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 <i_j} d_I

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 <i_j\leq n} d_I=0 \text{ for }j=0, \cdots, l-1\text{ and }\sum_{1\leq i_1<\cdots <i_{l}\leq n} d_I\neq 0

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.

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