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Given any square matrix $A$ and its eigenvalue $\lambda$, if $A - \lambda I = 0$, then $(A - \lambda I) \mathbf{x} = \mathbf{0}$ becomes

$$(A - \lambda I)\mathbf{x} = \begin{bmatrix} 0 & \dots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \dots & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} 0 \\ \vdots \\ 0 \end{bmatrix} \ . $$

What are the corresponding eigenvectors of $\lambda$? Is it all the possible $n \times 1$ vectors in $\mathbb{R}^n$? If so, is this possible?

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  • $\begingroup$ Note that $A = \lambda I$ so every vector is an eigenvector. $\endgroup$
    – Maffred
    Apr 12, 2021 at 18:49
  • $\begingroup$ If $A-\lambda I = 0$, the $\lambda I = A$ so that $Av = \lambda v$ for all $v$. Unless I am missing something here. $\endgroup$ Apr 12, 2021 at 18:49
  • $\begingroup$ Perhaps you are skeptical because that would be infinitely many eigenvectors and you are under the impression that there should be fewer... but recall that the set of eigenvectors (along with the zero vector) for a particular eigenvalue is itself a vector subspace of $\Bbb R^n$ and so clearly is infinite in size, and that for an infinite subspace there are infinitely many different choices of bases that could describe the space. $\endgroup$
    – JMoravitz
    Apr 12, 2021 at 18:51
  • $\begingroup$ I asked this question since I have a system of ODEs problem where choosing a constant $b$ to be zero would make $A - \lambda I = 0$, and choosing it to be non-zero would give me two, distinct eigenvalues and a pretty standard process to evaluate. I want to argue that $b = 0$ is not possible (because the latter option is easier to solve) but it seems that it is possible, based on your replies that the eigenvalue to have all vectors to be its eigenvector and hence, $\mathbf{x} = \mathbf{k}e^{\lambda t}$ is a solution to the system, where $\mathbf{k}$ could be any vector in $\mathbf{R}^n$. $\endgroup$
    – sinclair
    Apr 12, 2021 at 18:58

2 Answers 2

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Well, $A = \lambda I$. Then every vector $x$ fulfills the equation $Av = \lambda v$, so all vectors ($\neq 0$) are eigenvectors with eigenvalue $\lambda$.

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The relation between an eigenvector ($v$) and the correspondent eigenvalue (\lambda) is

$$Av=\lambda v\Leftrightarrow (A-\lambda I)v=0 \quad (1)$$

So, once you find the solution, $\lambda$, of $\det (A-\lambda I)=0$ you have to place it back in the linear system $(1)$ and find the correspondent solution, $v$, of it.

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