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As in subject: given a matrix $A$ of size $n$ with all elements equal exactly 1.

What are the eigenvalues of that matrix ?

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3 Answers 3

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Suppose $\,\begin{pmatrix}x_1\\x_2\\...\\x_n\end{pmatrix}\,$ is an eigenvector of such a matrix corresponding to an eigenvalue $\,\lambda\,$, then

$$\begin{pmatrix}1&1&...&1\\1&1&...&1\\...&...&...&...\\1&1&...&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\...\\x_n\end{pmatrix}=\begin{pmatrix}x_1+x_2+...+x_n\\x_1+x_2+...+x_n\\.................\\x_1+x_2+...+x_n\end{pmatrix}=\begin{pmatrix}\lambda x_1\\\lambda x_2\\..\\\lambda x_n\end{pmatrix}$$

One obvious solution to the above is

$$W:=\left\{\begin{pmatrix}x_1\\x_2\\..\\x_n\end{pmatrix}\;;\;x_1+...+x_n=0\right\}\,\,\,,\,\,\lambda=0$$

For sure, $\,\dim W=n-1\,$ (no need to be a wizard to "see" this solution since the matrix is singular and thus one of its eigenvalues must be zero)

Other solution, perhaps not as trivial as the above but also pretty simple, imo, is

$$U:=\left\{\begin{pmatrix}x_1\\x_2\\..\\x_n\end{pmatrix}\;;\;x_1=x_2=...=x_n\right\}\,\,\,,\,\,\lambda=n$$

Again, it's easy to check that $\,\dim U=1\,$ .

Now, just pay attention to the fact that $\,W\cap U=\{0\}\,$ unless the dimension of the vector space $\,V\,$ we're working on is divided by the definition field's characteristic (if you're used to real/complex vector spaces and you aren't sure about what the characteristic of a field is disregard the last comment)

Thus, assuming this is the case, we get $\,\dim(W+U)=n=\dim V\Longrightarrow V=W\oplus U\,$ and we've thus found all the possible eigenvalues there are.

BTW, as as side effect of the above, we get our matrix is diagonalizable.

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    $\begingroup$ what will be the characteristic polynomial of such matrix? $x^{n-1} (x-n)=0$ ? $\endgroup$
    – Marso
    Sep 11, 2013 at 9:41
  • $\begingroup$ Why is $dim W =n-1$? $\endgroup$
    – rose
    May 23, 2019 at 18:29
  • $\begingroup$ @math Because in the conditions defining $\;W\;$ you have $\;n-1\;$ free variables and one dependent in the other ones...or because $\;x_1+\ldots+x_n=0\;$ is a hyperplane in an $\;n-\,$ dimensional space, or because the kernel of any non zero linear functional on an $\;n\,$ dimensional vector space is always a maximal subspace...Choose your explanation. $\endgroup$
    – DonAntonio
    May 23, 2019 at 20:15
  • $\begingroup$ Sorry, that was straightforward. Somehow, it wasn't clear to me instantly. $\endgroup$
    – rose
    May 23, 2019 at 20:18
  • $\begingroup$ There are $n-1$ eigenvectors corresponding to the eigenvalue of $0$. But how do you actually choose the $n-1$ eigenvectors? Do you choose $n-1$ eigenvectors that span an $n-1$ subspace with the constraint that the eigenvector sum to zero (i.e., orthogonal to the 1 vector)? $\endgroup$
    – 24n8
    Jul 26, 2020 at 5:55
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HINT $$\begin{bmatrix}1 & 1 & \cdots & 1\\ 1 & 1 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & \cdots & 1\\\end{bmatrix}_{n \times n} = \begin{bmatrix} 1\\ 1 \\ \vdots \\ 1 \end{bmatrix}_{n \times 1} \begin{bmatrix} 1 & 1 & \cdots & 1\end{bmatrix}_{1 \times n}$$ What are the eigenvalues of $uv^T$, where $u$ and $v$ are column vectors?

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Since it looks like a homework question, here are a few hints: What is the rank of the matrix? What does this tell about the eigenvalues? Can you guess a certain eigenvector (also of a specific very simple structure)?

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