How to find the eigenvalues of this matrix, which has zeros on the diagonal and 1's elsewhere, without computing the characteristic polynomial,

Question

I don't think that computing the characteristic polynomial is the way to proceed for this problem. Also, the size of this matrix C is just given as "nxn".

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 determinant = 0.

Then I have that

$$det[I+C]$$ $$= det[C+I]$$ $$=det[C-(-1)I]$$ $$=0$$

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,

Answer 1

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 eigen value.

Multiplicity of $\lambda =-1$

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

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

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

Answer 2

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 $1$.