[講義] RMT2018 Spring Lectures (Percy A. Deift): Lecture 03

10-11

來自專欄隨機矩陣不隨機8 人贊了文章

Let $S_n = {M}$ denote the set of real n x n symmetric matrices. Suppose $M in S_n$ has eigenvalues ${lambda_j}_{j = 1}^n$ and associated eigenvectors $u(lambda_j) = (u_{1j}, cdots, u_{nj})^T$ .

Let $A_n = { M in S_n:$ for all eigenvectors $u=(u_1, cdots, u_n)^T$ of M, $u_1 e 0 }$

Claim 1:

$A_n$ is a dense, open set in $S_n$ of full measure, i.e., $meas(S_n ackslash A_n) = 0$ .

Note that

(42.1)

$A_n subset { M in S_n: lambda_j(M) e lambda_k(M), j e k }$

Indeed if $M in A_n$ and $lambda_j(M) = lambda_k(M) equiv lambda$ , then associated with $lambda$ there are at least two independent eigenvectors $M u(lambda_j) = lambda_j u(lambda_j) = lambda u(lambda_j)$ , $M u(lambda_k) = lambda_k u(lambda_k) = lambda u(lambda_k)$ , from which it follows that there is an eigenvector $u e 0$ , $Mu = lambda u$ , with $u(1) = 0$ . This is a contradiction and so (42.1) is true. $Box$

Let $C_n = A_n^prime = S_n ackslash A_n$

For an n x n matrix M, let $M_1$ denote the (n-1) x (n-1) matrix obtained by deleting the first row and column of M.

Claim 2:

$C_n = D_n = { M in S_n: M$ and $M_1$ have a common eigenvalue}

Clearly $C_n subset D_n$ for if $Mu = lambda u, u=(u_1, u_2, cdots, u_n)^T$ with $u_1 = 0$ , then $v = (u_2, cdots, u_n)^T e 0$ is an eigenvector for $M_1$ , $M_1 v = lambda v$ . Conversely, suppose that $M in D_n$ , but $M otin C_n$ . Let $lambda$ be a common eigenvalue of M and $M_1$ , then $(M - lambda) u = 0$ for some $u = (u_1, cdots, u_n)^T$ and as $M otin C_n$ , $u_1 e 0$ . Without loss, suppose $u_1 = -1$ . Write

$M - lambda = egin{pmatrix} a & b^T \ b & M_1 - lambda end{pmatrix}$

where $a in mathbb{R}$ and $b in mathbb{R}^{n-1}$ . As $(M - lambda) u = 0$ , we have in particular,

$b = (M_1 - lambda) egin{pmatrix} u_2 \ vdots \ u_n end{pmatrix}$

and so

$b^T = (u_2, cdots, u_n) (M_1 - lambda)^T$

$= (u_2, cdots, u_n) (M_1 - lambda)$

Now $exists w = (w_2, cdots, w_n)^T e 0$ such that

$(M_1 - lambda) w = 0$

and so

$b^T w = 0$

But then

$(M - lambda) egin{pmatrix} 0 \ w end{pmatrix} = egin{pmatrix} a & b^T \ b & M_1 - lambda end{pmatrix} egin{pmatrix} 0 \ w end{pmatrix} = 0$

and so M has an eigenvector $u = egin{pmatrix} 0 \ w end{pmatrix}$ with $u_1 = 0$ . This is a contradiction and so $D_n subset C_n$ . $Box$

Note: By the proof of Claim 2, we in fact see that if $M in D_n$ , then M and $M_1$ have a "common" eigenvector i.e., an eigenvector $w = (w_2, cdots, w_n)^T$ for $M_1$ , $M_1 w = lambda w$ , such that $v = (0, w_2, cdots, w_n)$ is an eigenvector for M, $Mv = lambda v$ . Note however that not every eigenvector $w$ of $M_1$ has the property that $v = egin{pmatrix} 0 \ w end{pmatrix}$ is an eigenvector of M.

For example

if $M = egin{pmatrix} a & b & c \ b & 0 & 0 \ c & 0 & 0 end{pmatrix}$ , $b^2 + c^2 e 0$ , then

$M_1 = egin{pmatrix} 0 & 0 \ 0 & 0 end{pmatrix}$ and $egin{pmatrix} u_2 \ u_3 end{pmatrix}$ is an eigenvector of $M_1$ . However $egin{pmatrix} 0 \ u_2 \ u_3 end{pmatrix}$ is an eigenvector of M if and only if $egin{pmatrix} u_2 \ u_3 end{pmatrix}$ is an multiple of $egin{pmatrix} -c \ b end{pmatrix}$

We now show that $D_n$ has measure 0. Let

$p(lambda) = det(lambda - M)$

$q(lambda) = det(lambda - M_1)$

Write

$p(lambda) = lambda^n + p_{n-1} lambda^{n-1} + cdots + p_0$

$q(lambda) = lambda^{n-1} + q_{n-2} lambda^{n-2} + cdots + q_0$

and consider the resultant R of p and q

$R = detegin{pmatrix} p_0 & p_1 & cdots & p_{n-1} & 1 & & & \ & p_0 & cdots & & p_{n-1} & 1 & \ & ddots & & & & & & \ & & & p_0 & cdots & & & 1 \ q_0 & q_1 & cdots & q_{n-2} & 1 & cdots & & \ & q_0 & cdots & & & 1 & \ & ddots & & & & & & \ & & q_0 & & & & & 1 end{pmatrix}$

where $(p_0, cdots, p_{n-1}, 1)$ is repeated n-1 times and $(q_0, cdots, q_{n-2}, 1)$ is repeated n times. Clearly R is a determinant of size n+(n-1) = 2n-1. By standard theory (exercise) for any 2 polynomials p and q

$R = 0 Leftrightarrow p(lambda)$ and $q(lambda)$ have a common root

Now R = R(M) is a real analytic function (in fact a polynomial) in the entries of M and hence if it vanishes on a set of positive measure in $mathbb{R}^{frac{n(n+1)}{2}} approxeq S_n$ , it is identically zero (exercise). But

$M = egin{pmatrix} 1 & 1 & & cdots & 1 \ 1 & 2 & & & \ & & 3 & & \ & & & ddots & \ 1 & & & & n end{pmatrix}$

does not have a common eigenvalue with $M_1 = diag(2, 3, cdots, n)$ , specially $M_1 = {2, 3, cdots, n}$ . Indeed, if $Mu = ju$ , $u = (u_1, cdots, u_n)^T e 0$ , for $2 le j le n$ , then $0 = (e_j, (M - j)u) = u_1$ . But the $(k - j)u_k = 0$ , $2 le k le n$ , $k e j$ , so $u = (0, cdots, 0, u_j, 0, cdots, 0)^T$ . But then from the first row of (M - j) we see that $u_j = 0$ . Thus $R otequiv 0$ . Hence we must have $meas(D_n) = 0$ .

example

Suppose

$p(z) = p_0 + p_1 z + z^2$

$q(z) = q_0 + z$

2n-1 = 4 - 1 = 3

Then

$R = det egin{pmatrix} p_0 & p_1 & 1 \ q_0 & 1 & 0 \ 0 & q_0 & 1 end{pmatrix} = p_0 - p_1q_0 + q_0^2$

so if $z = -q_0$ is a root of $q$ and also of $p(z) = p_0 + p_1(- q_0) + (-q_0)^2$ , we see that R = 0.

Exercise:

Generalize R(p, q) for any 2 polynomials of arbitrary order.

Exercise:

Instead of R, we can consider associated tensors.

For a polynomial $p(z) = p_0 + p_1 z + cdots + p_{n-1}z^{n-1} + z^n$ , the companion matrix for p is defined as follows,

$C_p = egin{pmatrix} 0 & & cdots & 0 & -p_0 \ 1 & 0 & cdots & 0 & -p_1 \ 0 & 1 & & & \ & ddots & & & vdots \ & 0 & & 1 & -p_{n-1} end{pmatrix}$ , $C_p$ is n x n.

Then (exercise) p(z) = 0 if and only if z is an eigenvalue of $C_p$ , i.e., $det(z - C_p) = 0$ .

If $q(z) = q_0 + cdots + q_{m-1} z^{m-1} + z^m$ is a second polynomial, with companion matrix $C_q$ , $C_q$ is m x m, set

$A = C_p otimes I_m - I_n otimes C_q$

Show that

$det A = prod_{1 le i le n \ 1 le j le m} (lambda_i - mu_j)$

and so det A = 0 $Leftrightarrow$ p and q have a common root.

If $p(z) = det(z - M)$ , $q(z) = det(z - M_1)$ , we see that det A is a polynomial in the entries of M, and the argument proceeds as before for R.

Finally, we show that $A_n$ is dense and open, which completes the proof of Claim 1.

If $M_0 in A_n$ , then $M_0$ has distinct spectrum $lambda_1 < lambda_2 < cdots < lambda_n$ (see Claim 1) and $u_1 e 0$ for all eigenvectors $u = (u_1, cdots, u_n)^T$ of M. But as the spectrum of $M_0$ is simple it follows by standard spectral theory that the eigenvalues and eigenvectors are continuous for M in a neighborhood of M. This shows that $A_n$ is open. On the other hand, as $meas(A_n^prime) = 0$ , $A_n$ is certainly dense. We are done. $Box$

It follows from the above calculations that the spectral theorem

$M = O Lambda O^T$ , $Lambda = diag(lambda_1, cdots, lambda_N)^T$

induces a well-defined and smooth map

$varphi: M mapsto (lambda_1 < cdots < lambda_N, O)$

from $A_N$ into $Lambda_N^T imes O_1^T$ ,

where $O_1^T$ is the self-orthogonal matrices whose columns $o_j = Oe_j$ have positive first entries, $o_j(1) > 0, j = 1, cdots, N$ ,

and where $Lambda_N^T$ is the vector in $mathbb{R}^N$

${ lambda_1 < cdots < lambda_N }$ .

In fact $varphi$ is a bijection with a smooth inverse $varphi^{-1}$ : Indeed if $varphi(M) = varphi( ilde{M})$

then $ilde{lambda_i} = lambda_i, i = 1, cdots, N$ and $ilde{O} = O$ and

$M = O Lambda O^T = ilde{O} ilde{Lambda} ilde{O}^T = ilde{M}$

so $varphi$ is $1 ightarrow 1$ . Also if $Lambda = { lambda_1 < cdots < lambda_N } in Lambda_N^T$ and $O in O_1^T$ , then $M equiv O Lambda O^T$ is a real symmetric matrix with simple spectrum.

Moreover the column of O are clearly the eigenvectors of M and $Oe_j(1) e 0$ . Hence $M in A_n$ and clearly $varphi(M) = (Lambda, O)$ . This $varphi$ is a bijection. Clearly the inverse of $varphi$ is given by

$varphi^{-1}: (Lambda, O) ightarrow O Lambda O^T$

and $varphi^{-1}$ is smooth.

Now as $A_n$ is an open set of full measure, we can use $varphi$ as a change of variables to compute probabilities. The first order of business is to compute the Jacobian

$det(frac{partial(M)}{partial(Lambda, O)})$

on $A_n$ .

Fix $M_0 in A_n$ and let $(Lambda_0, O_0) = phi(M_0)$

Let $p = (p_1, cdots, p_l)$ , $l = frac{N(N-1)}{2}$ , be local co-coordinates on $O_1^T$ in a small neighborhood of $O_0$ , $sum_{i=1}^l p_i^2 < epsilon^2$

where

$(p_1, cdots, p_l) mapsto O(p_1, cdots, p_l)$

$(0, cdots, 0) mapsto O_0$

For sufficiently small $epsilon > 0$ , $lambda_1, cdots, lambda_N, p_1, cdots, p_l$ with

$lambda_1 < cdots < lambda_N, sum_{i=1}^l p_i^2 < epsilon^2$

are co-coordinates for an open neighborhood of $M_0$

$O_{M_0} = varphi^{-1}( (lambda_1 < cdots < lambda_N), sum_{i=1}^l p_i^2 < epsilon^2)$

$= { M = M(Lambda, p) = O(p_1, cdots, p_l) Lambda O^T(p_1, cdots, p_l), lambda_1 < cdots < lambda_N, sum_{i=1}^l p_i^2 < epsilon^2 }$

Differentiating $M = M(Lambda, p) = O(p) Lambda O(p)^T$ w.r.t. $p_j$ ,

we find

$M_{p_j} = O_{p_j} Lambda O^T + O Lambda O_{p_j}^T, 1 le j le l$ .

But $O^TO =I$ . Hence $O^T_{p_j} O + O^T O_{p_j} = 0$ and so

(53.1)

$S_j = O^T O_{p_j}$

is skew-symmetric. It follows that for $1 le j le l$

(53.2)

$O^T M_{p_j} O = S_j Lambda - Lambda S_j = [S_j, Lambda]$

Similarly,

(53.3)

$O^T M_{lambda_j} O = Lambda_{lambda_j}, j = 1, cdots, l$

Consider the map

(53.4)

$V_O: C mapsto V_O(C) = O^TCO$

mapping real symmetric matrices to real symmetric matrices. Let f denote the bijection of real symmetric N x N matrices onto $mathbb{R}^{N(N+1)/2}$ given by (cf $M mapsto ilde{M}$ , 8.29)

$f: M mapsto (M_{11}, cdots, M_{N, N}, M_{12}, M_{13}, cdots, M_{N-1, N})^T equiv vec{M} = f(M)$

Define the inner product on $mathbb{R}^{N(N+1)/2}$ as before by

(54.1)

$(vec{M}, vec{M})_{tr} equiv tr M^2$

where $vec{M} = f(M)$ . We find for $M = M^T = ar{M}$ ,

$(vec{V_OM}, vec{V_OM})_{tr} = tr (V_OM)^2 = tr(O^TM^2O) = tr M^2 = (vec{M}, vec{M})_{tr}$

It follows that

(54.2)

$vec{V_O}(vec{M}) = vec{V_O(M)}$

where $vec{M} = f(M)$ is orthogonal and as before

(54.3)

$det vec{V_O} = pm1$

More precisely, if the matrix $T_O$ represents $vec{V_O}$ in an orthonormal basis ${e_i}_{i=1}^{frac{N(N+1)}{2}}$ for $(mathbb{R}^{frac{N(N+1)}{2}}, (cdot, cdot)_{tr})$ , i.e., $hat{V_O}e_j = sum_{i=1}^{N(N+1)/2} (T_O)_{ij}e_i$ , then $T_O$ is orthogonal and so $det T_O = pm1$ . Now (53.2) (53.3) can be written in the form

(54.4)

$V_O(M_{lambda_1}, cdots,M_{lambda_N}, M_{p_1}, cdots, M_{p_l}) = (Lambda_{lambda_1}, cdots, Lambda_{lambda_N}, [S_1, Lambda], cdots, [S_l, Lambda])$

where the elements on the RHS are N x N matrices,

(55.1)

$vec{V}_O(vec{M}_{lambda_1}, cdots,vec{M}_{lambda_N}, vec{M}_{p_1}, cdots, vec{M}_{p_l}) = (vec{Lambda}_{lambda_1}, cdots, vec{Lambda}_{lambda_N}, [S_1, vec{Lambda}], cdots, [S_l, vec{Lambda}])$

where the elements on the RHS are column vectors of size N(N+1)/2.

Hence by (54.3) and the representation $vec{V}_O(vec{M}) = T_O (vec{M})$ ,

(55.2)

$lvert det (vec{M}_{lambda_1}, cdots, vec{M}_{lambda_N}, vec{M}_{p_1}, cdots, vec{M}_{p_l}) vert = lvert det (vec{Lambda}_{lambda_1}, cdots, vec{Lambda}_{lambda_N}, [S_1, vec{Lambda}], cdots, [S_l, vec{Lambda}]) vert$

Now,

(55.3)

$vec{Lambda}_{lambda_j} = (0, cdots, 0, 1, 0, cdots, 0)^T in mathbb{R}^{N(N+1)/2}$

where 1 is at the $j^{th}$ place. Also $[S_q, Lambda]_{ij} = (lambda_j - lambda_i)(S_q)_{ij}$ for $q = 1, cdots, l$ and so for $1 le q le l$

(55.4)

$[S_q, vec{Lambda}] = (0, cdots, 0, (lambda_2 - lambda_1)(S_q)_{12}, (lambda_3 - lambda_1)(S_q)_{13}, cdots, (lambda_N - lambda_{N-1})(S_q)_{N-1, N})$

with 0s in the first N entries. Thus graphically we have

(56.0)

$(vec{Lambda}_{lambda_1}, cdots, vec{Lambda}_{lambda_N}, [S_1, vec{Lambda}], cdots, [S_q, vec{Lambda}]) = left(egin{array}{c|c} I_N & O \ hline O & X end{array} ight)$

where X denotes the $frac{N(N-1)}{2} imes frac{N(N-1)}{2}$ matrix

(56.1)

$egin{pmatrix} (lambda_2 - lambda_1)(S_1)_{12} & (lambda_2 - lambda_1)(S_2)_{12} & cdots & (lambda_2 - lambda_1)(S_l)_{12} \ (lambda_3 - lambda_1)(S_1)_{13} & (lambda_3 - lambda_1)(S_2)_{13} & cdots & (lambda_3 - lambda_1)(S_l)_{13} \ vdots & vdots & ddots & vdots \ (lambda_N - lambda_{N-1})(S_1)_{N-1, N} & (lambda_N - lambda_{N-1})(S_2)_{N-1, N} & cdots & (lambda_N - lambda_{N-1})(S_l)_{N-1, N} end{pmatrix}$

ans so

$lvert det X vert = prod_{1 le i < j le N} lvert lambda_i - lambda_j vert f_1(O)$

which implies by (55.2) (56.0)

(56.2)

$lvert det frac{partial vec{M}}{partial (Lambda, O)} vert = prod_{1 le i < j le N} lvert lambda_i - lambda_j vert f_1(O)$

As $varphi$ and $varphi^{-1}$ are smooth, $f_1(O) > 0$ (why?)