If there is an $n\times n$ matrix with one eigenvalue, that must mean that the equation $(A-λI)=0$ vector has only one nontrivial solution. But how would I find a matrix $A$ like that?
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$\begingroup$ Maybe play with diagonal matrices first? $\endgroup$– angryavianMay 3, 2018 at 4:22
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3 Answers
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the $0$ matrix works.${}{}{}{}{}$
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$\begingroup$ why was I downvoted so harshly? please be kind with ur fellow human $\endgroup$– AsinomásMay 3, 2018 at 4:51
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A diagonal matrix with only $\lambda$ on the diagonal. But you should be more general with $(A-\lambda I)^n=0$ instead. Then you might be able to put some $1$'s right above the diagonal. Try a bunch of those.
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Let's say we are working in $n$ dimensional space. We need to be clear about what we mean by one eigenvalue. Do you mean a single eigenvalue with geometric multiplicity $n$? In which case the matrix $cI$ fits the bill: it has only one eigenvalue, $c$, and the corresponding eigenspace is dimension $n$. I believe this is what you mean because you wrote that you want $A-\lambda I=0$ to have only one solution in $\lambda$, in which case $A=\lambda I$, where we just get to pick the constant. However, note that we don't find eigenvalues by finding solutions to $A-\lambda I = 0$, but rather finding solutions to $\det(A-\lambda I)=0$, this is an important distinction.
Or do you mean a single eigenvalue with geometric multiplicity $1$, in which case the matrix with $c$ along diagonal and $1$ along super diagonal is what we want. It's only going to have a one-dimensional eigenspace and there will be no other eigenvalues.
