# Finding the number of eigenvector(s) of non-zero linear transformation $T$ to $R$

I was thinking about the following problem but do not know how to progress with it.The problem says:
A non-zero linear transformation $T$ to $R$
$1.$ may not have any eigenvector.
$2.$ has exactly one eigenvector.
$3.$ has more than one (but finitely many eigenvectors).
$4.$ has infinitely many eigenvector.

EDIT: As City of God suggested "Any linear transformation on $\mathbb{R}$ is of the form $T(x)=cx$ for some fixed $c\in\mathbb{R}$ and $x$ is any real number." But I am not sure how I make a conclusion from it.

Can someone explain a little bit further?Thanks in advance for your time.

• I take it $R$ is the reals, and $T$ is linear from $R$ to $R$? Feb 9 '13 at 11:29
• Yes sir.You are right. Feb 10 '13 at 5:59
• OK, next question: what do you make of the equation $T(x)=cx$, in the light of the definition of eigenvector? Feb 10 '13 at 8:48
• @learner I have added some details. Feb 12 '13 at 17:25

Hint:Any linear transformation on $\mathbb{R}$ is of the form $T(x)=cx$ for some fixed $c\in\mathbb{R}$ and $x$ is any real number.

Try to prove the following things these might help you to understand!These are standard results! Proof of the respective results can be found in any standard text on Linear Algebra, Functional Analysis, Real Analysis respectively.

$1.$ $\mathbb{R}$ is a one dimensional vector space over the field $\mathbb{R}$ itself!

$2.$ Any Linear Map on a finite dimensional vector space is continuous!

$3.$ A continuous function on $\mathbb{R}$ which satisfies $T(x+y)=T(x)+T(y)\forall x,y\in\mathbb{R}$ must be of the form $T(x)=cx$ for some fixed $c\in\mathbb{R}$

$x\ne 0$ is called an eigenvector if $T(x)=\lambda x$ for some $\lambda$, As we know that a non-zero linear transformation from $\mathbb R$ to $\mathbb R$ has the form $Tx=cx$ for some $c\ne 0$ then any non-zero real number will be an eigenvector. Thus, any such non-zero linear transformation will have infinitely many eigenvectors.

Maybe it will help to first see a way that a linear transformation can have more than one eigenvalue. Let's take a countable vector space $V$ over $\mathbb{R}$, that is $V$ is spanned by $\{e_i:i\in \mathbb{N}\}$, where $e_i$ is the vector with a $1$ in the $i$th position and a $0$ everywhere else. Linear transformations over $V$ are in the form of countable infinite matrices. For example we can make infinite diagonal matrices: $$T:=\left(\begin{array}{ccccc} a_1 & 0 & 0 & \cdots \\ 0 &a_2& 0 & \cdots \\ 0 & 0 & a_3&\cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\right)$$ In this case, $Te_i=a_ie_i$. So if we picked all different numbers for the $a_i$'s, we would have a linear transformation with infinitely many eigenvectors and eigenvalues.

So that is a way that you can have a linear transformation with infinitely many eigenvalues. Now I am not sure if your linear transformation into $R$ is into a vector space $R$ or into the real numbers $\mathbb{R}$. If that is the case, consider that $\mathbb{R}$ can be viewed as a one-dimensional vector space over itself. We can't have a construction like the above - instead every linear transformation is going to look like a one dimensional matrix. $$T:=\left(\begin{array}{c}a\end{array}\right)$$ The vector $(1)$ spans $\mathbb{R}$ as an $\mathbb{R}$-vector space. So we can write $(x)=x(1)$ and thus $T(x(1))=ax(1)$ (which is just the same as writing $T(x)=ax$). This is the same $a$ for any $x$ we choose - it's defined by $T$ - so we can't have more than one possible eigenvalue. Similarly, there is no way we can have zero eigenvalues.