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

  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.

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    $\begingroup$ 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. $\endgroup$ – hunter Oct 4 '18 at 0:27
  • $\begingroup$ @hunter I couldn't get the change of basis matrix. Can you elaborate more please? Thank you. $\endgroup$ – Zouhair El Yaagoubi Oct 4 '18 at 0:41
  • $\begingroup$ The image of $A$ is obviously one-dimensional. What is the nullity of this matrix and how does that relate to its eigenvalues? $\endgroup$ – amd Oct 4 '18 at 1:05
  • $\begingroup$ 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 $$\sigma(A)=\{0,n\}$$ $\endgroup$ – Chinnapparaj R Oct 4 '18 at 1:41
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$(1)$ 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) .

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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 $$ \begin{pmatrix} n & 0 & 0 & \ldots & 0 \\ 1 & 0 & 0 & \ldots & 0 \\ 1 & 0 & 0 & \ddots & 0 \\ 1 & 0 & 0 & \ldots & 0 \\ \end{pmatrix}. $$

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