# A linear Trans. prob.

1. Jul 23, 2012

### 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)

$A= \begin{bmatrix} 0 & 0&... & 0 &0\\ 1 & 0 & 0&....&0 \\ 0&1&0&.....&0\\ .\\ .\\ 0&0&....&1&0 \end{bmatrix}$
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 $B'=${$a_i, Sa_i,S^2a_i,....,S^{n-1}a_i$}. Note if we proved $B'$ is a bases, then $_{B'}=A$.i.e. the matrix of $S$ relative to the bases $B'$ is A.
Unfotunatly, I could not get to an end with both ways.
Am I doing the right thing? Any suggestions?.

Last edited: Jul 23, 2012
2. Jul 23, 2012

### tiny-tim

Welcome to PF!

Hello Abuattallah! Welcome to PF!

Hint: use the fact that Sn-1 ≠ 0

3. Jul 23, 2012

### Abuattallah

Thank you tiny-tim.
This is my try:

consider the base $B'=${$a,Sa,....,S^{n-1}a$}, for some $a$ where $S^ia≠0, \forall i≤n-1$, now for the sake of simplicity lets 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 $B'$ is linearly independent.

4. Jul 23, 2012

### tiny-tim

Hi Abuattallah!

Yes, that's right.

Just a few "tweaks" needed …

Start with "Since Sn-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"

5. Jul 23, 2012

### HallsofIvy

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

6. Jul 25, 2012

### Abuattallah

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

7. Jul 25, 2012

### HallsofIvy

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