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Diagonalize this nxn matrix

$\begin{bmatrix} 0&1&0&&&...&&&0\\ 0&0&1\\ &&0&1\\ &&&0&1\\ &&&&0&.\\ &&&&&.&.\\ &&&&&&.&.\\ &&&&&&&.&1\\ 1&&&&&&&&0\\ \end{bmatrix} $

I have so far found that it is unitary, and by the spectral theorem it is conjugate to some diagonal matrix. But I have no idea what that matrix is.

Det$\begin{bmatrix} -\lambda&1&0&...&0\\ 0&-\lambda&1\\ &&-\lambda&1\\ &&&-\lambda&1\\ &&&&-\lambda&.\\ &&&&&.&.\\ &&&&&&.&.\\ &&&&&&&.&1\\ 1&&&&&&&&-\lambda\\ \end{bmatrix} =0=\lambda^n-1$

$\therefore \lambda^n=1$ then I get that the only eigenvector is v=(1...1) which makes the matrix non-diagonalizable

I figure that I am not supposed to be trying to solve this literally but I don't know any relevant parts of spectral theorem to create a diagonalization

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    $\begingroup$ $\lambda^n=1$ has $n$ solutions in the complex numbers. $\endgroup$ – Lord Shark the Unknown Feb 3 at 6:29
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    $\begingroup$ I found the eigenvectors for each eigenvalue in general form. where $x_j=(1, \lambda_j, \lambda_j^2, ... , \lambda_j^{N-1})$ How do I use this to diagonalize the matrix? $\endgroup$ – Big_Tubbz Feb 3 at 6:51
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Let's solve this problem first over the field of complex numbers. You have successfully found the characteristic polynomial $\lambda^n=1$, so we know that the the matrix has $n$ complex eigenvectors $\lambda_k=e^{2\pi ik/n}$ and you have also found the eigenvectors of form $v_k = (1,\lambda_k,\ldots,\lambda_k^{n-1})$.

So that gives you a diagonal matrix $D$ and transformation matrix $S$: $$ D = \begin{pmatrix}\lambda_1&0&\cdots&0\\0&\lambda_2&\cdots&0\\ \vdots& \vdots& \ddots& \vdots\\0&0&\cdots&\lambda_n \end{pmatrix}, \qquad S=\frac1{\sqrt{n}}\begin{pmatrix}1&\lambda_1&\lambda_1^2&\cdots&\lambda_1^{n-1}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&\lambda_n&\lambda_n^2&\cdots&\lambda_n^{n-1} \end{pmatrix} $$

When you consider this problem over the field of real numbers, the fact that $\lambda^n=1$ has complex solution gives you an idea, that the matrix is not diagonalizable.

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