Rank and characteristic/minimal Polynomial

• January 18th 2011, 07:37 AM
skyking
Rank and characteristic/minimal Polynomial
I am having trouble with this question and need some help.

Let $A$ be a square matrix of order $n$ over $F$ with a rank of 1.

1. Show that the characteristic polynomial of $A$ is $P(t)=t^{n}-tr(A)t^{n-1}$ and find the minimal polynomial of $A$.
2. Let $T:F^{n}\rightarrow F^{n}$ be the linear map defined by $Tv=Av$ for every $v\in F^{n}$. Show that for every $1\leq k\leq n$ there exists a subspace of $F^{n}$ of dimension $k$ that is $T-$reserved.

I can show the first part of 1. If the rank is 1 then the nullity of $A$ is of dimension $n-1$ and therfore the geometric multiplicity of the eignvalue $0$ is $n-1$, so we get that the algebraic multiplicity of $0$ is at least $n-1$. Therfore the characteristic polynomial is of the form $t^{n-1}(t-a)=t^{n}-at^{n-1}$ (where $a$ is the other eignvalue). We have shown before in a previous lin alg class that the coefficient of $t^{n-1}$ of the characteristic polynomial of any square matrix of order $n$ is $-tr(A)$, so we get then that the characteristic polynomial is indeed $t^{n}-tr(A)t^{n-1}$.

Now for the minimal polynomial I know that it has to be of the form $t^{k}(t-tr(A))$ where $k\leq n-1$, but I am not sure how to proceed.

As for 2 I would appreciate some direction.

Thanks,

SK
• January 18th 2011, 08:57 AM
FernandoRevilla
We have

$\dim (\ker A)=n-\textrm{r}(A)=n-1$.

This implies that $\lambda=0$ is an eigenvalue of $A$.

The algebraic multiplicity of $\lambda=0$ can't be $n$ (why?).

So, the characteristic polynomial is

$\chi(t)=t^{n-1}(t-a),\;\;(a\neq 0)$.

On the other hand:

$\chi(t)=t^n-(\textrm{tr}A)t^{n-1}+\ldots\Rightarrow a=\textrm{tr}A$

As $\lambda=a$ is simple then, $\dim\ker(A-aI)=1$ , so $A$ is diagonalizable on $F$ .

Then, the minimal polynomial $\mu(t)$ is:

$\mu(t)=t(t-\textrm{tr}A)$ .

Fernando Revilla

Edited: Sorry, now I see you solved the first part.
• January 18th 2011, 09:09 AM
FernandoRevilla
Quote:

Originally Posted by skyking
Show that for every $1\leq k\leq n$ there exists a subspace of $F^{n}$ of dimension $k$ that is $T-$reserved.

What does $T$-reserved mean?.

Fernando Revilla
• January 18th 2011, 09:56 AM
skyking
[QUOTE=FernandoRevilla;606818]What does $T$-reserved mean?.

I wasn't sure I was translating this correctly, I wasn't sure what the term in english is. But if $W$ is " $T-$reserved" it means that for every $u\in W$ we get $Tu\in W$.
• January 18th 2011, 10:18 AM
FernandoRevilla
Quote:

Originally Posted by skyking
I wasn't sure I was translating this correctly, I wasn't sure what the term in english is. But if $W$ is " $T-$reserved" it means that for every $u\in W$ we get $Tu\in W$.

All right, then you meant $T$-invariant subspace .

Hint :

Use the fact that in a determined basis $B$ of $F^{n}$ the matrix of $T$ is:

$D=\textrm{diag}(a,0,\ldots,0)$

Fernando Revilla