Characteristic polynomial of a matrix of $1$'s

Solution 1:

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 $\lambda \neq 0$.

Solution 2:

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.

Solution 3:

Using the matrix determinant lemma,

$$\det \left( s \, \mathrm I_n - \mathbb 1_n \mathbb 1_n^\top \right) = s^n \left( 1 - \frac{1}{s} \mathbb 1_n^\top \mathbb 1_n \right) = s^n - n s^{n-1} = s^{n-1} \left( s - n \right)$$

linear-algebra matrices determinant characteristic-polynomial

Solution 4:

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 $A\cdot v_j$?