# Calculate the real eigenvalues and affiliated eigenspace of a square matrix whose elements are all 1

Question: Let $A \in M(n,n,\mathbb{R}), a_{ij} = 1$ for all $i,j$. Calculate the real eigenvalues and the affiliated eigenspace of $A$.

So first of all what I would be trying to calculate are values of $\lambda$ such that $0=\text{det}\left( \begin{array}{cccc} 1- \lambda & 1 & \cdots & 1 \\ 1& 1- \lambda & & \vdots\\ \vdots& &\ddots& 1\\1&1&\cdots&1-\lambda \end{array}\right)$, right?

I've come to the conclusion that in terms of $n$ the eigenvalues will be equal to $(-1)^{n}\lambda^{n} + (-1)^{n-1}n\lambda^{n-1}$. This was the result of trial and error and eventually the comparison with the theorem that for a characteristic polynomial of a square matrix:

1) $\alpha_n=(-1)^{n}$

2) $\alpha_{n-1}=(-1)^{n-1}\text{trace}(A)$

3) $\alpha_0=\text{det}(A)$

That is, a matrix full of $1$'s is going to have det$=0$ and trace$(A)=n$... However I am now stuck, how do I justify the observation that the rest of the coefficients will be zero?

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Maybe it's easier if you think in terms of a system of equations that $AX=\lambda X$ that you have to solve. What do you know about the matrix $A$, obviously, it has rank $1$ while it is $n$-dimensional. This immediately implies it has a degenerate eigenvalue $0$ (degeneracy $n-1$) and another eigenvalue which then must be equal to the trace of the matrix, thus $n$. – Raskolnikov Jan 18 '11 at 21:52
@Raskolnikov, thanks for the comment, that sounds like a better way to think of it. I'm still a bit confused about why rank$1$ implies the eigenvalue which must be equal to the trace of the matrix... – ghshtalt Jan 18 '11 at 22:03
See Arturo's reply. But, if you know that all other $n-1$ eigenvalues are $0$ and the trace is equal to the sum of all eigenvalues which is also equal to $n$, then I guess you get it already. – Raskolnikov Jan 18 '11 at 22:07
@user3711: I'm confused about what you are confused about. The trace is equal to the coefficient of $\lambda^{n-1}$ in the characteristic polynomial (up to sign), and if the characteristic polynomial splits then the trace is the sum of the eigenvalues; the fact that $A$ has rank $1$ has something to do with figuring out what all the eigenvalues (with multiplicities) are, but not with the fact that the trace equals the coefficient of $\lambda^{n-1}$. – Arturo Magidin Jan 18 '11 at 22:08
You can check that $A^2=nA$, so $Ax=\lambda x$ implies $\lambda^2x=A^2x=nAx=n\lambda x$. If $x$ and $\lambda$ are nonzero, this shows that $\lambda=n$. – Jonas Meyer Jan 18 '11 at 22:20

There is an easier argument. Let's try to see directly what the eigenvalues and eigenvectors are. (I'll assume $n>1$. If $n=1$, $A=[1]$ and the only eigenvalue is 1 and any non-zero real(=$1\times 1$ matrix) is an eigenvector.)

First, since any two rows are dependent, the matrix $A$ is not invertible, so $0$ is an eigenvalue. Let $x=(a_1,\dots,a_n)^T$ be an eigenvector ($T$ denotes the transpose, so I do not have to write column vectors).

Then $Ax=0$ iff $a_1+\dots+a_n=0$. Note that, given any $a_1,\dots,a_{n-1}$, there is exactly one $a_n$ that makes this equation true. So: The space of eigenvectors corresponding to 0 has dimension $n-1$.

An easy way of finding a basis for this space is to take the vectors (1): $$(1,0,0,\dots,0,-1)^T,(0,1,0,0,\dots,-1),\dots,(0,0,\dots,0,1,-1)^T.$$

We are left to find the non-zero eigenvalues. Since the trace of $A$ is $n$, and we already located $n-1$ eigenvalues, the remaining one is $n$. Let $x=(b_1,\dots,b_n)^T$ be an eigenvector. Then $Ax=nx$, or $b_1+\dots+b_n=nb_1=nb_2=\dots=n b_n$, so we can take (2): $$x=(1,1,\dots,1)^T.$$

Any other eigenvector is either a (non-zero) linear combination of the vectors listed in (1), or a non-zero multiple of the vector in (2).

(Note that the case $n=1$ is a particular case of what we found.)

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First: because each row of $A$ adds up to $n$, then you know that $(1,1,\ldots,1)^T$ is an eigenvector with eigenvalue $n$. (This is an often useful shortcut: if each row of $A$ adds up to $k$, then $k$ is an eigenvalue with eigenvector $(1,1,\ldots,1)^T$; if every column of $A$ adds up to $\ell$, then $\ell$ is an eigenvalue, though we don't know a priori what an eigenvector may be).

Second, because $A$ is not invertible, you know that $0$ is an eigenvalue of $n$. The dimension of the eigenspace of $0$ is the nullity of $A$, which is easy to check is equal to $n-1$. That means that the multiplicity of $\lambda=0$ as an eigenvalue of $A$ is at least $n-1$; you also know that the eigenvalue $\lambda=n$ has multiplicity at least $1$. And you know that the sum of the multiplicities of all the eigenvalues is at most $n$.

So that means that you already have all the eigenvalues with their multiplicities: $\lambda=0$ is an eigenvalue with algebraic multiplicity $n-1$, and $\lambda=n$ is an eigenvalue with algebraic multiplicity $1$. That means that the characteristic polynomial is divisible by $\lambda^{n-1}(\lambda-n)$. Since the characteristic polynomial is of degree $n$ and leading coefficient $(-1)^n$, this means that the characteristic polynomial must be $(-1)^n\lambda^{n-1}(\lambda-n) = (-1)^n\lambda^n - (-1)^nn\lambda$.

Now you need to find the eigenspaces, of course...

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This all makes much more sense to me after looking up 'multiplicity' :) – ghshtalt Jan 18 '11 at 22:26
@user3711: The "algebraic multiplicity" of an eigenvalue is just its multiplicity as a root of the characteristic polynomial. The "geometric multiplicity" of an eigenvalue is the dimension of the eigenspace. It is a theorem that if $\lambda$ is an eigenvalue, then $1\leq\text{geom. mult.}(\lambda)\leq\text{alg. mult.}(\lambda)$. – Arturo Magidin Jan 19 '11 at 3:02