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Let $A \in \mathbb{R}^{n \times n}$ and $w \in \mathbb{R}^n$. Suppose that, $w_i>0$ and $a_{i,j} = w_i / w_j$ for all $i,j=1,\dots,n$.

Note that from the construction comes that $\operatorname{rank} A=1$.

Prove that the eigenvectors of $A$ are a basis in $\mathbb{R}^n$, or equivalently prove that $A$ is diagonalizable.

user153012
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    More precisely, one ought to say something like, "Prove that $\mathbb{R}^n$ admits a basis of eigenvectors of $A$"---the set of eigenvectors of $A$ is infinite and hence never forms a basis. – Travis Willse Apr 24 '15 at 01:36

2 Answers2

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Note that your matrix can be written as product of a column matrix and a row matrix: $$ A=\pmatrix{w_1\\w_2\\\vdots\\w_n}\pmatrix{1/w_1&1/w_2&\ldots&1/w_n}, $$ which explains why it has rank$~1$ (provided that $n>0$; the case $n=0$ is trivial and I will henceforth not consider it). Now as you can for instance read in this answer, such a matrix is diagonalisable if and only if it has nonzero trace (the part that the condition is sufficient is easy: due to the rank, eigenvalue $0$ already has geometric multiplicity $n-1$, and there must be a nonzero eigenvalue because of, and equal to, the trace). Here the diagonal entries are all$~1$, so the trace is $n\neq0$ and $A$ is diagonalisable.

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Marc van Leeuwen
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Here is a start -- perhaps you can fill in the details:

We have $w = (w_1,\ldots,w_n)$ with every entry positive. Let $v = (\frac{1}{w_1},\ldots,\frac{1}{w_n})$, inverting each entry. Observe that, as a linear transformation, $A$ is given by $A(x) = (x \cdot v) w$, where "$\cdot$" is the Euclidean dot product.

  • Find the kernel of $A$ and its dimension.
  • Find the eigenspace of $\lambda=n$ and its dimension.
Mike F
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  • Actually, I think there is a shorter argument: once you have guaranteed that there is a nonzero eigenvalue, it automatically has an eigenspace of positive dimension. So now you have one eigenvector with a nonzero eigenvalue and $n-1$ of them with the zero eigenvalue, so you're done. – Ian Apr 24 '15 at 01:58
  • @Ian: I don't disagree with you, but I think I will leave this hint as is. Nice to know we have a (rank-1) projection. – Mike F Apr 24 '15 at 01:59
  • There is no eigenvalue $\lambda=1$ (unless $n=1$). – Marc van Leeuwen Apr 25 '15 at 11:53
  • @MarcvanLeeuwen: Oops! Thanks for letting me know. – Mike F Apr 25 '15 at 16:09
  • @Ian: By Marc's comment, I should have actually said "n times a rank-1 projection". – Mike F Apr 25 '15 at 16:13