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3

Hint: Suppose that $B(r,s)$ is the matrix where $b_{r,s}=1$ and $b_{i,j}=0$ for $i\not=r$ or $j\not=s$. In other words, $B(r,s)$ is all zeros except for the one position $(r,s)$. Consider $AB(r,s)$. This matrix is all zeros except for the $s$-th column, which is the $r$-th column of $A$. Therefore, the trace of $AB(r,s)$ is $a_{s,r}$. Since the trace of ...


2

Alternatively, let $\lambda_1,\lambda_2,\ldots,\lambda_n\in\mathbb{C}$ be eigenvalues of $A$. Observe that $$\begin{align} n\,\text{trace}\left(A\,A^\top\right)&\geq n\,\sum_{i=1}^n\,\left|\lambda_i\right|^2 \geq\left(\sum_{i=1}^n\,\left|\lambda_i\right|\right)^2\geq \left|\sum_{i=1}^n\,\lambda_i\right|^2 ...


2

A seemingly natural way would be to define a new function that is the value of $v$ on $e_i$ and the value of zero everywhere else. Not really. If a function has value $v$ on $e_i$ that is non-zero on any compactly supported subset of $e_i$, and $0$ elsewhere, it does not have $H^{1/2}$ regularity on the whole boundary: $$ \int_{\partial ...


2

I prove only $y\ge \frac x2(3x-1)$. We know that $A+B+C\ge 0$, so let $\lambda_1,\dots,\lambda_n$ be its eigenvalues. Then $$ \frac 1n Tr(A+B+C)=\frac 1n \sum \lambda_i=\bar \lambda= 3x $$ and $$ Tr((A+B+C)^2)=\sum \lambda_i^2\ge n \bar \lambda^2=9nx^2. $$ But $$ (A+B+C)^2=A^2+B^2+C^2+AB+AC+BA+BC+CA+CB $$ and so $$ ...


2

Suppose that $b,c\ge 0$. Since $u=(a+d)/2$, we have $$ad-bc=\left(\frac{a+d}{2}\right)^2+v^2,$$ Multiplying the both sides by $4$ gives $$4ad-4bc=a^2+2ad+d^2+4v^2$$ i.e. $$-4bc=(a-d)^2+4v^2$$ The LHS is non-positive since $b,c\ge 0$, and the RHS is positive since $v\not=0$. This is a contradiction. Hence, either $b$ or $c$ has to be negative.


2

The given equality is equivalent to $AC=CA$. According to the Jacobson lemma, $C$ is nilpotent and we are done. cf. page 1 of https://jankobracic.files.wordpress.com/2011/02/on-the-jacobsons-lemma.pdf EDIT. to @ George R. Assume that we replace the underlying field $\mathbb{C}$ with a commutative ring $R$. Then the previous reference shows that there is ...


2

The statement is indeed wrong so let's construct a counter example: Let $A=B=\begin{pmatrix}0& 1\\ 0 &0\end{pmatrix}\in M^{2\times 2}(\mathbb R)$, then we have $$ AB=\begin{pmatrix}0& 1\\ 0 &0\end{pmatrix}\begin{pmatrix}0& 1\\ 0 &0\end{pmatrix}=0 $$ and therefore $tr(AB)=0$, we also have $tr(A)=tr(B)=0$ and of course $n=2$ and ...


1

We can indeed say that $$ |\operatorname{Tr}[PA]| \leq \operatorname{Tr}[A] $$ The same cannot be said for $R$. We do have, however, $$ |\operatorname{Tr}[RA]| \leq \|R\| \operatorname{Tr}[A] $$ Where $\|R\|^2$ is the largest eigenvalue of $R^*R$. Another interesting result in the case of $P$ is that we can say $$ |\operatorname{Tr}[PA]| \leq ...


1

Recall that for $K/\mathbb Q$, the trace is defined as $$\text{Tr}(x) = \sum_{\sigma} \sigma(x),$$ where $\sigma$ runs over all embeddings of $K$ into $\mathbb C$. You have already shown that $\text{Tr}\left(\frac{\sqrt 3}{\alpha}\right) = 4b$. Let me calculate it another way. There are four embeddings of $\mathbb Q(\alpha)$ into $\mathbb C$ given by ...


1

If $\sigma_1,\sigma_2\ge0$ then $$ \begin{pmatrix}\sigma_1 & b\\0 & \sigma_2\end{pmatrix},\quad b\ge0 $$ does it. Suppose $\sigma_1<0$. Since the trace has to be nonnegative, it must be that $\sigma_2\ge-\sigma_1$. You can choose any nonnegative $a,b,c,d$ such that $$ a+d=\sigma_1+\sigma_2,\quad b\,c=a\,d-\sigma_1\,\sigma_2. $$ A possible choice ...


1

Hint: Multiply the second equation by $a$ (or $d$, it doesn't matter) and substitute into the first equation.


1

We will use the following results of my first answer: $y\ge \frac x2(3x-1)$. The bound given by the OP is attained for every $n$ and every admissible $x$. Set $$ A_0=\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix},\quad B_0=\begin{pmatrix} 1/4&-\sqrt{3}/4\\ -\sqrt{3}/4&3/4 \end{pmatrix}\quad\text{and}\quad C_0=\begin{pmatrix} 1/4&\sqrt{3}/4\\ ...



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