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I am calculating the characteristic polynomial for this matrix:

$$A = \begin{pmatrix} 1 & 2 & \cdots & n \\ 1 & 2 & \cdots & n \\ \vdots & \vdots & \cdots & \vdots \\ 1 & 2 &\cdots & n \end{pmatrix}$$ First I was asked to figure out that $0$ is an eigenvalue, and since it is not invertible then $0$ is an eigenvalue, and its' geometric multiplicity is $n-1$. Now I need to calculate the characteristic polynomial but I am finding this determinant hard!

$$\mbox{det}\begin{pmatrix} \lambda - 1 & -2 & \cdots & -n \\ -1 & \lambda - 2 & \cdots & -n \\ \vdots & \vdots & \cdots & \vdots \\ -1 & -2 & \cdots & \lambda -n \end{pmatrix} = ? $$

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    $\begingroup$ Hint: You know $0$ is an eigenvalue o multiplicity $n-1.$ Calculating the trace should tell you the other eigenvalue. $\endgroup$
    – Geoff Robinson
    Commented Jun 1, 2013 at 14:55

2 Answers 2

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The last eigenvalue $\lambda$ is the trace of the matrix $A$ so $$\lambda=1+2+\cdots+n=\frac{n(n+1)}{2}$$ hence the chararcteristic polynomial is $$\chi_A=x^{n-1}\left(x-\frac{n(n+1)}{2}\right)$$

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  • $\begingroup$ Why is the last eigenvalue the trace of the matrix? $\endgroup$
    – TheNotMe
    Commented Jun 1, 2013 at 15:02
  • $\begingroup$ @TheNotMe: you are aware that the trace is equal to the sum of all the eigenvalues, and you also know that $n-1$ of the eigenvalues are zero. Thus... $\endgroup$
    – J. M. ain't a mathematician
    Commented Jun 1, 2013 at 15:17
  • $\begingroup$ Is that always valid? I mean, $A$ is not a diagonal matrix, is your theorem still valid? $\endgroup$
    – TheNotMe
    Commented Jun 1, 2013 at 15:41
  • $\begingroup$ @TheNotMe To determine the Characteristic polynomial it suffices to know the eigenvalues of the matrix whatever it is diagonalizable or not diagonalizable $\endgroup$
    – user63181
    Commented Jun 1, 2013 at 15:48
  • $\begingroup$ Well, if you don't mind me asking an extra question. Given that exact above matrix $A$. How do we figure out what the minimum ploynomial is? I mean surely I can't try $n-1$ options to figure out which option gives $m_A(A) = 0$. Any idea? $\endgroup$
    – TheNotMe
    Commented Jun 1, 2013 at 15:52
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Suggestion: add to the first column all the others. We get $\lambda-\frac{n(n+1)}2$ for each entry. Then use linearity with respect to this column, and finally do $L_j\leftarrow L_j+jL_1$.

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