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Let $\mathcal{M}$ be the set all symmetric $n \times n$ real matrices. Let $\lambda$ be a real number and let $m$ be a nonnegative integer. How is the set $\mathcal{W}$ of all symmetric $n \times n$ real matrices with $\lambda$ as an eigenvalue of multiplicity $m$ a smooth submanifold of $\mathcal{M}$?

Also, why does $\mathcal{W}$ have codimension $m(m+1)/2$?

I'm having trouble finding a nondegenerate system of equations. Can I avoid that with the following approach?

Let $\mathcal W' = \{W - \lambda I\ : W \in \mathcal W\}$. Then the eigenvalues of each $W - \lambda I$ are the eigenvalues of $W$ shifted by $-\lambda$. So $0$ is an eigenvalue of multiplicity $m$ for each matrix in $\mathcal W'$, and it follows that each matrix in $\mathcal W'$ has rank $n - m$.

The set of all $n \times n$ matrices with rank $n - m$ forms a smooth manifold (see Proposition 1.14 on page 133 of Optimization and Dynamical Systems by Helmke and Moore). Is there an easy way to show that $\mathcal W'$ is a submanifold of that manifold?

SorTheene
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    You can deal with $\lambda \neq 0$ the same way by just "transporting" your manifold to be around the origin, by subtracting $\lambda I$. (That said, I don't see why your situation implies $W$ is open in $M$.) – Ian Mar 08 '15 at 03:27
  • @Ian Thanks. Since each matrix $A$ has rank $k := m-n$, there is a $k \times k$ submatrix of $A$ whose determinant is nonzero. And since the determinant function is continuous, picking a small interval around $k$ not containing zero will give you an open around $A$ contained in $\mathcal{W}$. – SorTheene Mar 08 '15 at 05:42
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    Yes, but there should be other conditions on determinants of submatrices to have rank exactly $k$ (all determinants of smaller order should be zero). It's not an open property. – Evgeny Mar 08 '15 at 05:46
  • @Evgeny So is $\mathcal{W}$ not open when $\lambda = 0$? – SorTheene Mar 08 '15 at 05:50
  • As far as I'm concerned it's not. – Evgeny Mar 08 '15 at 05:52
  • @Evgeny Thanks. Do you know how $\mathcal{W}$ is a submanifold of $\mathcal{M}$? – SorTheene Mar 08 '15 at 05:53
  • Can you find the conditions on coefficients of matrix that make vanish coefficients of characteristic poly up to some order? This would be a system of equations for matrix entries, and systems of equations is also a way to describe submanifold. – Evgeny Mar 08 '15 at 05:59
  • @Evgeny This is just stated in a paper I'm reading. It says "consider the submanifold" $\mathcal{W} \subset M$, as described above. That's all I've got. – SorTheene Mar 08 '15 at 06:02
  • You know, the proof 1.15 from this book does it's job very well and I'd rather stick to it. I had almost the same idea as theirs but I'm not used to Lie groups techniques and couldn't write proof out of my head. – Evgeny Mar 14 '15 at 11:03
  • @Evgeny Can that proof be adapted to the set of symmetric $n \times n$ matrices of fixed rank? (I don't know much about Lie groups, either.) – SorTheene Mar 14 '15 at 14:12
  • Yes, I think that proof could be adjusted. You just have to change the group action. More precisely, each symmetric matrix from $\mathbb{R}^{n \times n}$ could be diagonalized via orthogonal change of basis. Consider the action of $O(n)$ on set of symmetric matrices via conjugation: matrix $R \in O(n)$ sends symmetric matrix $S$ to $R^{T} S R$. Then you can choose orbit space of this action equal to set of diagonal normal forms for symmetric matrices. However, I still have problems with calculating dimension. – Evgeny Mar 15 '15 at 07:49

1 Answers1

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As I've already mentioned, set of (symmetric) matrices of rank $k$ is not open. It's quite easy to construct counterexample: take a $n \times n$ diagonal matrix with exactly $k$ non-zero diagonal elements. If this set was open, then there would exist a small neighbourhood that consists of rank-$k$ matrices. But you can add arbitrary small perturbations to other diagonal elements of matrix and still change rank.

By $\chi_{M}$ let's denote the characteristic poly of matrix $M$, $$\chi_M(\lambda) = a_0 \lambda^n + a_1 \lambda^{n-1} + \dots + a_{n-1} \lambda + a_n .$$ It's well known that coefficients of characteristic poly are polynomial functions of matrix entries. For example, in case of $2 \times 2$ matrix characteristic equation can be written as $$(-1)^2 \lambda^2 + (-1)^1 \, {\rm tr}\, M \cdot \lambda^1 + (-1)^0 \det M \cdot \lambda^0 = 0.$$

Let's first consider the case when we're looking for conditions for $\lambda = 0$ being the root of multiplicity exactly $m$. Then the condition is simple: $a_{n-p} = 0$ for $0 < p < m$ and $a_{n-m} \neq 0$. If we remove the last condition then we'll have a set of matrices with root $\lambda = 0$ of multiplicity at least $m$.

The general idea behind saying that this set is a submanifold is simple. The first part of conditions defines system of $m$ equations. If Jacobi matrix of this system has full rank in each point, then the set of points which satisfy system is a submanifold of dimension $n-m$. However, singularities (points, where Jacobi matrix doesn't have full rank) might occur in systems of equations.

A quick example that illustrates the idea. Consider the set of symmetric $2 \times 2 $ matrices. Let's find how the set of matrices with simple root $\lambda = 0$ looks like. You can write any symmetric $2 \times 2$ matrix as

$$ \left ( \begin{array}{cc} x && y \\ y && z \end{array} \right )$$

In order to have simple root $\lambda = 0$ this matrix must have $\det = 0$ and ${\rm tr} \neq 0$, i.e. $xz - y^2 = 0$ and $x + z \neq 0$. We have a "system" of 1 equation; in this case Jacobi matrix is just a gradient:

$$ \frac{Da_2}{D(x,y,z)} = \left ( \begin{array}{c} z \\ -2y \\ x \end{array} \right ) $$

Note that this matrix has full rank iff any of its entry is non-zero. So, the only point of $f(x,y,z) = xz - y^2$ where Jacobi matrix doesn't have full rank is the point $(0,0,0)$. That's where the singularity occurs. If you visualize this set in $3D$ (for example, here ), you will see that this surface is a cone and that point $(0,0,0)$ is an apex of this cone. So, strictly speaking, set $xz - y^2$ is not a submanifold of $\mathbb{R}^3$ — the point $(0,0,0)$ messes everything up, it has no neighbourhood that is homeomorphic to disk in $\mathbb{R}^2$. It's time to use second condition, $x + z \neq 0$ : it removes the singularity and now we have a true submanifold of $\mathbb{R}^3$. However, now it consists of two disjoint parts.

I hope that this will make things clear. The main idea is a fact that non-degenerate system of equations (i.e., which has full-rank Jacobian in each point) defines a submanifold. The last condition $a_{n-m} \neq 0$ might be crucial for removing singularities in this problem (I haven't proven rigorously that it's so, but I think it might be true - I'll be glad if someone will point me to the counterexample).

Evgeny
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  • Why does $a_{n-p} = 0$ for $0 < p < m$ and $a_{n-m} \not= 0$ imply that $\lambda$ has multiplicity $m$? – SorTheene Mar 08 '15 at 20:53
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    If this condition holds, then characteristic poly equals $a_0 \lambda^n + a_1 \lambda^{n-1} + \dots + a_{n-m} \lambda^m = \lambda^m (a_0 \lambda^{n-m} + a_1 \lambda^{n-m-1} + \dots + a_{n-m})$ and $\lambda = 0$ is clearly a root of multiplicity $m$. For $\lambda^\ast \neq 0$ you might use condition $\chi_M (\lambda^\ast) = \chi_M^\prime (\lambda^\ast) = \dots = \chi_M^{(m-1)} (\lambda^\ast) = 0$ and $\chi_M^{(m)} \neq 0$. – Evgeny Mar 09 '15 at 03:55
  • Thanks, that helps a lot. Could you check what I've added to the question? – SorTheene Mar 10 '15 at 21:37
  • I'm not sure that it'll be easy in general case to go the way I've showed. To some extent it was a simple demonstration i low dimension. However, since coefficients of characteristic polynomial depend only on matrix minors, you might try to find formulas for their (in Matrix Cookbook, for example). – Evgeny Mar 11 '15 at 07:22