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I'm asking a sensitive question.

Suppose, you are a student who is taking an exam on linear algebra. Suppose you're encountering a question asking you to compute the eigenvalues and eigenvectors for $A\in M_n(\Bbb R)$. Such problem is stated as below:

Q: Compute the eigenvalues and eigenvectors for $\begin{bmatrix}3&2\\2&0\end{bmatrix}$.

Now, after some computation, you gain the eigenvalues are $-1,~4$. Next you wish to get the eigenvectors.

For $\lambda=-1:$ we are going to solve the linear system $\begin{bmatrix}3+1&2\\2&0+1\end{bmatrix}$. Then $\begin{cases}x=-t\\y=2t\end{cases}$.

However, what should the range of $t$ be here? Of course by definition we can't take $t=0$. But what is the exact range we should write?

$\begin{cases}x=-t\\y=2t\end{cases}(t\in\Bbb R\setminus\{0\})$?

Or $\begin{cases}x=-t\\y=2t\end{cases}(t\in\Bbb C\setminus\{0\})$?

Or $\begin{cases}x=-t\\y=2t\end{cases}(t\in\Bbb Q(\sqrt{2})\setminus\{0\})$

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    $\begingroup$ Sensitive? The eigenspace is $$\left\{t\pmatrix{-1\\2}:t\in\Bbb R\right\}.$$ $\endgroup$
    – Angina Seng
    Jan 28, 2018 at 10:39
  • $\begingroup$ @LordSharktheUnknown Why do you think it should not be $\left\{t\pmatrix{-1\\2}:t\in\Bbb C\right\}.$? $\endgroup$
    – Eric
    Jan 28, 2018 at 10:40
  • $\begingroup$ Your question asked for the eigenspace. An eigenspace is a set of vectors.... $\endgroup$
    – Angina Seng
    Jan 28, 2018 at 10:41
  • $\begingroup$ @LordSharktheUnknown Well, I see your point... But the really point I'm actually asking is the issue of whether the eigenspace/eigenvectors should consists of broadly as $\Bbb C^n$-vectors, or $\Bbb R^n$-vectors. $\endgroup$
    – Eric
    Jan 28, 2018 at 10:43
  • $\begingroup$ Anyway, now I edit the original word eigenspace to eigenvectors. $\endgroup$
    – Eric
    Jan 28, 2018 at 10:43

3 Answers 3

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The eigenvectors with eigenvalue $-1$ are the vectors of the type $(-t,2t)$ with $t\in\mathbb{R}\setminus\{0\}$. But the eigenspace which corresponds to the eigenvalue $-1$ is$$\left\{(-t,2t)\,\middle|\,t\in\mathbb R\right\}.$$Of course, in the general case we should replace $\mathbb R$ with the field $F$ that we're working with.

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  • $\begingroup$ Where justify that this answer should not be $\left\{(-t,2t)\,\middle|\,t\in\mathbb C\right\}$? $\endgroup$
    – Eric
    Jan 28, 2018 at 10:41
  • $\begingroup$ @Eric I have added another sentence to my answer. The field that we're working with should be part of the question. $\endgroup$
    – José Carlos Santos
    Jan 28, 2018 at 10:42
  • $\begingroup$ I think so! But many exercise or test doesn't clearly specify what the field we're concerning... If $A$ has any of imaginary entry, then it is more explicit. But if $A$ is a real matrix, the things going to be quite ambiguous. For example, $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ is diagonalizable in the $\Bbb C$ sense, but not $\Bbb R$ sense. But what if a T/F question just asking whether $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ is diagonalizable? $\endgroup$
    – Eric
    Jan 28, 2018 at 10:46
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    $\begingroup$ @Eric Believe me: my Linear Algebra students have never been able to find a single exercise proposed by me in which the base field is not clearly stated. $\endgroup$
    – José Carlos Santos
    Jan 28, 2018 at 10:48
  • $\begingroup$ @Eric Concerning the question from your last comment: without stating somewhere what the base field is, the assertion is neither true nor false. $\endgroup$
    – José Carlos Santos
    Jan 28, 2018 at 10:51
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You used the argument that $A∈M_n(\Bbb R)$, so you're talking about a linear mapping of the form $\Bbb R^n \to \Bbb R^n$. The eigenspace being a sub-space of the co-domain, you clearly need $t \in \Bbb R$. If we were dealing with a co-domain of vectors over $\Bbb C$, your argument would be valid. Keep in mind we're working with $M_n(\Bbb R)$ rather than $M_n(\Bbb C)$ or $M_n(\Bbb C, \Bbb R)$.

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  • $\begingroup$ Ya, but any matrix that is "$\in M_n(\Bbb R)$" is inevitably, necessarily, also "$\in M_n(\Bbb C)$". Like the matrix $\begin{bmatrix}3&2\\2&0\end{bmatrix}$ I proposed, in fact $\begin{bmatrix}3&2\\2&0\end{bmatrix}\in M_n(\Bbb C)$ as well. So this arises the annoying ambiguous problems.. $\endgroup$
    – Eric
    Jan 28, 2018 at 10:58
  • $\begingroup$ If that's true then anything in $\Bbb R^3$ is in $\Bbb R^{42}$ and then we should have an eigenvetor for an endomorphism over $\Bbb R^3$ that can belong to $\Bbb R^{42}$. See my point? If your co-domain is $\Bbb R^n$, the eigenspace has to be a subspace of that. $\endgroup$
    – Roulbacha
    Jan 28, 2018 at 11:04
  • $\begingroup$ Wait, form what I learnt, $\Bbb R^3$ is not a subset of $\Bbb R^{42}$ $\endgroup$
    – Eric
    Jan 28, 2018 at 11:06
  • $\begingroup$ It is a subspace and a subset. You compose any vector of $\Bbb R^3$ by combining 3 independent vectors of $\Bbb R^{42}$ $\endgroup$
    – Roulbacha
    Jan 28, 2018 at 11:14
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It depends on the field $F$ you're working over... You get all scalar multiples $t \begin{bmatrix}-1\\2 \end{bmatrix}$ for $t\in F\setminus \{0\}$...

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