# Tag Info

5

You can't get a nontrivial lower bound on $|\tr(A^2)|$, if that's what you meant, because it is quite possible to have $\text{tr}(A^2) = 0$ while $\text{tr}(A^T A) \ne 0$. For example, try $$A = \pmatrix{1 & -1\cr 1 & 1\cr}$$ As for $\tr(A^2)$ itself, from your bound you have $$\tr(A^2) \ge - \tr(A^T A)$$ with equality e.g. for $$\pmatrix{0 &... 5 This is false. Consider A = \pmatrix{1 & 0 \\ 0 & -1}. Edit after question edited: If A\in M_{n\times n} is rank 1 then A=uv^T for some u,v\in M_{n\times 1}. We also know that \operatorname{trace}(A) = u^Tv=u\cdot v (can you see why?). So the condition that this matrix is traceless is that u^Tv=0=v^Tu. Then$$A^2 = (uv^T)(uv^T) = ...

4

$$tr(ABA^TC)=\sum_{ijkl}A_{ij}B_{jk}A^T_{kl}C_{li}=\sum_{ijkl}A_{ij}B_{jk}A_{lk}C_{li}$$ $$(\nabla_{A}tr(ABA^TC))_{mn}=\frac{\partial}{\partial A_{mn}}\sum_{ijkl}A_{ij}B_{jk}A_{lk}C_{li}$$ $$=\sum_{ijkl}B_{jk}A_{lk}C_{li}\delta_{im}\delta_{jn}+\sum_{ijkl}A_{ij}B_{jk}C_{li}\delta_{lm}\delta_{kn}$$ $$=\sum_{kl}B_{nk}A_{lk}C_{lm}+\sum_{ij}A_{ij}B_{jn}C_{mi}$$ $$... 3 It's not A*A, it's A^*A where A^* is the transpose of A. If you calculate \operatorname{tr}(A^*A) (called the Frobenius norm^\dagger) you'll find that it equals \|A\|. ^\dagger: Technically this is the square of the Frobenius norm. 3 If you write the function in terms of the Frobenius Inner Product, then finding the differential and gradient is almost trivial$$\eqalign{ f &= C^TAB^T:A \cr\cr df &= C^T\,dA\,B^T:A + C^TAB^T:dA \cr &= (CAB + C^TAB^T):dA \cr\cr \frac{\partial f}{\partial A} &= CAB + C^TAB^T \cr\cr }$$Frobenius products can be rearranged in a variety of ... 2 Given A, B, C \in \mathbb R^{n \times n}, define f : \mathbb R^{n \times n} \to \mathbb R by$$f (X) := \mbox{tr} (A X B X^T C)$$The directional derivative of f in the direction of V at X is$$\begin{array}{rl} D_V f (X) &= \displaystyle\lim_{h \to 0} \frac{1}{h} \left( f (X + h V) - f (X) \right) \\\\ &= \mbox{tr} (A V B X^T C) + \mbox{...

2

You can use the characteristic polynomials. If $A$ is similar to $B$, then their characteristic polynomials $f_A(x)$ and $f_B(x)$ are identical. Since $$f_A(x)=x^n-tr(A)x^{n-1}+\ldots\pm\det(A)\\$$ and $$f_B(x)=x^n-tr(B)x^{n-1}+\ldots\pm\det(B)$$ it follows in particular that $tr(A)=tr(B)$.

2

Hint: By very definition, two matrices $A,B$ are similar iff there exists an invertible matrix $S$ such that $A=SBS^{-1}$. Now apply the trace on both sides, and conclude using associativity of the matrix product.

2

No. Let $A=I\le B=\pmatrix{1+\frac1t&1\\ 1&1}$ with $t>0$. Then $B^{-1}=\pmatrix{t&-t\\ -t&1+t}$ and hence $\operatorname{tr}(A^{-1})=2<1+2t=\operatorname{tr}(B^{-1})$ when $t$ is sufficiently large.

1

Your explanation for the choice of the function space is right. Note that $u \in H_0^2(\Omega)$ if and only if $u = 0$ and $\nabla u = 0$ on $\partial \Omega$. This, of course, is not the right function space for your Neumann problem!

1

Your bound is indeed correct in general. This question (with say the top answer) shows that the sum of the singular values is at least the sum of absolute values of the eigenvalues. On the other hand, a rank-$r$ matrix has exactly $r$ nonzero singular values $\sigma_1,\ldots,\sigma_r$. Since the trace of $A^TA$ is the sum of the squares of the singular ...

1

$1\iff4$ : Assume $1$. Given $x\in N_+$, there exists nonzero $y\in N_\tau^+$ with $y\leq x$. Now use Zorn to find a maximal ordered family $\{y_j\}\subset N_\tau^+$ with $y_j\leq x$ for all $j$. As the net is bounded, it has a sup, say $y=\lim_{sot}y_j$. Then $y=x$, because otherwise a nonzero element of $N+\tau^+$ below $y-x$ contradicts the maximality. ...

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