Prove that A is similar to B and what is the multiplicity of $0$ as characteristics root of $A$?

Question

Prove that the $n \times n$ matrix $$A=\left(\begin{matrix} 1 & 1 & ... & 1\\ 1 & 1 & ... & 1 \\ .& .& . & .\\ .& . & . & .\\ . & .& & . \\1 & 1 & ... & 1 \end{matrix}\right)$$ is similar to $$B= \left(\begin{matrix} n & 0 & ... & 0\\ 0 & 0 & ... & 0 \\ .& .& . & .\\ .& . & . & .\\ . & .& & . \\0& 0 & ... & 0\end{matrix}\right)$$

If the characteristics of $\mathbb{F}$ is $0$ or if it is $p$ and $p \nmid n$. what is the multiplicity of $0$ as characteristics root of $A$ ?

My trial : I was taking $n= 2$ ,$A = \begin{bmatrix} 1 &1 \\ 1& 1\end{bmatrix}$ and $B=\begin{bmatrix} 2 &0 \\ 0& 0\end{bmatrix}$

Here I did not find A is similar matrix to B because A is symmetric. But B is not, as both ranks are also not the same...

I don't know how to tackle this question

Any hints/solution

thanks u

Answer 1

In your $2\times 2$ case, let $v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $v_2 = \begin{bmatrix}1 \\ - 1\end{bmatrix}$.

Then \begin{align*} A v_1 &= \begin{bmatrix}1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \end{bmatrix} =2\begin{bmatrix} 1 \\ 1 \end{bmatrix} \\ A v_2 &= \begin{bmatrix}1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} =0\begin{bmatrix} 1 \\ 1 \end{bmatrix} \\ \end{align*} So if $$ P = \begin{bmatrix} v_1 & v_2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$ then $$ AP = PB $$ So $P^{-1}AP = B$ and the two are similar.

This has pretty much everything you need to answer the general case. You are looking for one eigenvector with eigenvalue $n$, and $n-1$ linearly independent eigenvectors with eigenvalue $0$.

Answer 2

Let $e_1=(1,1,\ldots,1)$ and let $(e_2,\ldots,e_n)$ be a basis of$$\left\{(x_1,\ldots,x_n)\in\mathbb{F}^n\,\middle|\,\sum_{k=1}^nx_k=0\right\}.$$Then $e_1$ is an eigenvector of $A$ with eigenvalue $n\neq0$. Furthermore$$(\forall j\in\{1,2,\ldots,n\}):A.e_j=0.$$So, the matrix of $A$ (seen as an endomorphism of $\mathbb{F}^n$) with respect to the basis $(e_1,e_2,\ldots,e_n)$ is $B$.