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I don't think that computing the characteristic polynomial is the way to proceed for this problem. Also, the size of this matrix $$C=\begin{pmatrix}0&1&\dots&1\\ 1&0&\dots &1\\ \vdots &\vdots &\ddots &\vdots\\ 1&1&\dots &0\end{pmatrix}$$ is just given as $n\times n$.

So, instead, I can look at the matrix $I+C$. Then this is a matrix of all $1$'s.

We know that this matrix is rank-deficient and so has $\det (C+I)= 0$.

Then I have that

\begin{align}\det[I+C] &= \det[C+I] \\&=\det[C-(-1)I] \\&=0\end{align}

The last equality shows that $-1$ is an eigenvalue of the original matrix $C$, which is what I wanted.

But, how do I know its multiplicity?

And, how do I know whether there are any more eigenvalues of $C$?

Thanks,

Bowei Tang
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User001
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2 Answers2

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Consider the following $$C\begin{pmatrix}1\\1\\\vdots\\1\end{pmatrix}=(n-1)\begin{pmatrix}1\\1\\\vdots\\1\end{pmatrix}.$$ Thus $n-1$ is also an eigenvalue.

Multiplicity of $\lambda =-1$

You already have that $\lambda=-1$ is an eigenvalue. Observe that $\text{rank }(C+I)=1$, thus null space of $C+I$ has dimension $n-1$. In other words, eigenspace corresponding to the eigenvalue $\lambda=-1$ is spanned by $n-1$ linearly independent vectors. Thus multiplicity is $n-1$.

Moreover $\text{trace }C=0$ implies that sum of the eigenvalues must be $0$.

Thus the eigenvalues are $\lambda =\underbrace{-1,-1, \ldots -1}_{n-1 \text{ times}}, n-1$.

Bowei Tang
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Anurag A
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  • Hi @AnuragA, I like this nice intuitive argument, but I have a few follow-up questions: since sum of eigenvalues = trace = sum of diagonal elements, then the multiplicity of the eigenvalues really is its algebraic multiplicity, i.e., the number of times it shows up on a diagonalized matrix, if the matrix were diagonalizable. – User001 Oct 16 '15 at 07:10
  • Here you showed the geometric multiplicity for -1 is (n-1). But geometric mult $\le$ algebraic multiplicity, so we haven't said for sure yet that it is definitely (n-1), unless the operator were diagonalizable in which case the multiplicities are equal. Also what can we say about the multiplicity of the other eigenvalue (n-1)? Not much, it seems like. But what you showed agrees with trace(C) = sum of eigenvalues = 0. – User001 Oct 16 '15 at 07:10
  • So, I'm pretty sure we have the multiplicities correct, although not fully justified, though? Feel free to comment further on this, if you'd like. I might have possibly just overlooked a simple fact. The end goal is to just compute determinant of C = product of eigenvalues. So, I think we pretty much have all information... except for the multiplicity of the eigenvalues.... – User001 Oct 16 '15 at 07:10
  • Actually, by your argument, there can't be any more eigenvectors, and hence eigenvalues. You have found (n-1) linearly independent eigenvectors for -1, and one linearly independent eigenvector for (n-1); this forms a basis for $R^n$ already, and so this is a full set of linearly independent eigenvectors. Then the (n-1) geometric multiplicity is forced to agree with the algebraic multiplicity for -1, I think. And so your answer is complete, I think :-). Thanks so much @AnuragA. – User001 Oct 16 '15 at 07:34
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Let $v=\pmatrix{1\\\vdots\\1}$. Then $C=vv^T-I$. So the characteristic polynomial of $C$ is $$ p(\lambda)=|\lambda I-C|=|(\lambda+1) I-vv^T|=(\lambda+1)^n-Tr(vv^T)(\lambda+1)^{n-1}=(\lambda+1)^n-n(\lambda+1)^{n-1}=(\lambda+1)^{n-1}(\lambda-n+1) $$ for $vv^T$ is a Rank-$1$ matrix and all principal minors above $2$ are $0$.

Thus eigenvalues of $C$ are $-1$ with multiplicity of $n-1$ and $n-1$ with multiplicity of $1$.

Eugene Zhang
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