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21

These answers require way too much machinery. By definition, the characteristic polynomial of an $n\times n$ matrix $A$ is given by $$p(t) = \det(A-tI) = (-1)^n \big(t^n - (\text{tr} A) \,t^{n-1} + \dots + (-1)^n \det A\big)\,.$$ On the other hand, $p(t) = (-1)^n(t-\lambda_1)\dots (t-\lambda_n)$, where the $\lambda_j$ are the eigenvalues of $A$. So, ...


13

$\newcommand{\tr}{\operatorname{tr}}$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 ...


13

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 ...


11

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 ...


11

If the matrices are non-singular, then writing $A=-BAB^{-1}$ and taking the trace, we get $\mathrm{tr}A=-\mathrm{tr}A$. Hence $\mathrm{tr}A=0$, and the procedure for $B$ is analogous. Next compute the determinant of both sides of $AB=-BA$: this yields $\mathrm{det}\,A\,\mathrm{det}\,B=(-1)^N\mathrm{det}\,B\,\mathrm{det}\,A$, where $N$ stands for size of ...


11

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 ...


11

This statement is false : take $S = I$ for example. It would give us $2^n = 1 + n$.


10

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} ...


9

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})\\ ...


8

Hint Compare the characteristic polynomials of $AB$ and $BA$. The determinant (whence characteristic polynomials) admits basis-free definitions. We have $$ \left(\matrix{I&A\\B&tI}\right)\left(\matrix{tI&-A\\0&I}\right)=\left(\matrix{tI&0\\*&tI-BA}\right) $$ and $$ ...


8

Let $A$ be a matrix. It has a Jordan Canonical Form, i.e. there is matrix $P$ such that $PAP^{-1}$ is in Jordan form. Among other things, Jordan form is upper triangular, hence it has its eigenvalues on its diagonal. It is therefore clear for a matrix in Jordan form that its trace equals the sum of its eigenvalues. All that remains is to prove that if ...


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 ...


7

As mentioned in the comments, the assertion "$\operatorname{tr}(e^{tA}) = e^{\operatorname{tr}(tA)}$" is simply false. On the other hand, the integration problem is straightforward. We have $\exp(tA) = \sum_{n\geq 0} A^n \frac{t^n}{n!}$ for any finite-dimensional matrix $A$, since $\exp$ has infinite radius of convergence. By linearity of integration, and ...


7

Let $A$ be symmetric positive definite matrix hence $\exists$ a diagonal matrix $D$ whose diagonal entries are nonzero and $A=P D P^{-1}$ so $A^{-1} = P D^{-1} P^{-1}$ and $Tr(A^{-1})= Tr(D^{-1})$. Now $D$ being diagonal matrix with non zero diagonal entries $D^{-1}$ has diagonal entries reciprocal of the diagonal entries of $D$ so $Tr(D^{-1})$ is sum of ...


6

Note that, when $D$ is diagonal: $$(DA)_{ii} = D_{ii} A_{ii}$$ So $tr(DA) = \sum_{i=1}^n D_{ii} A_{ii}$. About the best bound you can do for this is the Cauchy-Schwarz inequality, i.e. $$|tr(DA)| \leq \left ( \sum_{i=1}^n D_{ii}^2 \right )^{1/2} \left ( \sum_{i=1}^n A_{ii}^2 \right )^{1/2}$$ If you want a result in terms of traces, you can use the fact ...


6

I assume you want the trace of a matrix $A\in M_n(F)$ to be defined as the sum of the diagonal elements and that you take the coefficients in a (commutative) field $F$. Here is an approach using only basic facts about bases and matrices. 1) Recall the trace is commutative $\mathrm{tr}(AB)=\mathrm{tr}(BA)$, as shown by the usual computations. In particular ...


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 & ...


6

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*} ...


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

The reason that the equality $Tu=u_{|\partial \Omega}$ is stated for functions in $C(\overline{\Omega})$ is that for other functions it is not clear what $u_{|\partial \Omega}$ means. Of course, we can take any function $u\in W^{1,p}(\Omega)$ and extend its domain to $\overline{\Omega}$ by letting $u$ be equal to $Tu$ on the boundary. This will be an ...


6

The following is a simple combinatorial interpretation of this identity. Not exactly what you asked for, but still fun and relevant. Suppose we have two sets $S,T$ with functions $g: S \to T$ and $f : T \to S$. Then $f\circ g : S \to S$ and $g\circ f: T \to T$ are endo-functions of $S$ and $T$ respectively. Now consider $\text{Fix}(f\circ g) \subseteq S$, ...


6

Not true. Try $A = B = \pmatrix{1 & t\cr t & 1\cr}$ for $0 < t < 1$.


6

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 $t=1$, $\det(e^{A})= e^{\text{tr}(A)}$.


5

If $A=(a_{ij})$ and $B=(b_{ij})$ and $C=B^TA=(c_{ij})$ then $$(c)_{ij}=\sum_{k=1}^m b_{ki}a_{kj}$$ $$\mathrm{tr}(B^TA)=\sum_{i=1}^n c_{ii}=\sum_{i=1}^n\sum_{k=1}^m b_{ki}a_{ki}$$ so we see that $\langle.,.\rangle$ is an inner product (Euclidian) by identifying $\mathcal M_{m\times n}(\mathbb R)$ to $\mathbb R^{m\times n}$.


5

For every $A=(A_{ij}) \in \mathbb{R}^{m\times n}$ we have $$ \langle A,A\rangle=\text{tr}(A^TA)=\sum_{i=1}^n(A^TA)_{ii}=\sum_{i=1}^n\sum_{j=1}^mA^T_{ij}A_{ji}=\sum_{i=1}^m\sum_{j=1}^nA_{ij}^2 \ge 0, $$ and $$ \langle A,A\rangle=\sum_{i=1}^m\sum_{j=1}^nA_{ij}^2 = 0\iff (A_{ij}=0 \quad \forall i,j) \iff A=0 $$ Since $$ \text{tr}(X^T)=\text{tr}(X), \quad ...


5

The trace of a matrix $A\in\mathbb R^{n\times n}$ is nothing but the sum of its eigenvalues. So Ker(Trace) is the set of matrices with vanishing sum of eigenvalues, and its dimension is $n^2-1$. The quotient space ${\mathcal M}_n/$Ker(Trace) is a subspace of dimension $1$ isomorphic to $\mathbb R$. The elements of this quotient space are classes of ...


5

The trace of a matrix $M$ is $0$ if and only if the sum of the elements on the (main) diagonal of $M$ is $0$. Since the dimension of all $n\times n$ matrices is $n^2$ and the dimension of its image $\mathbb R$ is $1$ (see below), we know that the dimension of the kernel of $\mathrm{tr}$ is $n^2-1$. (That follows from the fact that the dimension of the image ...


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

I'll try to show it another way. We know that if we have a polynomial $x^n+b_{n-1} x^{n-1} + \dots +b_1 x+ b_0$, then $(-1)^{n-1} b_{n-1}$ is the sum of the roots of this polynomial. (So-called Vieta's formulas) In our case, the polynomial is $\det(tI-A)$ and we have $(-1)^{n-1} b_{n-1}=\lambda_1+\lambda_2+\dots+\lambda_n$. $\def\S{\mathcal{S}_n}$ Let ...



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