# Determine $(n-1)$ eigenvectors of $f$ associated to $0$?

$$A=\begin{pmatrix}1&1&\cdots&1\\1&1&\cdots&1\\\vdots&\vdots&\ddots&1\\1&1&\cdots&1\end{pmatrix}\in\mathcal M_n(\mathbb R).$$

1. Prove that $$0$$ and $$n$$ are eigenvalue of $$A$$.
2. Determine the characteristic polynomial. (Hint: You can use the basis $$(e_1, e_2 - e_1, \cdots, e_n - e_1)$$)
3. Determine the eigenvalues of $$A$$.
4. Prove that $$f(e_1)$$ is an eigenvector.
5. Determine $$(n-1)$$ eigenvectors of $$f$$ associated to $$0$$.
6. Prove that there exists a basis $$\mathcal{B_2}$$ in which the matrix is of the form: $$D = \begin{pmatrix} n & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{pmatrix}$$

1. Taking the vector $$v = \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix}$$

We get: $$Av = nv$$.

But I could not see how $$0$$ can be an eigenvalue.

1. $$P_A(X) = \begin{vmatrix} 1 - X & 1 & \cdots & 1 \\ 1 & 1 - X & \cdots & 1 \\ \vdots & \vdots & \ddots & 1 \\ 1 & 1 & \cdots & 1 - X \end{vmatrix} = (1 - X)\begin{vmatrix} 1 & 1 & \cdots & 1 \\ 1 & 1 - X & \cdots & 1 \\ \vdots & \vdots & \ddots & 1 \\ 1 & 1 & \cdots & 1 - X \end{vmatrix}$$

I could not proceed to get an expression of $$P_f(X)$$ as I could not see how to use the hint.

1. Computing $$Ae_1 = \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix}$$ which is not related to the eigenvalues $$0$$ or $$n$$.

2. I am stuck in this question too.

• Write the matrix $A$ in the basis from part (2) (the only difficult part is the first column). This will make the problem much easier. – hunter Oct 4 '18 at 0:27
• @hunter I couldn't get the change of basis matrix. Can you elaborate more please? Thank you. – Zouhair El Yaagoubi Oct 4 '18 at 0:41
• The image of $A$ is obviously one-dimensional. What is the nullity of this matrix and how does that relate to its eigenvalues? – amd Oct 4 '18 at 1:05
• First of all the rank of $A$ is $1$, so it is not invertible and so $0$ is an eigenvalues of $A$. By dimension theorem, null space of $A$ has dimention $n-1$ which is same as the eigen space corresponding to $0$, so $0$ is an eigenvalue with multipilicity $n-1$ and hence the other eigenvalue must the trace $n$ of $A$. Hence $$\sigma(A)=\{0,n\}$$ – Chinnapparaj R Oct 4 '18 at 1:41

$$(1)$$ For $$\lambda=0$$, try $$v=\begin{pmatrix}1\\1\\\vdots\\1\\1-n\end{pmatrix}$$.

$$(2)$$ For instance, use the change of basis matrix (whose columns are the $$-e_1+e_i$$).

$$(4)$$ $$f(e_1)$$ is an eigenvector for the eigenvalue $$n$$.

$$(5)$$ Take the eigenvector in $$(1)$$ and move the $$1-n$$ around to different coordinates, i.e. put $$1-n$$ in the $$i$$th coordinate, and $$1$$'s everywhere else. $$n-1$$ of these are linearly independent.

$$(6)$$ Since it has a $$n-1$$ linearly independent eigenvectors for $$\lambda =0$$, and $$1$$ for $$\lambda =n$$, there is such a basis (consisting of eigenvectors) .

Let $$f_1 = e_1$$, $$f_2 = e_2 - e_1$$, ..., $$f_n = e_n - e_1$$ as in the hint. Note that $$\sum f_i = (\sum e_i) - (n-1) e_1$$.

Then $$A$$ annihilates $$f_2$$, $$f_3$$, ..., and $$f_n$$. Meanwhile, \begin{align*} Af_1 &= \sum_i {e_i} \\ &= \sum_i f_i + (n-1) e_1 \\ &= nf_1 + \sum_{i \neq 1} f_i, \end{align*} so the matrix for $$A$$ in the new basis is $$\begin{pmatrix} n & 0 & 0 & \ldots & 0 \\ 1 & 0 & 0 & \ldots & 0 \\ 1 & 0 & 0 & \ddots & 0 \\ 1 & 0 & 0 & \ldots & 0 \\ \end{pmatrix}.$$