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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.

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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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