# Product of Linear Operators - A basic quesiton

• Oct 11th 2009, 12:53 AM
aman_cc
Product of Linear Operators - A basic quesiton
Let $T_1, T_2,....T_n$ be linear operators on a vector space V.

Let $v\in V$ such that one of $T_1(v),T_2(v),..,T_n(v)$ is 0.

Is it ok to say $T_1T_2....T_n(V)=0$?
• Oct 11th 2009, 01:34 AM
NonCommAlg
Quote:

Originally Posted by aman_cc
Let $T_1, T_2,....T_n$ be linear operators on a vector space V.

Let $v\in V$ such that one of $T_1(v),T_2(v),..,T_n(v)$ is 0.

Is it ok to say $T_1T_2....T_n(V)=0$?

i think you meant $T_1T_2....T_n(v)=0$ not $T_1T_2....T_n(V)=0$? if so, the answer is no. for example let $\{v_1,v_2 \}$ be a basis for $V$ and define $T_1(v_1)=0, \ T_1(v_2)=v_1, \ T_2(v_1)=v_2, \ T_2(v_2)=v_1.$

here's a much less trivial question: suppose that for every $v \in V$ there exists $1 \leq i \leq n$ such that $T_i(v)=0.$ would we necessarily have $T_1T_2 \cdots T_n = 0$?
• Oct 11th 2009, 01:44 AM
aman_cc
Quote:

Originally Posted by NonCommAlg
i think you meant $T_1T_2....T_n(v)=0$ not $T_1T_2....T_n(V)=0$? if so, the answer is no. for example let $\{v_1,v_2 \}$ be a basis for $V$ and define $T_1(v_1)=0, \ T_1(v_2)=v_1, \ T_2(v_1)=v_2, \ T_2(v_2)=v_1.$

here's a much less trivial question: suppose that for every $v \in V$ there exists $1 \leq i \leq n$ such that $T_i(v)=0.$ would we necessarily have $T_1T_2 \cdots T_n = 0$?

Yes I meant $T_1T_2....T_n(v)=0$

My book - says the following
If v is a eigenvector then one of
(T-c1I), (T-c2I),....,(T-ckI) sends v to 0.
where c1,c2,....ck are eigenvalues

I understand till this point.

Then it says, hence
[(T-c1I)(T-c2I)....(T-ckI)](v) = 0

This is where I got confused and asked the question above.

• Oct 11th 2009, 02:06 AM
NonCommAlg
Quote:

Originally Posted by aman_cc
Yes I meant $T_1T_2....T_n(v)=0$

My book - says the following
If v is a eigenvector then one of
(T-c1I), (T-c2I),....,(T-ckI) sends v to 0.
where c1,c2,....ck are eigenvalues

I understand till this point.

Then it says, hence
[(T-c1I)(T-c2I)....(T-ckI)](v) = 0

well, this is very different from your original question! suppose $Tv=c_j v.$ then $(T-c_kI)(v)=(c_j - c_k)v$ and so $(T-c_1I) \cdots (T-c_kI)(v)=(c_j-c_k)(T-c_1I) \cdots (T-c_{k-1}I)(v) = \cdots$

i'm sure you can finish the proof now!
• Oct 11th 2009, 09:03 PM
aman_cc
Quote:

Originally Posted by NonCommAlg
well, this is very different from your original question! suppose $Tv=c_j v.$ then $(T-c_kI)(v)=(c_j - c_k)v$ and so $(T-c_1I) \cdots (T-c_kI)(v)=(c_j-c_k)(T-c_1I) \cdots (T-c_{k-1}I)(v) = \cdots$

i'm sure you can finish the proof now!

Yes - I got it !! Thanks

Now working on your non-trivial poser, let's see.
• Oct 12th 2009, 01:40 AM
aman_cc
Quote:

Originally Posted by NonCommAlg
i think you meant $T_1T_2....T_n(v)=0$ not $T_1T_2....T_n(V)=0$? if so, the answer is no. for example let $\{v_1,v_2 \}$ be a basis for $V$ and define $T_1(v_1)=0, \ T_1(v_2)=v_1, \ T_2(v_1)=v_2, \ T_2(v_2)=v_1.$

here's a much less trivial question: suppose that for every $v \in V$ there exists $1 \leq i \leq n$ such that $T_i(v)=0.$ would we necessarily have $T_1T_2 \cdots T_n = 0$?

First - Thanks for your question. Here is my attempt (I hope I got it :))
Ans: Yes $T_1T_2 \cdots T_n = 0$ as I claim one of $T_i= 0$

Let $K_i$ be kernel of $T_i$
You tell me $\bigcup K_i = V$
I claim this is possible only if one of $K_i = V$ and hence corresponding $T_i= 0$

If I show you that $\forall i$, if $K_i \neq V$ then $\bigcup K_i \neq V$, I will be done. Correct?

I will prove this by induction over $i$. $i=1$or $i=2$ are trivial I am omitting them.

Let above be true for $i=n$. I claim it is true for $i=n+1$, for otherwise we have a situation where

$\bigcup K_i \neq V$ but $(\bigcup K_i)\bigcup K_{n+1} = V$. Note also $\forall i, K_i \neq V$

Under the above situation, I can find vectors
$a,b \in V$ such that $a \in K_{n+1}$ and $a \notin (\bigcup K_i)$. Similarly $b \notin K_{n+1}$ and $b \in (\bigcup K_i)$

Consider $b, a+b, a+2n, ..., a+(n-1)b$. No two of these can be in same $K_i$ for $i \in [1,n]$. As other wise we will have $a \in K_i$, contrary to our claim.

Therefor consider $a+nb \in V$. But, $a+nb \notin (\bigcup K_i)\bigcup K_{n+1}$.

A contradiction and hence we are done.

Thanks
• Oct 12th 2009, 04:25 AM
NonCommAlg
Quote:

Originally Posted by aman_cc

Consider $b, a+b, a+2n, ..., a+(n-1)b$.

considering the sentence after this i guess you meant: $b, b+a, b+2a, \cdots , b+(n-1)a.$

Quote:

$a+nb \notin (\bigcup K_i)\bigcup K_{n+1}$.

why?
• Oct 12th 2009, 05:17 AM
aman_cc
Quote:

Originally Posted by NonCommAlg
considering the sentence after this i guess you meant: $b, b+a, b+2a, \cdots , b+(n-1)a.$

why?

Yes - I meant $b, b+a, b+2a, \cdots , b+(n-1)a.$. Sorry about that

Any vector of the form $b+ka$ can't belong to $K_{n+1}$. Because if it does => $b+ka - (ka) = b \in K_{n+1}$, which is a contradiction. Note: $a \in K_{n+1}$ => $ka \in K_{n+1}$

So any vector of the form $b+ka \in \bigcup K_i, i \in [1,n]$

Also $b+k_1a, b+k_2a$ for two different $k1,k2$ cannot belong to same $K_i, i \in [1,n]$. Because if it does $(b+k_1a) - (b+k_2a) = (k_1-k_2)a \in K_i$. Which again is a contradiction as we have assumed $a \notin \bigcup K_i, i \in [1,n]$

Now consider $b+na$. Sorry for typo here in the earlier post.

By Pigeon-hole principle $b+na$ can't belong to any of the $K_i, i \in [1,n]$.
Also, $b+na \notin K_{n+1}$. as it is of the form $b+ka$

Thus $b+na \notin (\bigcup K_i)\bigcup K_{n+1}$

Thus $(\bigcup K_i)\bigcup K_{n+1}\neq V$
• Oct 12th 2009, 05:52 AM
NonCommAlg
Quote:

Originally Posted by aman_cc
Yes - I meant $b, b+a, b+2a, \cdots , b+(n-1)a.$. Sorry about that

Any vector of the form $b+ka$ can't belong to $K_{n+1}$. Because if it does => $b+ka - (ka) = b \in K_{n+1}$, which is a contradiction. Note: $a \in K_{n+1}$ => $ka \in K_{n+1}$

So any vector of the form $b+ka \in \bigcup K_i, i \in [1,n]$

Also $b+k_1a, b+k_2a$ for two different $k1,k2$ cannot belong to same $K_i, i \in [1,n]$. Because if it does $(b+k_1a) - (b+k_2a) = (k_1-k_2)a \in K_i$. Which again is a contradiction as we have assumed $a \notin \bigcup K_i, i \in [1,n]$

Now consider $b+na$. Sorry for typo here in the earlier post.

By Pigeon-hole principle $b+na$ can't belong to any of the $K_i, i \in [1,n]$.
Also, $b+na \notin K_{n+1}$. as it is of the form $b+ka$

Thus $b+na \notin (\bigcup K_i)\bigcup K_{n+1}$

Thus $(\bigcup K_i)\bigcup K_{n+1}\neq V$

it's correct now. good work! what you're saying is that for every $0 \leq i \leq n-1$ there exists a "unique" $1 \leq j \leq n$ such that $b + ia \in K_j.$ now $b + na \notin K_{n+1}$ because $a \in K_{n+1}$ and $b \notin K_{n+1}.$

also $b + na \notin K_j,$ for all $1 \leq j \leq n,$ because we already know that $b + ia \in K_j$ for some $0 \leq i \leq n-1.$ so we'd have $(n-i)a \in K_j$ and hence $a \in K_j,$ which is false.

Remark: as you might have noticed that your proof works only for vector spaces over fields with zero characteristic. this proof can be modified to work for all infinite fields. it is an interesting

question to see if there's any counter-example for vector spaces over "finite" fields! (Evilgrin)
• Oct 12th 2009, 07:43 AM
aman_cc
Quote:

Originally Posted by NonCommAlg
it's correct now. good work! what you're saying is that for every $0 \leq i \leq n-1$ there exists a "unique" $1 \leq j \leq n$ such that $b + ia \in K_j.$ now $b + na \notin K_{n+1}$ because $a \in K_{n+1}$ and $b \notin K_{n+1}.$

also $b + na \notin K_j,$ for all $1 \leq j \leq n,$ because we already know that $b + ia \in K_j$ for some $0 \leq i \leq n-1.$ so we'd have $(n-i)a \in K_j$ and hence $a \in K_j,$ which is false.

Remark: as you might have noticed that your proof works only for vector spaces over fields with zero characteristic. this proof can be modified to work for all infinite fields. it is an interesting

question to see if there's any counter-example for vector spaces over "finite" fields! (Evilgrin)

Thanks very much. Really appreciate you being patient and helpful. I really enjoyed doing this question.

Thanks again