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This is a followup to my question here.

Let $A$ be the $n \times n$ matrix over a field of characteristic 0, all of whose entries are 1. Is $A$ diagonalizable?

Community
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2 Answers2

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$A$ is always diagonalisable when the field has characteristic 0 or characteristic $>n$.

Let $e_i$ denotes the vector with a $1$ at the $i$-th position and zeros elsewhere. Let $P$ be the matrix whose first and second columns are respectivelly $\sum_je_j$ and $e_1-e_2$ and whose $j$-th column is $e_2-e_j$ when $j\ge3$. Clearly, the columns of $P$ are eigenvectors of $A$. It is not hard to show that $\det P=(-1)^{n+1}n$. Therefore $P$ is invertible and the aforementioned eigenvectors form an eigenbasis when the characteristic of the field is either $0$ or $>n$.

user1551
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Here is a matrix $P$ that I made up some time ago. Note that $P$ is not orthogonal, although the columns are pairwise orthogonal.

10: $$ \left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \end{array} \right). $$

3: $$ \left( \begin{array}{rrr} 1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 0 & 2 \\ \end{array} \right). $$

4: $$ \left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 2 & -1 \\ 1 & 0 & 0 & 3 \\ \end{array} \right). $$

5: $$ \left( \begin{array}{rrrrr} 1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 \\ 1 & 0 & 0 & 0 & 4 \\ \end{array} \right). $$

The columns of $P$ are of varying lengths; for the 10 by 10 case depicted, lengths $ \sqrt{10}, \sqrt{2}, \sqrt{6}, \sqrt{12},..$ All that is necessary to make an orthogonal matrix $Q$ out of this is to divide each column by its length. In turn, this tells us that $\det P = \pm n!,$ as $\det Q = \pm 1.$ The result is $Q^T Q = Q Q^T = I$ and $Q^T = Q^{-1}.$ Meanwhile, $Q^T A Q= Q^{-1}AQ$ is diagonal, in this case the diagonal elements are the eigenvalues.

Will Jagy
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  • I'm not sure how this fits the question. Maybe the question changed, but I see no note about this. – egreg Dec 08 '15 at 20:40
  • @egreg, given $A$ with all ones, and my $P,$ we get $P^T A P = D$ diagonal. There are two commonly confused types of diagonalization, for general matrices $R^{-1} A R,$ for symmetric matrices $S^T A S.$ The overlap is orthogonal diagonalization, my $Q^T A Q.$ If the OP had followed up with questions in comment, I could have explained. – Will Jagy Dec 08 '15 at 20:48
  • @egreg , compare http://math.stackexchange.com/questions/395634/given-a-4-times-4-symmetric-matrix-is-there-an-efficient-way-to-find-its-eige where the OP mistakenly thought he could find the eigenvalues. Either that, or the whole thing was in a numerical linear algebra course, but that was not my impression. I see, it was your answer there that said the eigenvalues were hopeless. Anyway, I like the answer by el.salvador and now know how to do his algorithm, my own way though. – Will Jagy Dec 08 '15 at 20:53
  • @egreg, well, anyway, the students here do not really know, is the problems they post, whether a symmetric matrix is being regarded as a linear transformation or as a quadratic form. I am happiest with quadratic forms. – Will Jagy Dec 08 '15 at 21:12
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    When I see “diagonalizable”, I always think to eigenvalues. – egreg Dec 08 '15 at 21:42