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4

In general, $\text{tr}(AB) = \text{tr}(BA)$, so $\text{tr}(Axx^T) = \text{tr}((Ax)x^T) = \text{tr}(x^TAx) = x^TAx$.


3

Lemma Given any matrix $A$ of order at least $2$, there exists an invertible matrix $E$ such that $(E^{-1}AE)_{11} = 0$. Proof: Let $A = [a_{ij}]_{n \times n}$, $n \ge 2$. If $a_{11} = 0$, the result holds with $E = I$, the identity matrix of order $n$. If $a_{12} = 0$, then let $E$ be the permutation matrix obtained by interchanging the first and second ...


2

Since the matrices are $2\times 2$, a direct computation is also successful. This is not the perfect solution, but shows that the claim is correct. Let $$ A=\begin{pmatrix} a_1 & a_3 \\ a_2 & a_4 \end{pmatrix},\quad B=\begin{pmatrix} b_1 & b_3 \\ b_2 & b_4 \end{pmatrix}. $$ Then $AB+BA=0$ and $\det(A+B)=0$ are equivalent to the system of ...


2

Let $$A=\begin{pmatrix} 1 & 0 \\ 0 & 1.1 \end{pmatrix}, B=\begin{pmatrix} 1.2 & 0 \\ 0 & 1 \end{pmatrix},C=\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}.$$ Can you conclude anything?


2

for $2\implies 3$: you have, since $a\geq0$, $$ \tau(uau^*)=\tau(ua^{1/2}a^{1/2}u^*)=\tau((a^{1/2}u^*)^*a^{1/2}u^*) =\tau(a^{1/2}u^*(a^{1/2}u^*)^*)=\tau(a^{1/2}u^*ua^{1/2})=\tau(a^{1/2}a^{1/2}) =\tau(a) $$ For $3\implies 1$: since the condition holds for positive $a$ and every operator is a linear combination of positives, the equality holds for arbitrary $...


1

We calculate $$ \DeclareMathOperator{\tr}{trace} \|X - XWG^T\|_F^2 = \\ \tr[(X - XWG^T)^T(X - XWG^T)] =\\ \tr(X^TX) - 2\tr(X^TXWG^T) + \tr(GW^TX^TXWG^T) $$ Note, however, that order matters in that last term.


1

the Laplacian is diagonalized in the orthonormal basis of eigenfunctions $(u_k)$ and $$\Delta \varphi = \sum_k u_k \langle u_k, \varphi \rangle \lambda_k$$ this operator is normal hence $\Delta^a$ means : $$\Delta^a \varphi = \sum_k u_k \langle u_k, \varphi \rangle (\lambda_k)^a$$ and for any complex number $t$ : $$e^{i t \Delta^{1/2}} \varphi = \sum_{m=0}^\...


1

Not sure what's the trick, but the following should work. Suppose the contrary that $\operatorname{tr}(A)\ne0$. As $A$ and $B$ anticommute, we have $AAB = -ABA = BAA$, i.e. $A^2$ commutes with $B$. Yet, $A^2=\operatorname{tr}(A)A - \det(A)I$. So, $A$ also commutes with $B$. Hence $AB=BA=0$ and both $A,B$ are singular (because they are nonzero by assumption)...



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