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I am trying to calculate the characteristic polynomial of the matrix $n\times n$, $A=\{a_{ij}=1\}$.

When $n=2$, I obtained $p(\lambda)=\lambda^2-2\lambda$ .

In the case $n=3$, $p(\lambda)=-\lambda^3+3\lambda^2$.

For $n=4$, $p(\lambda)=\lambda^4 - 4\lambda^3$.

I guess that for the general case, we have $p(\lambda)=(-1)^n\lambda^{n}+(-1)^{n-1}n\lambda^{n-1}$.

I tried to use induction, but it didn't work, unless I've done wrong

Somebody can help me or give me a hint

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  • $\begingroup$ I think it's easier to just compute all of the eigenvalues. The eigenvectors are easy to write down. (Also, you should be using a definition of the characteristic polynomial that makes it monic.) $\endgroup$ – Qiaochu Yuan Jun 3 '12 at 21:00
  • $\begingroup$ I really dont understand the problem, could any one explain me? $\endgroup$ – Marso Jun 6 '12 at 14:16
  • $\begingroup$ Also see math.stackexchange.com/questions/2853981/… $\endgroup$ – user538518 Jul 17 '18 at 0:55
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Note that the matrix $$A = e e^T$$ where $e = \begin{pmatrix}1\\1\\1\\\vdots\\1\\1 \end{pmatrix}_{n \times 1}$.

Hence, $A^2 = \left(ee^T \right) \left(ee^T \right)= e \left(e^T e \right) e^T = n ee^T = nA$.

This clearly indicates that the matrix is a rank one matrix. Hence it must have $n-1$ eigenvalues as $0$. The only non-zero eigen value if $\lambda =n$, since we have $\lambda^2 = n \lambda$ and $\lambda \neq 0$.

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The trace is $n$. The eigenvalue $0$ has multiplicity $n-1$. From this we can write down the characteristic polynomial without any computation. Or else we can pick up the eigenvalue of $n$ by noting that the all $1$'s vector times our matrix is the all $n$'s vector.

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  • $\begingroup$ span of 17 seconds, you, Dennis, Marvis. I was going to answer, but. $\endgroup$ – Will Jagy Jun 3 '12 at 21:05
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Hint: Denote $v=(1,1,...,1)$ and $v_j=(1,0,..,0,-1,0,...,0)$ (all zeroes except for the first, where there is $1$ and the $j$th, where there is $-1$) (all column vectors). What happens when you multiply $A\cdot v$ and $A\cdot v_j$?

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