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Let $A$ be an $n\times n$ matrix. Prove that $\dim(\mathrm{span}({I_n,A,A^2,...})) ≤ n$

I'm at a total loss here... Can someone help me get started?

mdp
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clueless
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  • Do you have a theorem that tells you anything about some sum of powers of $A$? – Eric Towers Mar 10 '14 at 09:36
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    This doesn't make any sense. – Git Gud Mar 10 '14 at 09:37
  • I'm also unsure what is meant here. $\mathrm{diag}$ normally means a (block, here) diagonal matrix. How is that $\leq n$? – Eric Towers Mar 10 '14 at 09:39
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    "diag" usually refers to a diagonal matrix, but it makes no sense for a matrix to be less than or equal to the number, $n$. Perhaps what's meant is dim, rather than diag? – Gerry Myerson Mar 10 '14 at 09:40
  • Assuming that what's meant is dimension, look up the Cayley-Hamilton Theorem. – Gerry Myerson Mar 10 '14 at 09:41
  • yeah, I wasn't sure if the question made sense. but if it's dimension it makes a lot more sense. thank you – clueless Mar 10 '14 at 09:52
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    Now that the typo is sorted out, if you have a math question that you still need answered then you should edit the question to fix the typos and get it reopened. – Noah Snyder Mar 10 '14 at 14:08
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    @clueless I decided to make a bold edit and replace the "diag" by "dim" (and $<$ by $\leq$ in the title), as it seems you agree that this is what it should be. It is better if anybody finds this page later if the question makes sense and agrees with the answer! However, if you don't agree with this edit, please roll it back (or if you don't know how, leave a comment and I'll do it myself.) – mdp Mar 11 '14 at 12:30

2 Answers2

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Let $F=\mathrm{span}({I_n,A,A^2,...})$. It suffices to prove that $(I_n, A,..., A^{n-1})$ is a family with cardinality $n$ that spans $F$.

By Cayley-Hamilton, one can write $A^n=a_{n-1}A^{n-1}+\ldots+a_0I_n$, hence $$A^n\in \text{span}(I_n, A,..., A^{n-1}).$$

More generally, it is proved by induction on $m$ that $\forall m \geq n, A^m \in \operatorname{span}(I_n, A,..., A^{n-1}) $.

Hence $(I_n, A,..., A^{n-1})$ spans $F$, thus $\dim F\leq n$.

Gabriel Romon
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  • In general having one dependent family is not sufficient to give an upper bound on the dimension of a vector space. Somehting should be said about higher powers $A^k$. – Marc van Leeuwen Mar 12 '14 at 12:10
  • @MarcvanLeeuwen You are very right. See my edit. – Gabriel Romon Mar 12 '14 at 16:11
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    But I still don't think there is much reason to keep $A^n$ in the arguments if 'span'. Cayley-Hamilton says $A^n$ is in $\operatorname{span}(I_n, A,..., A^{n-1})$, and the induction argument handling $A^m$ for $m\geq n$ will show it to be in $\operatorname{span}(I_n, A,..., A^{n-1})$, in fact for less effort (you now use the cited fact $A^n\in\operatorname{span}(I_n, A,..., A^{n-1})$ at each step, rather than $A^{n+1}\in\operatorname{span}(I_n, A,..., A^n)$). – Marc van Leeuwen Mar 20 '14 at 13:25
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The following observations suffice to prove the statement:

  1. A power $A^k$ is in the span of lower powers $A^0,\ldots,A^{k-1}$ if and only if there exists a (monic) polynomial$~P$ of degree$~k$ with $P[A]=0$.
  2. If this happens for some $k=m$, it also happens for all $k>m$, so that by an immediate induction argument, $\operatorname{span}(A^0,\ldots,A^{m-1})$ contains all powers of$~A$ (and of course it has dimension${}\leq m$).
  3. There exists a monic polynomial$~P$ of degree${}\leq n$ with $P[A]=0$.

Point 1. is fairly obvious; it suffices to consider an equation expressing $A^k$ as linear combination of $A^0,\ldots,A^{k-1}$, and to bring all its terms to the same side of the equation as$~A^k$. Point 2. is also easy, since it suffices to multiply the polynomial $P$ by a power$~X^d$ to raise is degree; the result still annihilates$~A$, since $(X^dP)[A]=A^d\circ (P[A])=0$ by linearity of$~A$. (Alternatively one could argue for the remainder $R$ of $X^k$ after division by$~P$ that $A^k=R[A]\in\operatorname{span}(A^0,\ldots,A^{m-1})$, since $\deg R<m$.) For point 3. the Cayley-Hamilton theorem says the characteristic polynomial $P=\chi_A$, which has degree $n$, can be chosen.

Added. When using Cayley-Hamilton, I always wonder if one could also do it by more elementary means (because the proof of C-H is somewhat subtle). For point 3. this is is in fact the case.

One can use strong induction on the dimension. The inductive hypothesis takes the following concrete form: for any $A$-stable subspace $V$, there exists a monic polynomial $P\in K[X]$ with $\deg P\leq\dim V$ such that $V\subseteq\ker(P[A])$. By expressing the restriction of $A$ to$~V$ on a basis, this is just point 3. for that restriction. So assume this result holds for any proper subspace. If $n=0$ one can take $P=1$, so also assume $n>0$ and choose a nonzero vector $v\in K^n$. The $n+1$ vectors $v,Av,A^2v,\ldots,A^nv$ are certainly linearly dependent, so one can take $d$ minimal such that $v,Av,A^2v,\ldots,A^dv$ are linearly dependent. Then $A^dv$ is a linear combination of the preceding vectors, and this gives a polynomial $R$ of degree$~d$ such that $R[A]v=0$. But $\ker(R[A])$ is $A$-stable so $v,Av,A^2v,\ldots,A^{d-1}v\in\ker(R[A])$, and since these vectors are linearly independent, one obtains $\dim\ker(R[A])\geq d$. Putting $V=\operatorname{Im}(R[A])$ the rank-nuullity theorem gives $\dim V\leq n-d$, and since $V$ is $A$-stable, the induction hypothesis gives a polynomial $Q$ with $\deg Q\leq n-d$ such that $V\subseteq\ker(Q[A])$. The latter means that $0=Q[A]\circ R[A]=(QR)[A]$; then taking $P=QR$ works, since one gets $\deg P=\deg Q+\deg R\leq(n-d)+d=n$.

Marc van Leeuwen
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