Linear Operator T on Vector Space V: Unique Matrix A in Ordered Bases B

[Abuattallah]

Hello,
I am a grad student preparing for a quals. I am using H. and Kunze book.
the problem is:
let V be a n-dim vector space over F. and let B={a_1,a_2,..., a_n} be an ordered bases for V.
a- According to them 1, there is a unique Linear operator T on V such that

Ta_i=a_{(i+1)} , i=1,...,n. and Ta_n=0.
what is the matrix A of T in the ordered bases B.
b- prove that T^n=0, and\ \ \ T^{n-1}≠ 0.
c- Let S be any linear operator on V such that S^n=0\ \ \ but\ \ \ S^{n-1}≠0. Prove that there is an ordered bases B' such that the matrix of S in the bases B' is the matrix A of part (a).

Solution Attempt.
Obviously we have for (a)A= \begin{bmatrix}<br /> 0 &amp; 0&amp;... &amp; 0 &amp;0\\<br /> 1 &amp; 0 &amp; 0&amp;...&amp;0 \\<br /> 0&amp;1&amp;0&amp;...&amp;0\\<br /> .\\<br /> .\\<br /> 0&amp;0&amp;...&amp;1&amp;0<br /> \end{bmatrix}
for (b) its obvious.
my problem is with the last question. I tried to justfy it by two ways, the first one is to find an ivertible linear transformation U:V→V such that S=UTU^{-1}, then we will be done and such a bases exists. The second way is that I am trying to show the follwing:
there exist at least on vector in the bases B such that S^ia_i≠0 for i=1,..,n-1. and I am considering the set B&#039;={a_i, Sa_i,S^2a_i,...,S^{n-1}a_i}. Note if we proved B&#039; is a bases, then <s>_{B&#039;}=A</s>.i.e. the matrix of S relative to the bases B&#039; is A.
Unfotunatly, I could not get to an end with both ways.
Am I doing the right thing? Any suggestions?.
Thank you in Advance.

 

[tiny-tim]

Welcome to PF!

Hello Abuattallah! Welcome to PF!

Hint: use the fact that S^n-1 ≠ 0 ​

 

[Abuattallah]

Thank you tiny-tim.
This is my try:

consider the base B&#039;={a,Sa,...,S^{n-1}a}, for some a where S^ia≠0, \forall i≤n-1, now for the sake of simplicity let's call them as follow a=a_1, Sa=a_2,...,S^ia=a_{i+1},...,S^{n-1}=a_n,Consider the follwing

c_1a_1+...+c_na_n=0 ... (1)
Aplly S^{n-1} to both sides of (1), we get
c_1a_n=0→c_1=0,
now apply S^{n-2} to (1), we get
c_2a_n=0→c_2=0, Continueing this way, we will get all c_i=0\ \ , \ \forall i=1,...,n, Hence the set B&#039; is linearly independent.
Any comments on my method?

Thanks in Advance.

 

[tiny-tim]

Hi Abuattallah!

Yes, that's right.

Just a few "tweaks" needed …

Start with "Since S^n-1 ≠ 0, there exists an a such that …"

Then "So define the ordered set B' = {…}" (you can't call it a basis yet )

And end "So B' is an independent ordered subset whose cardinality is the same as the dimension of V, and is therefore a basis for V"

 

[HallsofIvy]

By the way, "bases" is the plural of "basis". We speak of a basis for a vector space.

 

[Abuattallah]

Thank you so much Tiny-Tim.
and thanks for word correction, English is my second language ; ).
You have all a wonderful day,
Abuattallah

 

[HallsofIvy]

Your English is far better than my (put just about any language here!).