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Suppose that $T : V \to V$ is a linear operator on an $n$-dimensional vector space $V$.

(a) Show that for all $i$, $\ker T^i \subset \ker T^{i+1}$.

(b) Show that if $\ker T^k = \ker T^{k+1}$, then $\ker T^k = \ker T^{k+j}$ for all $j \geq 1$.

(c) Show that if $T^k=0$ for some $k$, then $T^n=0$.

My question is about (c), I do not understand what the question is asking, I was told "if some power $k$ of the operator is the zero operator, then the smallest such power must be no larger than $n$", if so, can someone help me with this?

Thanks in advance.

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Bob
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6 Answers6

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Let $k'$ be the smallest integer such that $T^{k'}=0$. Then $k'$ must be less than or equal to $n$, since if it were greater than $n$ we'd have the following chain of strict inclusions $$\ker(T)\subset \ker(T^2)\subset \cdots \subset \ker(T^{k'}) .$$ The inclusions are strict (meaning that "$\subset$" means "contained in but not equal to") since if one inclusion weren't strict, then by (b) all inclusions to its right would also be strict, and this would contradict the minimality of $k'$. Now since all the inclusions are strict, $\dim\ker(T^i)<\dim\ker(T^{i+1})$ for all $i<k'$. This in turn implies that $k'\leq n$, since if $k'$ were greater than $n$ then $\dim \ker T^{k'}$ would be greater than $n$, which can't happen since it must be a subspace of $V$. Now that we know that $k'\leq n$, we know that $T^n=0$ by (b).

dezign
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  • +1. I think though it would be a bit clearer is you said "if one inclusion weren't strict, then by (b) all inclusions to its right would also be strict, and this would contradict the minimality of $k′$". – Marc van Leeuwen Sep 23 '13 at 14:34
  • Yes I agree. Thanks! – dezign Sep 24 '13 at 02:50
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$T^k = 0$ is equivalent to $\ker T^k=V$. Now if $\ker T^n \ne V$ then according to (a) and (b) you have $\ker T^i = \ker T^{i+1}$ for some $1 \leqslant i < n$ and hence $\ker T^k \ne V$ for all $k$.

njguliyev
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  • ths for help, but I don't quite get how from (a) and (b) we get kerT$^i$ =kerT$^{i+1}$ – Bob Sep 23 '13 at 11:42
  • $\dim\ker T^i \leqslant \dim\ker T^{i+1}$, so if $\ker T^i \ne \ker T^{i+1}$ then $\dim\ker T^i < \dim\ker T^{i+1}$. But $1 \leqslant \dim\ker T^1 < \dim\ker T^2 < \ldots < \dim\ker T^n \leqslant n-1$ is impossible. – njguliyev Sep 23 '13 at 12:44
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Let $F$ be the field the vector space is defined on. Assume $T\neq 0$, which is trivial.

Define $M=\{f\in F[X]|f(T)=0\}$.

Note that $M$ satisfies two conditions below: (i) $M$ is nonzero (Cayley-Hamilton theorem shows that the characteristic polynomial of $T$ is an element of it) (ii) For any $f\in F[X]$ and $g\in M$, $fg\in M$.

Using "Polynomial Division Algorithm", you can show that there exists a unique monic $p\in M$ such that $M=\{pf\in F[X]|f\in F[X]\}$.

Note that $p$ is a monic polynomial with positive degree.

Since $T^k=0$, $p$ divides $X^k$.

Thus, there exists $m\in\mathbb{Z}^+$ such that $p=X^m$.

Since the characteristice polynomial is in $M$ and its degree is $n$ and $p$ divides it, $\deg(p)≦n$. So that $m≦n$.

Thus, $T^n=0$.

(By the way,i think "set-theory" tag is inappropriate)

Jj-
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Perhaps it will help if you translate the question into matrix form. Item three is asking you to prove that if $A$ is any $n\times n$ matrix and if some natural numbers $k$ exists for which $A^k=0$, then it must be the case that $A^n=0$. Low values of $n$ will clarify further. For $n=1$ the matrix $A$ can safely be thought of just being a scalar (i.e., a real number), say $a$. Then the claim is that if $a^k=0$ for some $k$, then $a^1=0$, namely $a=0$. This is a well known property of real numbers. For $n=2$ this is already less clear. The claim is then that if $A$ is a $2\times 2$ matrix for which, e.g., $A^9$ is the zero matrix, then $A^2$ must already be the zero matrix. This is not so easy to show directly, and in fact the exercise is asking you to prove the more general result. I hope this helps.

Ittay Weiss
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All you have is $T^k=0$ with out loss of generality i assume $k>n$.

Now, Assuming $ker(T^n)\neq V$

you know that $Ker(T)\subseteq Ker(T^2)\subseteq Ker(T^3)\dots \subseteq Ker(T^n)$

As $Ker(T^n)\neq V$, we have $dim(ker(T^n))<n$

But then the no of subspaces in above chain is "$n$"

So, this imply there would be atleast one equality in the chain

i.e., for atleast one $i : 1\leq i\leq n$ we have $Ker(T^i)=Ker(T^{i+1})$

But (by thanking $(b)$) we see that $Ker(T^i)=Ker(T^{m})$ for any $m>i$.

in particular $Ker(T^i)=Ker(T^k)$ As $Ker(T^i)\subset Ker(T^n)\neq V$,

we end up with the case that $Ker(T^k)=Ker(T^i)\neq V$

This contradicts the condition of $T^k=0$, which is equivalent to the condition of $Ker(T^k)=V$

So, I would then left with the case of $Ker(T^n)=V$ i.e., $T^n=0$

  • Hi ,thanks for help, should it be 1≤i<n instead? – Bob Sep 23 '13 at 11:55
  • and, after "But (by thanking (b)) we see that Ker(Ti)=Ker(Tm) for any m>i." can we directly say Ker(Tk)=Ker(Tn)=Ker(Ti) as and k are both bigger than i thence get the contradiction? – Bob Sep 23 '13 at 12:03
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The question only makes sense if $T : V\to V$, however I don't think it's true for a general (non-linear) operator. Consider $T : \mathbb{R}^2 \to \mathbb{R}^2$ defined by $$ T \left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} x+1 \\ y\end{bmatrix} $$ such that $$ T^i \left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} x+i \\ y\end{bmatrix} $$ and $\ker T^i = (-i, 0)^T$. Clearly (a) fails since $\ker T^i$ are disjoint.

Likewise for (c) Consider $T : \mathbb{R} \to \mathbb{R}$ defined by $$ T(x) = \left\lbrace \begin{array}{ccc} 1 &:& |x| > 2 \\ 0 &:& |x| \le 2 \end{array} \right. $$ Edit: Didn't see the linear algebra tag

sn6uv
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