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How can I reduce a matrix like this? The goal is to prove that the dimension of the generated space is 1. It's the matrix of the homogeneous system.

$\left( \begin{array}{cccccc}1-n&1&\dots &\dots&\dots&1 \\ 1 & 1-n & 1 & \dots & \dots & 1 \\ \vdots & 1 & \ddots & \ddots & & \vdots \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ \vdots & \vdots && 1 & 1-n & 1 \\ 1 &1 & \dots & \dots & 1 & 1-n\end{array} \right)$.

Allonsy
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1 Answers1

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The matrix diagonalizes to $$\begin{pmatrix} 0&0&0&0\\ 0&-n&0&0\\ 0&0&-n&0\\ 0&0&0&-n\\ \end{pmatrix}$$

Let $A=\begin{pmatrix} 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ \end{pmatrix}$ And if $X=\begin{pmatrix} x_1\\ \vdots \\ x_n \end{pmatrix}$ let $\sum X=x_1+\cdots x_n$.

Then we have $$AX=\sum X \begin{pmatrix} 1\\ \vdots \\ 1 \end{pmatrix}$$

Now consider the eigenvalue problem $$(A-nI)X=\lambda X$$ or equivalently, $$AX=(n+\lambda) X=\sum X \begin{pmatrix} 1\\ \vdots \\ 1 \end{pmatrix}$$ Now if $\lambda +n \neq 0$ then $X$ is a multiple of $\begin{pmatrix} 1\\ \vdots \\ 1 \end{pmatrix}$ and we have that $\lambda +n=n$ so $\lambda$ is zero with an eigenspace of dimension one. Otherwise $\lambda=-n$ and this has an eigenspace of dimension $n-1$ as can be easily checked the vectors $$\begin{pmatrix} 1\\ 0\\ \vdots \\ 0\\-1\\ 0\\ \vdots\\ 0\\ \end{pmatrix}$$ are eigen vectors.

Finally it would be nice to check the minimal polynomial which should be $x(x+n)$

Rene Schipperus
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  • Mmmm I just found the characteristic polynomial, which is $(-1)^nx^n+(-1)^{n-1}nx^{n-1}=x^{n-1}((-1)^nx+(-1)^{n-1}n)$ which tell me that the roots are $x=0,x=n$ with multiplicity $n-1$ and $1$ respectively. So according to theory, $dim(ker(f-0Id))=dim(ker(f))$ should be $n-1$, and $dim(ker(f-nId))$ should be 1. Your solution seems right but it doesn't fit with what I've been told. – Allonsy Jun 23 '14 at 02:39
  • $(A-nI)(A-nI+nI)=A^2-nA==0$ so $x(x+n)$ is the minimal polynomial, you probably made a sign mistake in getting the characteristic poly. How did you calculate it anyway ? – Rene Schipperus Jun 23 '14 at 02:49
  • With the determinant of A-xId. Turns out that is pretty simple, just $(-1)^nx^n+(-1)^{n-1}tr(A)x^{n-1}$ where tr(A) is the trace of A, which is n. All the other terms goes to 0. – Allonsy Jun 23 '14 at 02:53