I'm feeling dumb even asking this. But there might be a definition for this somewhat like why $1$ is not a prime number. Therefor this might be the right place to ask this question anyway.

Given the matrix $$\begin{bmatrix} 4 & 0 \\ 0 & 4 \\ \end{bmatrix}$$

One sees immediately that the eigenvalues are $4$ and $4$ and the corresponding eigenvectors $$\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$$ and $$\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$$

Assuming one doesn't see that or one tries to program this he would use $(A-\lambda_i E)v_i=0$ to calculate the eigenvectors. But using this in this really simple example leads to $$\begin{gather} \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}v=0 \end{gather}$$ So every solution would be correct for $v$.

Where is my mistake? I hope it is something obvious. I really hate it when there are special cases and one can not always use one scheme for all related problems.

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    $\begingroup$ In your particular example, every vector is an eigenvector with EV 4 - you're not doing anything wrong. $\endgroup$ Commented Jun 23, 2019 at 7:58
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    $\begingroup$ Every solution is valid for $v$ but we can easily see that each value within the eigenvectors are independent so we typically use the eigenvectors $[0,1]^T$, $[1,0]^T$ to represent this. $\endgroup$ Commented Jun 23, 2019 at 7:59
  • $\begingroup$ @PeterForeman that was one of those things which made me suspicious. I could choose two time the same eigenvector since every solution is correct, right? For example v1 and v2 = [1,1]. Wouldn't this mean they are not independent? I assumed the eigenvectors should in this case always be orthogonal to each other to show this independency. $\endgroup$
    – Mr.Sh4nnon
    Commented Jun 23, 2019 at 8:22
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    $\begingroup$ Yes, if you choose the same one twice, or a multiple of an eigenvector, then they’re not linearly independent. So? That doesn’t prevent you from finding some other pair of eigenvectors that is linearly independent. $\endgroup$
    – amd
    Commented Jun 23, 2019 at 17:22
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    $\begingroup$ If $\lambda$ is an eigenvalue then there is no distinguished eigenvector corresponding to $\lambda$. What you have is the corresponding eigenspace which is the kernel of $A-\lambda I$. This eigenspace is the often represented by a basis, but the choice of the basis is not unique. $\endgroup$
    – Claire
    Commented Jun 23, 2019 at 17:32

2 Answers 2


There is nothing wrong with your calculations. Note that if $v_1$ and $v_2$ are eigenvectors corresponding to an eigenvalue $\lambda $, so is $c_1v_1+c_2v_2$. In your case, note that $e_1$ and $e_2$ are basis elements for $\mathbb R^2$.


This happens for any $n \times n$ identity matrix since the eigenvectors are always orthogonal and hence they span the entire $\mathbb{R}^n$ space. Thus, any vector in the space is an eigenvector. Therefore, there is no mistake in your solution.

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    $\begingroup$ The question here is: Where is Mr. Sh4nnon's mistake? $\endgroup$
    – 311411
    Commented Aug 22, 2022 at 16:50
  • $\begingroup$ There isn’t one. But the matrix isn’t an identity matrix. I do like that one of the comments mentions eigenspaces, I think that would be a really nice term to see in an answer, $\endgroup$
    – fuzzyTew
    Commented Jan 28 at 4:01

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