3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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'x 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 $\lambda=0$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – RideTheWavelet Dec 17 '17 at 2:27
  • $\begingroup$ @RideTheWavelet Good point, thanks! It's always interesting to have multiple perspectives on the problem. $\endgroup$ – eranreches Dec 17 '17 at 11:42
2
$\begingroup$
  • 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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.