Hot answers tagged trace
9
The trace of an endomorphism $f : X \to X$ of a dualizable object $X$ in a monoidal category is the composition $1 \xrightarrow{\eta} X \otimes X^* \xrightarrow{f \otimes \mathrm{id}} X \otimes X^* \cong X^* \otimes X \xrightarrow{\epsilon} 1$. This coincides with the usual definition in the category of vector spaces. There is a more general categorical ...
8
Both sides are continuous. A standard proof goes by showing this for diagonalizable matrices, and then using their density in $M_n(\mathbb{C})$.
But actually, it suffices to triangularize
$$
A=P^{-1}TP
$$
with $P$ invertible and $T$ upper-triangular. This is possible as soon as the characteristic polynomial splits, which is obviously the case in ...
7
Let $A$ be the positive definite square root of $X$ and $B$ the positive definite square root of $Y$.
You have
$$
\mbox{tr}(XY)=\mbox{tr}(AABB)=\mbox{tr}(BAAB)=\mbox{tr}((AB)^*AB)>0.
$$
Indeed, the latter is the sum of all $c_{i,j}^2$ where $c_{i,j}=(AB)_{i,j}$.
So it is nonnegative.
And if it were zero, this would imply $AB=0$ hence $A=B=0$ since ...
6
Hint: Prove that $AB$ and $BA$ have the same characteristic polynomial.
This can be done using only the determinant. And the determinant can be defined without reference to any basis, as you can see here: Formula of the determinant
Now write, for every $n\times m$ matrix $A$ and every $m\times n$ matrix $B$:
$$
...
6
What is true is that the expansion of the characteristic polynomials is given by traces of powers of the matrix $A$; explicitly, the characteristic polynomial $\chi_A(T)=\det(T\cdot{\rm Id}-A)$ is given by
$$T^n-{\rm tr}(A)T^{n-1}+\frac{{\rm tr}(A)^2-{\rm tr}(A^2)}{2}T^{n-2}-\frac{{\rm tr}(A)^3-3{\rm tr}(A){\rm tr}(A^2)+2{\rm tr}(A^3)}{6}T^{n-3}+\cdots $$
...
6
Let $S_1$, $S_2$ be two positive definite matrices.
Let $\Delta = S_2 - S_1$ and for $t \in [0,1]$, let
$$\phi = (S_1 + t\Delta)^{-1} = ((1 - t) S_1 + t S_2)^{-1}$$
We have:
$$\begin{align}
& \frac{d}{dt} \phi \;= - \phi \Delta \phi\\
\implies & \frac{d}{dt} \phi^2 \;= - \phi \Delta \phi^2 - \phi^2 \Delta \phi\\
\implies & ...
5
This is false. By the (finite-dimensional) spectral theorem, every symmetric matrix is diagonalizable, so we can write $S=QDQ^{-1}$. Thus we have
$$\det(I+S)=\det(Q^{-1}(I+S)Q)=\det(I+D)=\prod\limits_{i=1}^n (1+d_i)\\\ne 1+\sum\limits_{i=1}^n d_i=1+\mathrm{trace}(D)=1+\mathrm{trace}(S).$$
Perhaps the formula $\prod\limits_{i=1}^n (1+d_i)$ may be of use, ...
5
$\newcommand{\tr}{\operatorname{tr}}$Here is an exterior algebra approach. Let $V$ be an $n$-dimensional vector space and let $\tau$ be a linear operator on $V$. The alternating multilinear map $$
(v_1,\dots,v_n) \mapsto \sum_{k=1}^n v_1 \wedge\cdots\wedge \tau v_k \wedge\cdots\wedge v_n
$$ induces a unique linear operator $\psi: \bigwedge^n V \to ...
5
Hint:
Use that every complex matrix has a jordan normal form and that the determinant of a triangular matrix is the product of the diagonal.
use that $\exp(A)=\exp(S^{-1} J S ) = S^{-1} \exp(J) S $
And that the trace doesn't change under transformations.
\begin{align*}
\det(\exp(A))&=\det(\exp(S J S^{-1}))\\
&=\det(S \exp(J) S^{-1})\\
...
5
You don't need to choose any particular basis; all you need is the fact that $\wedge V$ is graded and $L_b$ raises the degree by $1$, so if you choose any basis that respects the gradation, then $L_b$ sends any basis vector to a different subspace, so the diagonal elements in such a basis all vanish.
4
Yes, it is. Consider $S(t) = A + t B$ where $A$ is symmetric positive definite and $B$ is symmetric. It is enough to show that $$\left.\dfrac{d^2}{d t^2} \text{Tr}(S(t)^{-1})\right|_{t=0} \ge 0$$ Now
$$ S(t)^{-1} = (A (I + t A^{-1} B))^{-1} = A^{-1} - t A^{-1} B A^{-1} + t^2 A^{-1} B A^{-1} B A^{-1} + \ldots$$
so $$ \left. \dfrac{d^2}{\partial t^2} ...
4
Hint: the trace is multiplicative, i.e. $\mbox{Tr}(AB)=\mbox{Tr}(BA)$, as soon as the scalar field (or ring) is commutative. In particular, for every invertible matrix $P$ and every matrix $A$ in $M_n(\mathbb{R})$ or $M_n(\mathbb{C})$, we have
$$
\mbox{Tr}\;(PAP^{-1})=\mbox{Tr}\; A.
$$
4
You may do it by first computing matrix powers and then you may calculate whatever you want. Now question is how to calculate matrix power for a given matrix, say $A$? Your goal here is to develop a useful factorization $A = PDP^{-1}$, when $A$ is $n\times n$ matrix.The matrix $D$ is a diagonal matrix (i.e. entries off the main diagonal are all zeros).
Then ...
4
The proof in Martin Brandenburg's answer may look scary but it is secretly about moving beads around on a string. You can see all of the relevant pictures in this blog post and in this blog post. The proof using pictures is the following:
In the first step $g$ gets slid down on the right and in the second step $g$ gets slid up on the left.
You can also ...
3
If there's a basis $|n \rangle$ in our vector space, you can expand the identity with $1 = \sum_n |n \rangle\langle n| $
\[tr(AB) = \sum_n \langle n|AB |n \rangle =\sum_{m,n} \langle n|A|m \rangle \langle m|B |n \rangle \]
and then you can run it backwards:
\[ = \sum_{m,n} \langle m|B |n \rangle\langle n|A|m \rangle
= \sum_{m} \langle m|B A|m \rangle ...
3
For clarification, following the question, let us emphasize some notations first.
Let $M_n$ be the collection of $n\times n$ complex valued matrices, let $H_n$ be the collection of $n\times n$ complex valued Hermitian matrices, and given $X,Y\in H_n$, denote $X\ge 0\Leftrightarrow$ $X$ is positive semidefinite, and $X\ge Y\Leftrightarrow X-Y\ge 0$. Given ...
3
Yes, it holds true. Let $A$ be a $n\times m$ and $B$ be a $m \times n$ matrix over the commutative ring $R$, we have
\begin{align*}
\mathrm{tr}(AB) &= \sum_{i=1}^n (AB)_{ii}\\
&=\sum_{i=1}^n \sum_{j=1}^m A_{ij}B_{ji}\\
&= \sum_{j=1}^m \sum_{i=1}^n B_{ji}A_{ij}\\
&= \sum_{j=1}^m (BA)_{jj}\\
&= \mathrm{tr}(BA)
\end{align*}
...
3
If you want to make the trace appear, you can use the following formula:
$$
y^Tz=\mbox{Tr} (yz^T).
$$
Then
$$
\|y-X\beta\|^2+\lambda \beta^TS\beta=(y-X\beta)^T(y-X\beta)+\lambda \beta^TS\beta
$$
$$
=\mbox{Tr}((y-X\beta)(y-X\beta)^T)+\lambda \mbox{Tr}(\beta \beta^TS^T)
$$
$$
=\mbox{Tr}((y-X\beta)(y-X\beta)^T+\lambda \beta \beta^TS^T))
$$
3
I have considered two different paths to this problem, one requires the diagonalization of $A$ (so may be not so useful depending on $A$) and the other uses the Kronecker product properties. One other note at the end gives an equivalent skew form of the equation. The Kronecker seems to have the most usefulness for finding the desired trace. The rest is maybe ...
3
There's no failure, unless you count assuming that such a matrix $X$ exists for any $A$. You've in fact given a proof that if $\det{A}\leq0$, there is no $X$ (with real trace) such that $e^X=A$.
If you allow $\operatorname{Tr}(X)$ to be complex (so it is possible for $e^{\operatorname{Tr}(X)}$ to be negative), then there does exist $X$ with $e^X=A$ for any ...
3
Let $f(t)= \det(e^{tA})$. Then $f'(t)=D \det(e^{tA}) \cdot Ae^{tA}=\text{tr} \left(^t \text{com}(e^{tA})Ae^{tA} \right)$. But $A$ and $e^{tA}$ commute, and $^t\text{com}(e^{tA})e^{tA}=\det(e^{tA}) \operatorname{I}_n$. Therefore, $f'(t)=\text{tr}(A)f(t)$ and $f(0)=1$, hence $f(t)=e^{\text{tr}(A)t}$. For $t=1$, $\det(e^{A})= e^{\text{tr}(A)}$.
3
The claim results from the following statement:
(*) If $E/F$ is a degree $n$ separable extension, then the trace
$$
t(E/F):E\to F
$$
of $E/F$ is nonzero.
To prove (*), denote by
$$
t(A/K):A\to K
$$
the trace of $A$ over $K$, whenever $K$ is a field and $A$ a finite dimensional $K$-algebra.
As
$$
t((K\otimes E)/K)=K\otimes t(E/K): K\otimes E\to K
$$
...
2
Unless some contextual details are missing, your book is wrong. For a counterexample, consider $R=\begin{pmatrix}1&1\\1&2\end{pmatrix}$ and $U=\begin{pmatrix}0&0\\0&1\end{pmatrix}$. Clearly $R$ is symmetric and $tr(U^TU)=1$. You may verify that
\begin{align*}
Q_1&=tr(URRU^T)-N^{-1}\left(tr(UR)\right)^2
...
2
I would do it as follows:
For $(x,y) \in H \times H$ define an element of $N(H)$ by $x \,\tilde{\otimes}\, y = \langle \cdot , y \rangle x$ and note that $\|x \, \tilde{\otimes}\,y\|_{1} = \|x\|\,\|y\|$.
For a functional $\varphi \in N(H)'$ define a sesquilinear form on $H$ by $B_{\varphi}(x,y) = \varphi(x \,\tilde{\otimes}\, y)$. The map $\varphi ...
2
Let $F$ be the field of $2^N$ elements. If $Tr(ax)=0$ for all $x\in F$, then the polynomial
$$
p(x)=Tr(ax)=ax+a^2x^2+a^4x^4+\cdots+a^{2^{N-1}}x^{2^{N-1}}
$$
has (at least) $2^N$ distinct zeros in the field $F$, namely all its elements. If $a\neq0$, then this polynomial has degree $2^{N-1}$. Therefore $\ldots$
2
Note that $\text{tr}(AB) = \sum_i \sum_j A_{ij} B_{ji}$ so $\dfrac{\partial}{\partial A_{ij}} \text{tr}(AB) = B_{ji}$, i.e. $\text{grad}_A \text{tr}(AB) = B'$. Thus
$ \text{grad}_A \text{tr}(ABDC) = (BDC)'=C'D'B'$, while $\text{grad}_A \text{tr}(DBA'C) =
\text{grad}_A \text{tr}(CDBA') = \text{grad}_A \text{tr}(AB'D'C') = (B'D'C')' = CDB$.
So (substituting ...
2
As commenters say, $L^* = \hom_K(L,K)$ is a vector space over $K$ of dimension $n$, the same dimension as $L$ over $K$.
You have a $K$ linear map $T : L \to L^*$, such that $T(x)$ is the linear form defined by $T(x)(y) = Tr(xy)$.
Since there is a $y \in L$ such that $Tr(y) \neq 0$, $T(1)$ is not the zero map of $L^*$.
Since $L$ is a field, for any $\alpha ...
2
Your $\wedge^k A^m$ is just the polarization:
$\wedge^k A^m = {N \choose m} \wedge^k(\underbrace{A,\dots,A}_m,1,\dots,1)$,
the right-hand side refers to $\wedge^k$ as a $k$-linear form. As we know, $\wedge^N$ is the determinant, so $\wedge^N A^m$ are just the coefficients of the characteristic polynomial. Those are elementary symmetric functions of the ...
2
Minimizing $f(X)$ with respect to $X$ is equivalent to first minimizing $f(X)$ with respect to $X$ under the constraint $g(X)=c$ and then minimizing the result with respect to $c$. Thus the minimum with respect to $X$ must be the minimum with respect to $X$ under the constraint $g(X)=c$ for some value of $c$, and to find that you can introduce a Lagrange ...
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