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How many distinct eigenvalues are there in the matrix.

$$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ \end{bmatrix} $$

I was wondering that is there any specific eigenvalues for matrices like this??

I hope I wouldn't have to find the determinant of this 4×4 matrix.

ArsenBerk
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demon
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  • Well, the determinant is clearly 0. How clear that is depends on what topics you've already covered, but basically since the rows are linearly dependent then the determinant is 0. –  Jan 04 '18 at 03:57
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    Moreover the multiplicity of the eigenvalue $0$ can be determined to be $3$ since this map collapses $\mathbb{R}^4$ onto a one-dimensional subspace. – Tob Ernack Jan 04 '18 at 04:00
  • https://math.stackexchange.com/questions/217521/what-are-the-eigenvalues-of-matrix-that-have-all-elements-equal-1 – Olivier Oloa Jan 04 '18 at 04:01
  • Once you know the fact from Tob Ernack, it's clear the other eigenvalue is 4 since the trace is the sum of eigenvalues. – JessicaK Jan 04 '18 at 05:11

3 Answers3

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Note that $$\left(\begin{matrix}1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\end{matrix}\right)\left(\begin{matrix} a\\ b\\ c\\ d\end{matrix}\right)=\left(\begin{matrix} a+b+c+d\\ a+b+c+d\\ a+b+c+d\\ a+b+c+d\end{matrix}\right)$$ If $a=b=c=d$, then $$\left(\begin{matrix}1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\end{matrix}\right)\left(\begin{matrix} a\\ a\\ a\\ a\end{matrix}\right)=4\left(\begin{matrix} a\\ a\\ a\\ a\end{matrix}\right)$$ Hence $4$ is an eigenvalue.

Also see that since all the columns of the matrix are same, the rank of the matrix is $1$. So $4$ is the only non zero eigenvalue. $0$ is the other eigenvalue, with eigenvector, for example $(a~-a~a~-a)^t$.

QED
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1

If $A$ is an $n$ by $n$ all-one matrix, then $A^2=nA$. If $\newcommand{\la}{\lambda}\la$ is an eigenvalue, with an eigenvector $v$ then $A^2v=nAv$, that is $\la^2 v=n\la n$. From this we get $\la=0$ or $n$. The other solutions show how to find eigenvectors for these eigenvalues. As the trace of $A$ is $n$, the eigenvalue $n$ appears with multiplicity one.

Angina Seng
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0

Let $e=(1,1,1,1)^T$ then $A=e e^T$.

Note that $\dim\ker A = 3$ and so $A$ has three eivenvalues at zero.

Suppose $Av=\lambda v$ for some non zero $v$ and $Av \neq 0$. We see that we can take $v=e$ (since ${\cal R} A = \operatorname{sp} \{ e \}$), and so $Ae = 4 e$ and so $\lambda = 4$.

Hence the eigenvalues are $0,0,0,4$.

In general, matrices of the form $uv^T$ are called dyads, and a similar analysis shows that they have $n-1$ zero eigenvalues and a single eigenvalue of $v^Tu$ (which could also be zero) corresponding to eigenvector $u$.

copper.hat
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