post. Let's integrate directly from the equations which describe the change of variables. Consider first $x_1$ as independent. Then, since $...
Andreas's user avatar
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$\int_{x_i>0} f(x_1+x_2, x_3+x_4)^* f(x_1+x_3, x_2+x_4) \geq0$ for any function $f$?

We consider the change of variables to $x_1, a, b, y$, where $$a = x_1 + x_2, b = x_1 + x_3, y = x_1 + x_2 + x_3 + x_4.$$ The Jacobian of this change of variable is $1$. The inverse of this change of ...
abacaba's user avatar
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Proving positive semidefiniteness of a matrix multiplication

Suppose $\lambda$ is an eigenvalue of $M$. Then $$Mv = \lambda v\implies Qv = \lambda(I+\alpha Q)v \implies (1-\lambda \alpha)Qv = \lambda v\implies Qv = \frac{\lambda}{1-\lambda \alpha} v $$ so $\mu ...
Exodd's user avatar
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Proving positive semidefiniteness of a matrix multiplication

We notice that the matrix $I+aQ$ is positive-semidefinite, since $x^T(I+aQ)x = x^TIx + x^T(aQ)x$, a sum of two non-negative numbers. The inverse is also a positive semi-definite, since the eigenvalues ...
Eemil Wallin's user avatar
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Definiteness of quadratic form with block matrix structure

This is only an incomplete answer, to better illustrate the kind of calculation I want to do. Here is a possible transformation one can do: $$ x=A\xi,\quad y=B\eta $$ Then, $$f = \hat{f}(\xi,\eta) = \...
a06e's user avatar
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Showing positive definiteness of joint process with cross-correlation

It’s a little presumptuous to write $LL^T$ for the matrix which we want to prove is PSD. Let’s just call the matrix $M$ and look at the product $\textbf{x}^T M \textbf{x}$. For ease of dealing with ...
Lee Fisher's user avatar
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Showing positive definiteness of joint process with cross-correlation

Notice that $K^\intercal K$ is always positive semidefinite since $x^\intercal K^\intercal K x = \|Kx\|^2$ and this will be $>0$ provided $Kx \neq 0$ and therefore, this will be $>0$ for all $x \...
William M.'s user avatar
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Showing positive definiteness of joint process with cross-correlation

I use positive definite below to mean non-negative eigenvalues, because that's how I think the OP uses it. Your matrix is $\rho K^TK +(1-\rho) D$ with $K:=(L_1 \; L_2)$ and $D$ being block diagonal ...
Bananach's user avatar
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