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,
Best Answer
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 $\lambda =\underbrace{-1,-1, \ldots -1}_{n-1 \text{ times}}, n-1$.