## Invariant under similarity

In this post I construct some quantities on symmetric matrices which are invariant under similarity. I don’t know if this is well-known in linear algebra, perhaps Zorn Leung will know about this.

Let ${A}$ be a ${n \times n}$ matrices. The determinant of ${A}$ is defined to be

$\displaystyle \det A=\sum_{\sigma \in S_n}(-1)^{sign(\sigma)}A_{1 \sigma(1)}A_{2 \sigma(2)}\cdots A_{n \sigma (n)}$

(${S_n}$ is the permutation group of ${n}$ numbers) and its trace is defined to be

$\displaystyle \displaystyle tr (A)=\sum_{i=1}^n A_{ii}.$

It is well-known (especially the former one) that the determinant and the trace of a ${n \times n}$ matrices under similarity. i.e. for any invertible ${P\in M_n}$, ${\det (P^{-1}AP)=\det A}$ and ${tr(P^{-1}AP)=tr(A)}$. The later identity has an easy proof: first note that ${tr(AB)=tr(BA)}$, apply this to ${P^{-1}}$ and ${AP}$, we get ${tr(P^{-1}(AP))=tr(AP(P^{-1}))=tr(A)}$.

Several months ago I was thinking about the geometry of rank ${k}$ symmetric ${n \times n}$ matrices and which led me to think about which function “characterizes” those matrices. Consequently, after discussing with Zorn Leung and Bean, I come up with some invariants which are quite interesting themselves.

The idea is actually simple. For a symmetric matrix ${A}$, it is of rank ${k=n-l}$ if and only if the characteristic polynomial ${p(t)=\det (tI-A)}$ has 0 as a root of multiplicity ${l}$ (this is not true in general when ${A}$ is not symmetric, e.g. ${A=\begin{pmatrix} 0 &1\\ 0 &0 \end{pmatrix} }$) and a polynomial ${p(t)=t^n+a_{n-1}t^{n-1}+\cdots a_1 t+ a_0}$ has 0 as a root of multiplicity ${l}$ if and only if ${a_0=a_1=\cdots a_{l-1}=0}$. So we are led to find those coefficient ${a_i}$ of the characteristic polynomial ${p(t)}$. Two of them are easy to spot: ${a_0=(-1)^n \det A}$ and ${a_{n-1}=-tr (A)}$, i.e.

$\displaystyle \det(tI-A)=t^n -tr(A)t^{n-1}+\cdots +(-1)^n \det (A)$

This can be seen by observing the power of ${t^{n-1}}$ in the expression ${\det(tI-A)}$ for example, and also by putting ${t=0}$ on both sides. Alternatively, it can be seen by the fact that (up to a sign) ${a_{n-1}}$ = sum of roots of the characteristic polynomial = sum of eigenvalues = trace of ${A}$. Similarly the determinant = product of the eigenvalues = product of the roots of the characteristic polynomial = ${(-1)^n a_0}$

What about other terms? It is not difficult either. By observing the power of ${t^k}$ in the expression ${\det(tI-A)}$, or by differentiating w.r.t. ${t}$ of ${\det (t-AI)}$ at ${t=0}$, and using the formula ${\displaystyle\frac{d}{dt}\left|\begin{matrix} a &b \\ c & d \end{matrix}\right|=\left|\begin{matrix} a' &b' \\ c & d \end{matrix}\right|+\left|\begin{matrix} a &b \\ c' & d' \end{matrix}\right|}$ etc, we arrive at

$\displaystyle a_{j}=(-1)^j \sum_{i_1<\cdots

where ${I}$ is the muli-index ${(i_1, \cdots , i_j)}$ and ${d_I}$ is the (sub) determinant of the submatrix obtained by deleting all the ${i_k}$-th row and ${i_k}$-th columns of ${A}$, ${1\leq k \leq j}$. When ${j=0}$, we recover ${a_0}$ to be the original determinant and when ${j=1}$, we obtain the trace, since it is the sum of the ${1\times 1}$ subdeterminant (i.e. the number itself) along the diagonal.

In particular this proves that

Theorem 1 (Zorn-Bean’s theorem, 2010) A ${n\times n}$ real symmetric matrix is of rank ${n-l}$ if and only if

$\displaystyle \sum_{1\leq i_1<\cdots

where ${d_I}$ is as defined above.

For example, if ${A= \begin{pmatrix} a &b\\ b &c \end{pmatrix} }$, then it is of rank 2 if and only if ${ac-b^2\neq 0}$. It is of rank 1, if and only if ${ac-b^2\neq 0}$ and ${tr(A)=a+c\neq 0}$. Finally it is of rank 0 (i.e. ${A=0}$) if and only if ${ac-b^2=0}$ and ${a+c=0}$. In the last case, it is easy to see that the last equation forces ${a=b=c=0}$.

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### 2 Responses to Invariant under similarity

1. kingleunglee says:

Or you may use the following fact.
Lemma: If $\lambda$ is an eigenvalue of a matrix $A$, then, it is an eigenvalue of $B^t A B$, where $B^t=B^{-1}$.
Pf: $det(tI-B^t A B)=det(B^t(tI-A)B)=det(tI-A)$.
Thus, all coefficient of characteristic polynomial remains unchanged.

With this, rank of a symmetric matrix equals number of roots of characteristic polynomial by diagonalize the matrix, then done.

2. KKK says:

Well… that’s actually the proof(s) that my $a_j$ are invariants under similarity and also that A is of rank $n-l$ if and only if $p(t)$ has 0 as a root with multiplicity $l$ (which I have actually omitted… oops!), however (for the first “fact”), $B$ need not be orthogonal, so perhaps you should replace $B^t$ by $B^{-1}$ in your proof above.

What I have done a little bit more is just writing down an explicit formula for those $a_j$.