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Suppose F(A) is a quadratic function of a real symmetric matrix, A. This means that there are numbers $f_{ijkl}$ so that F(A) = $\sum_{ijkl}f_{ijkl}a_{ij}a_{kl}$.

Suppose that $F(A) = F(QAQ^t)$ for every orthogonal matrix, Q. Show that there are numbers c and d so that $F(A) = cTr(A^2) + d(Tr(A))^2$. Here, Tr(A) is the trace of A.

Edited work:

Since A is symmetric, then we know that A is orthogonally diagonalizable, so that there's an orthogonal matrix Q such that $QAQ^t$ = D, where D is a diagonal matrix with the eigenvalues of A on the main diagonal.

Using the assumption that F is invariant under all orthogonal similarity transformations, including (of course) the transformations that diagonalize A, I have that:

$$F(A) = F(QAQ^t)$$

$$= F(D)$$ $$=\sum_{i,j,k,l} f_{i,j,k,l}d_{ij}d_{kl}$$ $$=\sum_{i,k} f_{iikk}d_{ii}d_{kk}$$ $$= F(Q_1DQ_1^t)$$ $$= F(Q_2DQ_2^t)$$ $$= F(Q_3DQ_3^t)$$ $$....$$

$$=\sum_{i,k} f_{iikk}d_{ii}d_{kk},$$

where $Q_i$ is an orthogonal, permutation matrix, so that $Q_iDQ_i^t$ is swapping the eigenvalues on the diagonal, resulting in a permuted diagonal matrix, still with the eigenvalues of A on the main diagonal.

Now, the problem reduces to proving the equation for diagonal matrices.

As whacka stated in his answer below, considering only the permuted, diagonal matrices, then F defines a quadratic form in the variables $\lambda_i$, for $1\le i \le n$.

So, we have $$ F(\lambda_1, ... \lambda_n) = \sum_{i,k} f_{iikk}\lambda_i \lambda_k $$

$$ = \sum_{i=k} \alpha_i (\lambda_i)^2 + \sum_{i,k} \beta_{ik}\lambda_i \lambda_k $$

$$ = \sum_{i=k} \alpha_i (\lambda_i)^2 + \sum_i \sum_k \beta_{ik}\lambda_i \lambda_k$$

$$ ?? = \sum_{i=k} \alpha_i (\lambda_i)^2 + \beta_{ik} (\sum_i\ \lambda_i\sum_k \lambda_k) $$

$$ = \alpha_i Tr(A^2) + \beta_{i,k}(Tr(A))^2$$

$$=F(A)$$

And, since F is invariant under any permutation of the eigenvalues on the diagonal, then this quadratic polynomial is also invariant; hence, the coefficients $\alpha_i$ and $\beta_{i,k}$ exist, are well-defined, and unique.

How is my proof? I don't feel confident about the equality that I labeled (??). But somehow, I have got to make that number, $\beta_{i,k}$ not dependent on the indices, in order to pull it outside of the summation, so that I can get my $(tr(A))^2$.

Any hints or suggestions are welcome and greatly appreciated.

Thanks,

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    $\begingroup$ There's still a lot more that can be deduced about $F$. By choosing $Q$ to be a permutation matrix, you should be able to show that many of the values of $f_{ij}$ must be equal to one another. $\endgroup$ – Erick Wong Aug 13 '15 at 23:53
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    $\begingroup$ The simplest example of a permutation matrix is one that swaps a single pair of entries. Typically these types are all you ever need to consider (since the full permutation group is generated by all the swaps). The goal at this stage is to extract more information about $F$ from the (very powerful) fact that it is invariant under ALL orthogonal transformations, not just the ones that diagonalize $A$. Also, note that applying a permutation matrix to a diagonal matrix preserves diagonality. $\endgroup$ – Erick Wong Aug 14 '15 at 0:27
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    $\begingroup$ I agree the diagonalizing matrix is still very useful for easily relating $\text{Tr}(A)$ and $\text{Tr}(A^2)$ to $\text{Tr}(D)$ and $\text{Tr}(D^2)$ without any tedious calculation. But right now you need to deal with the large number of independent variables $f_{ij}$, which can be trimmed down to a very small number (you need to do this because the final result is expected to have only two parameters). $\endgroup$ – Erick Wong Aug 14 '15 at 0:32
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    $\begingroup$ A quadratic form looks like $\sum e_{ij}\lambda_i\lambda_j$. We can split this up into two parts, $\sum e_{ii}\lambda_i^2$ and $\sum_{i\ne j} e_{ij}\lambda_i\lambda_j$. And then the second we can split further up into $\sum_{i<j}c_{ij}\lambda_i\lambda_j$ and $\sum_{j<i}c_{ij}\lambda_i\lambda_j$. The latter may be written instead as $\sum_{i<j}c_{ji}\lambda_j\lambda_i$. But $\lambda_i\lambda_j=\lambda_i\lambda_j$, so we may collect terms and write $\sum_i c_{ii}\lambda_i^2+\sum_{i<j}(c_{ij}+c_{ji})\lambda_i\lambda_j$. Even simpler is $\sum_i a_i\lambda_i^2+\sum_{i<j}b_{ij}\lambda_i\lambda_j$. $\endgroup$ – whacka Aug 14 '15 at 6:24
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    $\begingroup$ I do believe you would be able to see what's going if you see past the summation symbols. For instance, $e_{11} \lambda_1 \lambda_1+e_{12}\lambda_1\lambda_2+e_{21}\lambda_2\lambda_1+e\lambda_2\lambda_2$ you would know can be written much more simply as $\square \lambda_1^2+\square\lambda_2^2+\square\lambda_1\lambda_2$. You would see it even quicker if we wrote, say, $axx+bxy+cyx+dyy$ - you would know a simpler form for this is $ax^2+by^2+cxy$. Try it for $n=3$ with variables $x,y,z$ and coefficients $a,b,c,\cdots$ and see if you can see how this generalizes to what I said in my above comment. $\endgroup$ – whacka Aug 14 '15 at 6:27
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You've shown it suffices to prove $\exists c,d\in\Bbb R$ s.t. $F(A)$ is of the form $c\,{\rm tr}(A^2)+d\,{\rm tr}(A)^2$ for all diagonal matrices $A$, as $F$ and ${\rm tr}$ are both ${\rm O}_n$-invariant (the action being conjugation).

One may identify diagonal $n\times n$ real matrices with $\Bbb R^n$. A quadratic function $F:\Bbb R^n\to\Bbb R$ must be of the form $F(\lambda_1,\cdots,\lambda_n)=\sum_ia_i\lambda_i^2+\sum_{j<k} b_{jk}\lambda_j\lambda_k$. Two polynomials define the same function of $\Bbb R$ iff they are the same polynomial, two polynomials are identical iff their monomials have the same coefficients, and $F$ defines the same function for any permutation of its variables.

The key here is that $S_n$ acts transitively and $2$-transitively (in the language of group actions), so that we may permute all of the $a$ coefficients in front of the $\lambda_i^2$ terms, and permute any two $b$ coefficients in front of the $\lambda_i\lambda_j$ terms. If you are unfamiliar with the concept of a group action, you should still be able to take all this in stride I think.

Keeping in line with the language of group actions, the contragredient action of $S_n$ on functions is actually given by $(\sigma\cdot F)(\lambda_1,\cdots,\lambda_n)=F(\lambda_{\sigma^{-1}(1)},\cdots,\lambda_{\sigma^{-1}(n)})$ (the inverses are necessary for this to define a left action instead of a right action, and it has a sensible explanation in terms of graphs of functions). Explicitly writing out the polynomials, we have

$$\begin{array}{ll} \displaystyle \sum_i a_i\lambda_i^2+\sum_{j<k}b_{jk}\lambda_j\lambda_k & \displaystyle = \sum_i a_i\lambda_{\sigma^{-1}(i)}^2+\sum_{j<k}b_{jk}\lambda_{\sigma^{-1}(j)\sigma^{-1}(k)} \\ & \displaystyle = \sum_i a_{\sigma(i)}\lambda_i^2+\sum_{j<k}b_{\sigma(j)\sigma(k)}\lambda_j\lambda_k \end{array} $$

We must interpret $b_{\bullet\bullet}$ as symmetric to make sense of the above when $\sigma(j)>\sigma(k)$. Note the substitution of index that implicitly occurred. If we identify the coefficients of $\lambda_i^2$ above we get $a_i=a_{\sigma(i)}$ regardless of which $\sigma$ we picked, hence all $a$s are equal, and similarly equating coefficients of $\lambda_i\lambda_j$ yields $b_{jk}=b_{\sigma(j)\sigma(k)}$, hence all $b$s are equal: given any two pairs $j<k$ and $l<m$ there is a $\sigma\in S_n$ s.t. $\sigma(j)=l,\sigma(k)=m$ (this is what $2$-transitivity speaks to).

Now, given $F(\lambda_1,\cdots,\lambda_n)=a\left(\sum_i\lambda_i^2\right)+b\left(\sum_{i<j}\lambda_i\lambda_j\right)$ for some $a,b\in\Bbb R$, can you conclude from this that it's of the form $F(\lambda_1,\cdots,\lambda_n)=c\left(\sum_i\lambda_i^2\right)+d\left(\sum_j\lambda_j\right)^2$ for some $c,d\in\Bbb R$?

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    $\begingroup$ Yes you have it right so far. (1) Nobody writes e.g. $xy+yx$ on paper in a commutative setting; we simply write $2xy$. Same idea with any number of variables; we have $\lambda_i\lambda_j=\lambda_j\lambda_i$ so there's no reason to include that monomial twice. (2) The numbers $a$ and $b$ can be anything in general, because $a(\cdots)+b(\cdots)$ is a quadratic form no matter what $a$ and $b$ are. $\endgroup$ – whacka Aug 14 '15 at 6:17
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    $\begingroup$ @LebronJames It is not justified. Your $\beta_{ik}(\sum_i \lambda_i\sum_k\lambda_k)$ doesn't even make sense. Nor does $\alpha_i\,{\rm Tr}(A^2)$. The correct idea is that if all of the $b_{ij}$s are equal, you can simply use a single letter $b$ to denote the value, in which case $\sum_{i<j} b\lambda_i\lambda_j=b\sum_{i<j}\lambda_i\lambda_j$ by linearity. $\endgroup$ – whacka Aug 14 '15 at 6:30
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    $\begingroup$ Read the very last line in my answer. First you'll want to go from the desired form $c(\cdots)+d(\cdots)^2$ to the known form $a(\cdots)+b(\cdots)$, and from there figure out how to reverse the process. What happens if you expand the $(\sum \lambda_i)^2$ that appears in the desired form? $\endgroup$ – whacka Aug 14 '15 at 6:49
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    $\begingroup$ It's hard to tell what you're saying. If you mean $$c\left(\sum_i\lambda_i^2\right)+d\left(\sum_i\lambda_i\right)^2=\underbrace{ (c+d)}_a \left( \sum_i\lambda_i^2\right)+\underbrace{(2d)}_b \left(\sum_{j<k}\lambda_j\lambda_k\right)$$ then yes. Equating $a=c+d,b=2d$ and solving $d=b/2,c=a-b/2$ tells us that $$a\left(\sum_i\lambda_i\right)+b\left(\sum_{i<j}\lambda_i\lambda_j\right)=(a- \frac{1}{2}b)\left(\sum_i\lambda_i^2\right)+\frac{b}{2}\left(\sum_i\lambda_i \right)^2.$$ $\endgroup$ – whacka Aug 14 '15 at 7:38
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    $\begingroup$ I finally, finally got it, @whacka, and I can finally leave the library and go to bed. Thank you so, so, so much for your help and for your patience. Oh, gosh, what a great learning experience. I really appreciate it. Have a great night!! Thanks again!! $\endgroup$ – User001 Aug 14 '15 at 8:15

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