# Thread: Rank and characteristic/minimal Polynomial

1. ## Rank and characteristic/minimal Polynomial

I am having trouble with this question and need some help.

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

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

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

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

As for 2 I would appreciate some direction.

Thanks,

SK

2. We have

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

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

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

So, the characteristic polynomial is

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

On the other hand:

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

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

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

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

Fernando Revilla

Edited: Sorry, now I see you solved the first part.

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

Fernando Revilla

4. [QUOTE=FernandoRevilla;606818]What does $\displaystyle T$-reserved mean?.

First thanks, for the reply.

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

5. 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 $\displaystyle W$ is "$\displaystyle T-$reserved" it means that for every $\displaystyle u\in W$ we get $\displaystyle Tu\in W$.
All right, then you meant $\displaystyle T$-invariant subspace .

Hint :

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

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

Fernando Revilla

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