# Eigenvalues of same-row matrices

It has previously been discussed here that the eigenvalues of an all-ones $n \times n$ matrix $A$ such as the following are given by $0$ with multiplicity $n - 1$ and $n$ with multiplicity $1$, hence a total multiplicity of $n$ which means that the given matrix is diagonalizable. $$A = \begin{bmatrix} 1 & 1 & \cdots & 1 \\ 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \\ \end{bmatrix}$$

I recently wrote an exam that asked us to diagonalize a matrix with multiple (3) rows that contained the same entries, so I was wondering if there was some general case to apply.

Thus the question I am asking is given the following $n \times n$ matrix A, what are its eigenvalues? $$A = \begin{bmatrix} a_1 & a_2 & \cdots & a_n \\ a_1 & a_2 & \cdots & a_n \\ \vdots & \vdots & \ddots & \vdots \\ a_1 & a_2 & \cdots & a_n \\ \end{bmatrix}$$ For the sake of simplicity, lets first assume that $a_1, a_2, \ldots, a_n \in \mathbb{R} - \{0\}$; however, what happens if any (or all) are zero?

It seems logical that there be the eigenvalue $0$ with $n - 1$ multiplicity since the rank of this matrix will be $1$ (assuming at least one nonzero entry), and that the other eigenvalue be the sum of entries on the diagonal by observation $a_1 + a_2 + \cdots + a_n$ with $1$ multiplicity. I could not, however, write a formal proof for that second statement.

You can say a lot more about the matrix you presented. Lets define the following two vectors

$$u=\begin{pmatrix}1\\1\\\vdots\\1\end{pmatrix} \:{\rm and}\:v=\begin{pmatrix}a_{1}\\a_{2}\\\vdots\\a_{n}\end{pmatrix}$$

Then your matrix is exactly

$$A=uv^{T}$$

Assume $$v\neq 0$$ (because the zero matrix is a trivial case)

First Case: $$\sum_{i=1}^{n}a_{i}\neq 0$$.

You can easily see that

• The eigenvalue $$\lambda=0$$ is of multiplicity $$n-1$$, with $$n-1$$ linearly independent eigenvectors given by any basis of the subspace $${\rm Span}\left\{v\right\}^{\perp}$$ (that's true because $${\rm Span}\left\{v\right\}\oplus{\rm Span}\left\{v\right\}^{\perp}=\mathbb{R}^{n}$$ and $${\rm Span}\left\{v\right\}$$ is of dimension $$1$$).
• The eigenvalue $$\lambda=\sum_{i=1}^{n}a_{i}$$ corresponds to the eigenvector $$u$$, since

$$Au=uv^{T}u=\left(v^{T}u\right)u=\left(\sum_{i=1}^{n}a_{i}\right)u$$

Therefore, in this case you have $$n$$ linearly independent eigenvectors and the matrix is diagonalizable.

Second Case: $$\sum_{i=1}^{n}a_{i}=0$$.

In this case the first point still applies, but the matrix is not diagonalizable since it has only $$n-1$$ linearly independent eigenvectors (you can't have $$n$$ linearly independent eigenvectors with eigenvalue $$\lambda=0$$, unless the matrix is zero). It does, however, admits the following Jordan's canonical form

$$A\sim\begin{pmatrix}J_{2}\left(0\right)&0_{2\times\left(n-2\right)}\\0_{\left(n-2\right)\times2}&0_{\left(n-2\right)\times\left(n-2\right)}\end{pmatrix}=J_{2}\left(0\right)\oplus 0_{\left(n-2\right)\times\left(n-2\right)}$$

where

$$J_{2}\left(0\right)=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$

is a $$2\times 2$$ Jordan's block of eigenvalue $$\lambda=0$$.

• Since $\sum a_i=u^T v,$ you could view the failure in the second case to be a result of the fact that the would-be $n$th eigenvector $u$ already belongs to $\mathrm{span}(v)^{\perp}$, which is spanned by the first $n-1$ eigenvectors. – RideTheWavelet Dec 17 '17 at 2:27
• @RideTheWavelet Good point, thanks! It's always interesting to have multiple perspectives on the problem. – eranreches Dec 17 '17 at 11:42
• If all the columns are zero, it is the zero matrix which of course must be diagonalizable since the zero matrix is a diagonal matrix. $$0=I\cdot 0\cdot I$$

• If at least one of the column is non-zero, then the rank of the matrix is $1$ and the nullity is $n-1$. We check that the all-$1$ vector is an eigenvector and the eigenvalue is $\sum a_i$. Hence if $\sum_i a_i \neq 0$, then the matrix is diaognalizable since the geometry multiplicity is equal to the algebraic multiplicity.

• However, if one of the column is non-zero and $\sum_i a_i$ is equal to $0$. Since $\operatorname{tr}(A)=0=\sum_i \lambda_i$ and the nullity is $n-1$, we know the remaining eigenvalue must also be $0$. Suppose on the contrary that it is diagonalizable, then it is similar to the zero matrix, which shows that the matrix itself is the zero matrix, which is a contradiction since we assume at least one of the column is non-zero.

For example $A=\begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}$ is not diagonalizable. Let the two eigenvalue be $\lambda_1$ and $\lambda_2$, we know that $\lambda_1+\lambda_2=0,$ and we know that at least one of them is zero, hence both of them must be zero. If it is diagonalizable. then the matrix $A=P^{-1}\cdot 0 \cdot P = 0$ which is a contradiction.

Since $\operatorname{tr}\left(A\right) = \sum_i \lambda_i$, the only nonzero eigenvalue must be the sum of the diagonal elements.