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The problem goes as follows-
Let $A\in M_{10}(\Bbb{C})$, the vector space of $10\times 10$ matrices with entries in $\Bbb{C}$. Let $W_A$ be the subspace of $M_{10}(\Bbb{C})$ spanned by $\{A^n\mid n\ge0\}$. Choose the following statement(s).
(a) For any $A$, $\dim(W_A)\le10$.
(b) For any $A$, $\dim(W_A)<10$.
(c) For some $A$, $10<\dim(W_A)<100$.
(d) For some $A$, $\dim(W_A)=100$.
Intuitively, it looks like $W_A$ is not finite dimensional, but it's not.
I can't extract a linearly independent subset from $\{A^n\mid n\ge0\}$.
Can any one help find out a proper solution to that problem. Thanks for assistance in advance.

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Let $p(t) = a_0 + a_1t + \dots a_{9}t^9 +t^{10}$ be the characteristic polynomial of $A$. By Cayley-Hamilton theorem, $p(A) =0$ hence $$ a_0 A^0 + \dots a_9A^9 + A^{10} = 0. $$ Multiplying both sides by powers of $A$ it follows that $A^n$ for $n\geq 20$ is in the span of $\{ A^{10}, \dots, A^{19}\}$.

If $A$ is a diagonal matrix with pairwise distinct (and $\neq 0$) eigenvalues, then $p(t)$ is also the minimal polynomial, hence $\mbox{dim}W_A = 10 $ in this case. This is the maximum possible. In fact, if $\mbox{dim}W_A < 10 $ then we can take $$ b_0 A^{10} + \dots b_9A^{19} = 0, $$ and, using that $A$ is invertible, see that $b_0I + \dots b_9A^{9} = 0$ which is a contradiction.

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  • $\begingroup$ Check the definition of $W_A$: it is spanned by $\{A^n:\color{red}{n\ge10}\}$. $\endgroup$ – egreg Sep 8 '18 at 18:06
  • $\begingroup$ Thanks. I'll correct the answer. $\endgroup$ – Alan Muniz Sep 8 '18 at 18:08
  • $\begingroup$ But now the OP has changed the question to say $n\geq0$ instead of $n\geq 10$. $\endgroup$ – Andreas Blass Nov 3 '18 at 23:14
  • $\begingroup$ @AndreasBlass the argument is the same. The space is generated by $\{A^0 = I, A, \dots, A^9\}$ by virtue of the characteristic polynomial. $\endgroup$ – Alan Muniz Nov 4 '18 at 3:07
  • $\begingroup$ I know; even though I didn't check your original (before egreg's comment) answer, I assumed that it handled the $n\geq0$ situation analogously. In fact, your argument works fine with 10 treated as a variable, so in that sense it's immune even to future changes in the lower bound on $n$. $\endgroup$ – Andreas Blass Nov 4 '18 at 12:41
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By Cayley-Hamilton, $A^{10}$ is a linear combination of $\{A^0,A^1,\dots,A^9\}$ and the same is true for $A^n$, for any $n\ge 10$: just multiply by $A$ the dependence relation for $A^{10}$ and substitute $A^{10}$ with the linear combination, then do induction.

Therefore the dimension of $W_A$ is certainly $\le10$, because it is contained in a subspace spanned by $10$ elements. Now the problem is to find $A$ such that $\dim W_A=10$.

You can certainly find a basis of $W_A$ among the powers $A^n$ with $n\ge10$. If we find $A$ such that $\{A^{10},A^{11},\dots,A^{19}\}$ is linearly independent, then we're done.

Suppose the minimal polynomial of $A$ equals the characteristic polynomial; then $\{A^0,A^1,\dots,A^9\}$ are linearly independent. Suppose also that $A$ is invertible. Then from $$ a_0A^{10}+a_1A^{11}+\dots+a_9A^{19}=0 $$ we derive, by multiplying by $A^{-10}$, $$ a^0A^0+a_1A^1+\dots+a_9A^9=0 $$ and therefore $a_0=a_1=\dots=a_9=0$.

Take a diagonal matrix $A$ with diagonal entries $1,2,\dots,10$ and you're done.

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