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I'm trying to find eigenvector that dotproduct < 0, but without so luck,try many different matrixs in wolfram alpha, then I find out, that if two vectors $v_1$ and $v_2$ 's dotproduct < 0, I can always negate one of them, since −$v_1$ would also be eigenvector of a matrix. so my question is:

1.for the wolfram alpha eigenvector, when the result eigenvectors dotproduct>0, is it happen because the algorithm just picks the >0 eigenvectors? If it finds dotproduct<0,then it just flip the sign of one eigenvector? Or the algorithm is just random, It can generate dotproduct>0 or dotproduct<0, eigenvectors, I'm just happen to see all the dotproduct>0 case?

Edit:I change the subject,sorry for the confusion. suppose I have a set of eigenvectors {$v_1$,$v_2$,$v_3$,...} which have some dotproduct between two elements < 0, is it possible to always find a way to reconstruct the eigenvectors set to make it all dotproduct >=0. take (1,2,0),(5,0,1),(-1,1,1) as examples, this 3 eigenvectors have some negative dotproduct between them, but if they're the eigenvectors of I, then I can always change the eigenvectors set to another set which satisfy dotproduct>0 condition.

Then the same eigenvector under different eigenvalue condition,w.r.t eigenvalue = 1,2,3 yields

$\left[ \begin{array}{ccc}\frac{27}{13}&-\frac{7}{13}&-\frac{5}{13}\\-\frac{4}{13}&\frac{15}{13}&\frac{20}{13}\\-\frac{2}{13}&\frac{1}{13}&\frac{36}{13}\\\end{array} \right]$

since the eigenvalues are all different, I have no freedom to change any of these 3 eigenvectors so the result is a not-modifiable set with some negative dotproduct.

But then what happened If I have some geometric multiplicity at my disposal, what if whenever I have some problematic $v$, I have at least one more dimension to choose from, say I constructed $v_1$,$v_2$, but at $v_3$ I can't satisfy $v_1$ $\cdot$ $v_3$ >=0 and $v_2$ $\cdot$ $v_3$ >=0 simultaneously, but $v_3$ and $v_4$ have same $\lambda$ and form a eigenspace, I can change $v_3$ and $v_4$ to $v_3'$ and $v_4'$ to make it both satisfy dotproduct>=0 condition.

let's start with the simplest geometric multiplicity of 2, whenever I have some problematic $v$, its eigenspace is at least 2 if it's fresh,

for a 4-dimensional vector, that would mean $v_3$ and $v_4$ form dim-2 eigenspace, and then I choose $v_3'$ instead $v_3$, then for $v_4$, it's not fresh(fresh I mean compose with new vector with eigenspace, here since $v_3$ already form eigenspace with $v_4$, $v_4$ can only be chosen from this eigenspace).

femto
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  • Related. Better show us your matrix instead of telling us a saga about WA. – Kurt G. Mar 18 '24 at 07:14
  • For question 1, many matrix would do, just type eigenvalues {{1,2},{3,4}} in wolfram alpha or eigenvalues {{2,2},{3,4}} For question 2, I can't find a set of eigenvectors that have dotproduct < 0 – femto Mar 18 '24 at 07:25
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    You can if you refrain from clinging only onto WA: If $v_1,v_2$ are eigenvectors that have positive dot product then what are $-v_1,v_2,?$ – Kurt G. Mar 18 '24 at 07:33
  • I don't quite understand you, If 1,2 are eigenvectors that have positive dot product then what are −1,2? then they become negative, but that's not my question. my question is I have a proposition: Ha, all eigenvectors can be expressed as dotproduct>=0 set, and then I'm trying to find a counter-example myself. – femto Mar 18 '24 at 07:41
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    Answer to my hint: $-v_1,v_2$ are eigen vectors that have a negative dot product. Show me an interesting $3\times 3$-matrix now. – Kurt G. Mar 18 '24 at 07:45

2 Answers2

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Let $V$ be any $n\times n$ matrix such that $$ V^TV=P:=\pmatrix{n&-1&\cdots&-1\\ -1&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&-1\\ -1&\cdots&-1&n}. $$ (Such a matrix exists because $P=(n+1)I_n-ee^T$ is positive definite.) Denote the $j$-th column of $V$ by $v_j$. By construction, each $v_j$ is an eigenvector of $A=I$ and $\langle v_i,v_j\rangle=-1$ whenever $i\ne j$.

user1551
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  • thanks for providing an example, I think you mean $\langle v_i,v_j\rangle=-n-2$, since there are two n(-1) ,and then (n-2)(-1)*(-1) which results -n-2 < 0.But I mean, provided I have the freedom the change the eigenvector,can I construct all positive dotproduct. – femto Mar 18 '24 at 10:27
  • @femto No. I did mean $-1$. Note that $\langle v_i,v_j\rangle=v_i^Tv_j$ is the $(i,j)$-th entry of $V^TV$, which is $-1$. If you want all dot products to be positive, you may simply take $V$ as any triangular matrix such that $v_{ij}>0$ when $i\le j$ and $v_{ij}=0$ when $i>j$. Then the columns of $V$ are eigenvectors of the identity matrix. If you want to assign different eigenvalues to the eigenvectors, you may also consider $A=VDV^{-1}$ where $D$ is a diagonal matrix with your desired eigenvalues. – user1551 Mar 18 '24 at 11:14
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Too long for a comment.

When you have three orthogonal eigenvectors $$ v_1,v_2,v_3 $$ to the same eigenvalue you can set $$ w_1:=v_1\,,\;w_2:=v_2-v_1-v_3\,,\;w_3=v_3-v_2-v_1\,. $$ The $w_1,w_2,w_3$ are eigenvectors to that eigenvalue. Calculate the dot products between them and realize that there will be simple condition on the length of the $v_i$ that makes those dot products all negative. Changing the length of $v_i$ obviously does not change the fact that they (hence also the $w_i$) are eigenvectors.

Kurt G.
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