# Characteristic polynomial of a matrix of $1$'s

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

• Case $$n=2$$: I obtained $$p(\lambda)=\lambda^2-2\lambda$$ .

• Case $$n=3$$: I obtained $$p(\lambda)=-\lambda^3+3\lambda^2$$.

• Case $$n=4$$: I obtained $$p(\lambda)=\lambda^4 - 4\lambda^3$$.

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. Can somebody help me? Or give me a hint?

• 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.) Jun 3 '12 at 21:00
• I really dont understand the problem, could any one explain me? Jun 6 '12 at 14:16
• Jul 17 '18 at 0:55

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 $\lambda \neq 0$.

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.

• span of 17 seconds, you, Dennis, Marvis. I was going to answer, but. Jun 3 '12 at 21:05

Using the matrix determinant lemma,

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

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 $A\cdot v_j$?

We wish to compute the determinant of the $$n\times n$$ matrix $$M=\begin{bmatrix} 1-\lambda&1&1&\cdots&1\\ 1&1-\lambda&1&\cdots&1\\ 1&1&1-\lambda&\cdots&1\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&1&1&\cdots&1-\lambda \end{bmatrix}\tag1$$ Here are two approaches, the second of which uses a generalization of the lemma cited in Rodrigo de Azevedo's answer.

Finding a Similar Matrix

Note that the $$n-1$$ dimensional subspace orthogonal to $$\begin{bmatrix}1&1&1&\cdots&1\end{bmatrix}^\text{T}$$ is multiplied by $$-\lambda$$. (If the subspace is orthogonal to a given vector, we can subtract that vector from each row of the matrix when operating on that subspace.)

$$\begin{bmatrix}1&1&1&\cdots&1\end{bmatrix}^\text{T}$$ is multiplied by $$n-\lambda$$. (Just compute it.)

Thus, $$M$$ is similar to $$\begin{bmatrix} -\lambda&0&0&\cdots&0\\ 0&-\lambda&0&\cdots&0\\ 0&0&-\lambda&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&n-\lambda\\ \end{bmatrix}\tag2$$ Therefore, $$\det(M)=(-\lambda)^{n-1}(n-\lambda)\tag3$$

$$\boldsymbol{\det(\lambda I_n-AB)=\lambda^{n-m}\det(\lambda I_m-BA)}$$

$$\det(\lambda I_n-AB)=\lambda^{n-m}\det(\lambda I_m-BA)\tag4$$
Let $$A=\begin{bmatrix}1&1&1&\cdots&1\end{bmatrix}^\text{T}$$ and $$B=A^\text{T}$$, then $$M=AB-\lambda I_n$$.
Furthermore, $$m=1$$ and $$BA=\begin{bmatrix}n\end{bmatrix}$$.
$$(4)$$ then says that \begin{align} \det(-M) &=\det(\lambda I_n-AB)\tag{5a}\\[2pt] &=\lambda^{n-1}\det(\lambda I_1-BA)\tag{5b}\\[2pt] &=\lambda^{n-1}\det(\lambda I_1-\begin{bmatrix}n\end{bmatrix})\tag{5c}\\ &=\lambda^{n-1}(\lambda-n)\tag{5d} \end{align} which, since $$\det(-M)=(-1)^n\det(M)$$, becomes $$\det(M)=(-\lambda)^{n-1}(n-\lambda)\tag6$$