diagonalize the following n×n matrix. I am wondering if my solution for the characteristic polynomial is valid or if there is a better way to do it.

Question

The following is the question:

The following is my answer.

Answer 1

$x(x-n) = 0$ is not a polynomial, this is an equation.

$p(x)= x(x-n)$ might be a candidate, but it is also not the characteristic polynomial unless $n=2$ -- the characteristic polynomial will have degree $n$. (It's the minimal polynomial).

Note, though, that this will give you an idea to find the characteristic polynomial -- quite obviously, $n$ is an eigenvalue, and $0$ is another one. It's easy to see which rank $A$ has, so you can determine the multiplicity of $0$...

(The approach to find a polynomial for which $A$ is a root is nice, though).

Answer 2

In the characteristic polynomial, the coefficient of $\lambda ^{n-i}$ is $(-1)^i$(sum of principal minors of order i of A)for$1\le i\le n.$A principal minor of order i is the determinant of an ixi sub-matrix of A formed from the same i rows and columns, e.g. one of the principal minors of order 2 would be the determinant of the 2x2 submatrix $$\begin{bmatrix}a_{33}&a_{35}\\a_{53}&a_{55}\end{bmatrix}$$. For the given matrix A, the prinipal minors of order 1 are all 1 and the principal minors of order i for i>1 are all 0,so the characteristic polynomial is $\lambda ^n-n \lambda^{n-1}=\lambda ^{n-1}(\lambda -n)$Thus the eigenvalues are 0, with multiplicity $n-1$ and $n$ with multiplicity 1.For $1\le j\le n-1$ let $v_j$ be the $n$-element column vector with 1 in position $j$, -1 in position $j+1$and other elements all 0. Then $\{v_1,v_2, ... ,v_{n-1}\}$ is a basis for the eigenspace corresponding to eigenvalue 0. Let $v_n$ be the n-element column vector whose elements are all 1. Then $\{v_n\} $ is a basis for the 1-dimensional eigenspace corresponding to eigenvalue n. Let P be the n x n matrix with columns $v_1,v_2,...,v_{n-1},v_n. $. Let D be the diagonal matrix with $n$ -th entry $n$ and all other entries 0, i.e. D is the matrix with $n$ in the bottom right corner and 0 elsewhere. Then $$AP=PD,P^{-1}AP=D.$$ Let $Q=P^{-1}$. Then $$QAQ^{-1}=D$$