Trace of linear operator on $\mathbb R^n$ as an integral ?

Question

Let $T : \mathbb R^n\to \mathbb R^n$ be a linear map. Is it true that $Trace (T)=\dfrac 1{\mu(B)}\int_{x \in B} \langle Tx, x\rangle d\mu(x)$,

where $B$ is the closed unit ball in $\mathbb R^n$ and $\mu$ is the usual Lebesgue measure and $\langle.,.\rangle$ is the usual inner-product on $\mathbb R^n$?

Answer 1

The normalization of the formula is wrong, but let me propose a fix. Let

$$K_n=\int_{B\subset\Bbb{R^n}}x_1^2\,d\mu$$

be the integral we would get with $T=e_{11}$, the matrix with $1$ at position $(1,1)$ and zeros elsewhere. I think that we then have for all linear transformations $T:\Bbb{R}^n\to\Bbb{R}^n$ that $$ tr(T)=\frac1{K_n}\int_{B}\langle Tx,x\rangle\,d\mu. $$ I will think of $T$ as a matrix (use the natural basis). We can equally well use the symmetrized version of $T$, i.e. $(T+T^t)/2$ because the quadratic form $\langle Tx,x\rangle$ stays the same, and the symmetrized version shares the same trace.

Given that $T$ is symmetric the rest is easy. By linear algebra there exists an orthogonal matrix $P$ such that $P^tTP=D$ is a diagonal matrix. Furthermore, $P^t=P^{-1}$ and $\det P=1$. Also, the linear substitution $x\mapsto Px$ preserves the unit ball, because $P$ is length preserving. What this means is that it suffices to prove the formula for a diagonal matrix $D=diag(d_1,d_2,\ldots,d_n)$.

Obviously $$I_i=\int_{B\subset\Bbb{R^n}}x_i^2\,d\mu=K_n$$ for all $i=1,2,\ldots,n$ by the symmetries of the sphere. By linearity of the integral we then get that $$ \begin{aligned} \int_{B\subset\Bbb{R^n}}\langle Dx,x\rangle\,d\mu &=\int_{B\subset\Bbb{R^n}}(d_1x_1^2+d_2x_2^2+\cdots+d_nx_n^2)\,d\mu\\ &=\sum_{i=1}^nd_i I_i\\ &=tr(D) K_n, \end{aligned} $$ and we are done.

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In other words, instead of the measure of the $n$-dimensional ball you should factor out the integral of $x_1^2$ over that ball. I'm sure the exact value of $K_n$ is known. May be a friendly physicist calculated the moments of inertia of the $n$-dimensional homogeneous ball? Or a probability person has calculated the expected value of $x_1^2$ of a random point uniformly distributed over that ball?

Answer 2

This is the kind of nice formula one wishes were true, but it's not a stated. Here's a simple counterexample:

Suppose

$n = 2, \tag 1$

and

$T = I, \tag 2$

the identity map. Then

$\text{Tr}(T) = \text{Tr}(I) = 2, \tag 3$

and $B$ is the unit disk, so

$\mu(B) = \pi, \tag 4$

whence

$\mu(B) \text{Tr}(T) = 2\pi; \tag 5$

for $\vec r = (x, y) \in B, \tag 6$

$\langle T\vec r, \vec r \rangle = x^2 + y^2 = r^2, \tag 7$

and

$\displaystyle \int_B r^2 d\mu = \int_B r^2\; r \; dr \; d\theta = \dfrac{2\pi \cdot 1^4}{4} = \dfrac{\pi}{2} \ne 2\pi, \tag 8$

as we see the formula fails for $n = 2$, $T = I$.

Answer 3

This is not quite true but it is true up to a scale factor. Let me first prove that it must be true up to a scale factor, and then figure out what the factor is. Let us write $$A(T)=\dfrac 1{\mu(B)}\int_{ B} \langle Tx, x\rangle d\mu(x).$$ Note that $A$ is linear, and is also invariant under conjugation by orthogonal matrices. Also, if $T$ is antisymmetric, then $$\langle Tx,x\rangle=\langle x,T^tx\rangle=\langle x,-Tx\rangle=-\langle Tx,x\rangle$$ for all $x$ so $A(T)=0$. Now fix a rank $1$ orthogonal projection $T_0$. Note that every symmetric matrix is a linear combination of conjugates of $T_0$ by orthogonal matrices, by the spectral theorem. Moreover, every matrix is a linear combination a symmetric matrix and an antisymmetric matrix by writing $T=\frac{T+T^t}{2}+\frac{T-T^t}{2}$. So, we can write an arbitrary matrix as a linear combination of antisymmetric matrices and conjugates of $T_0$ by orthogonal matrices.

Now, there exists a scalar $c$ such that $A(T_0)=c\operatorname{tr}(T_0)$ (in fact, $c=A(T_0)$ since $\operatorname{tr}(T_0)=1$). Since $A$ and $\operatorname{tr}$ both are linear and are preserved by conjugation by orthogonal matrices and vanish on antisymmetric matrices, it follows that $A(T)=c\operatorname{tr}(T)$ for all $T$.

To determine the constant $c$, we can just take $T=I$ to be the identity. We then have $$A(I)=\dfrac 1{\mu(B)}\int_{ B} \|x\|^2d\mu(x).$$ Letting $s$ denote the surface area of the unit $(n-1)$-sphere, we can compute the integral using polar coordinates: $$\int_{B} \|x\|^2d\mu(x)=\int_0^1 sr^2\cdot r^{n-1}\, dr=\frac{s}{n+2}.$$ On the other hand, we can also compute $\mu(B)$ from $s$ by polar coordinates to get $$\mu(B)=\int_0^1 s\cdot r^{n-1}\,dr=\frac{s}{n}.$$ Combining these results, we conclude that $A(I)=\frac{n}{n+2}.$ Since $\operatorname{tr}(I)=n$, this means $c=\frac{1}{n+2}$ and $$A(T)=\frac{1}{n+2}\operatorname{tr}(T)$$ for all $T$.

(To be ridiculously pedantic, several steps of this only work assuming $n>0$. Of course, the final conclusion is still true for $n=0$ since in that case the only possible value of $T$ is $0$.)