# If every vector is an eigenvector, the operator must be a scalar multiple of the identity operator? [duplicate]

I am posed with the following question:

Suppose that $T\in \mathcal{L}(V)$, where $V$ is a finite-dimensional vector space, is such that every vector in $V$ is an eigenvector of $T$. Prove that T is a scalar multiple of the identity function.

My attempt goes as follows:

If $T$ is a scalar multiple of the identity function, then $Tv=av$ for all $v\in V$, where $a$ doesn't depend on $v$. We will start off assuming that that may not necessarily be true, and work our way to the result.

From the problem statement, we know that $Tv=a_j v$, where $a_j$ may depend on the choice of $v$. To show that it doesn't, consider two non-zero vectors $v_1$ and $v_2$ both in $V$ (It would be pointless to consider zero vectors). Consider $T(v_1+v_2)$. Also, let us first consider the case where $v_2$ is not a scalar multiple of $v_1$ (so $v_1$ & $v_2$ form a linearly independent set). On one hand we have

\begin{align*} T(v_1+v_2)&=\alpha (v_1+v_2)\\ &= \alpha v_1 + \alpha v_2 \end{align*}

Which is true because of the the assumption that every vector is an eigenvector. On the other hand, we have

\begin{align*} T(v_1+v_2)&=T(v_1)+T(v_2)\\ &=a_1 v_1 + a_2 v_2 \end{align*}

So we are left with the following equality

$$a_1 v_1 + a_2 v_2=\alpha v_1 + \alpha v_2$$ $$\implies (a_1 - \alpha)v_1 + (a_2 - \alpha)v_2 = 0$$

Because $v_1$ & $v_2$ are linearly independent, $a_1=a_2=\alpha$.

Now consider $\beta v$, a scalar multiple of $v$.

\begin{align*} T(\beta v)&= \beta T(v)\\ &= \beta (a v)\\ &= a (\beta v) \end{align*}

Because every vector in $V$ that is not equal to an arbitrary vector $v$ is either a scalar multiple of it or the sum of $v$ and some other vector in $V$, $Tv=av$ for all $v\in V$.

Is my prove valid? Please criticize my proof holistically.

P.S: Is the same proof valid for infinite-dimensional vector spaces? I ask because such truth would lie in the validity of the last paragraph, and I don't know if it is true in infinite-dimensional vector spaces.

• Is $V$ finite-dimensional? Jan 4, 2015 at 3:07
• Yeah it is. I'll add that to the problem statement. Jan 4, 2015 at 3:09
• I don't see anything in the proof that requires $V$ to be of finite dimension! Am I missing something? Jan 4, 2015 at 3:16
• Hey, just realize any two vectors are linearly dependent, or they're not! Jan 4, 2015 at 6:16
• If Anyone else, like me, stumbles upon this first, there is a much more clear and concise proof here: math.stackexchange.com/questions/146416/… Nov 24, 2016 at 2:16

Here's as holistic approach as I've got to the present problem:

This very same question actually came up in a research project on differential equations I last worked on a couple of years ago, so it is something with which I am well familiar. And my proof was essentially the same as OP Arturo Don Juan's, so I believe I can affirm the correctness of his argument with confidence.

Just for completeness, here's the way I wrote it up (from my research notes):

Let $$\Bbb F$$ be any field, and let $$V$$ be any vector space over $$\Bbb F$$ (note we don't require $$\dim V < \infty$$.). If $$T \in \mathcal L(V)$$ is such that every non-zero $$v \in V$$ is an eigenvector, then there is an $$\alpha \in \Bbb F$$ such that $$T(v) = \alpha v$$ for all $$v \in V$$.

Proof: The hypothesis that every (non-zero) $$v \in V$$ is an eigenvector of $$T$$ may be expressed by the equation

$$Tv = \phi(v) v, \tag{1}$$

where $$\phi(v) \in \Bbb F$$ for every $$v$$; that is $$\phi:V \to \Bbb F$$ is a function taking vectors in $$V$$ to scalars in $$\Bbb F$$. If $$u \in V$$ is related to $$v \in V$$ by the equation $$u = \beta v$$ for some $$\beta \in \Bbb F$$, we have

$$\phi(u)u = \phi(\beta v)(\beta v) = T(\beta v) = \beta T(v) = \beta \phi(v) v = \phi(v) \beta v, \tag{2}$$

or in short form,

$$\phi(\beta v)(\beta v) = \phi(v) \beta v; \tag{3}$$

when $$\beta v \ne 0$$, (3) shows that

$$\phi(\beta v) = \phi(v); \tag{4}$$

i.e., $$\phi(v)$$ only depends on the one-dimensional subspace $$v$$ generates; on $$\text{span}\{v\}$$; this implies that, for $$v_1, v_2$$ linearly dependent, $$\phi(v_1) = \phi(v_2)$$. In the event that $$v_1, v_2$$ are linearly independent, we may write in manner analogous to our OP

$$T(v_1 + v_2) = \phi(v_1 + v_2) (v_1 + v_2) = \phi(v_1 + v_2) v_1 + \phi(v_1 + v_2) v_2 \tag{5}$$

and

$$T(v_1 + v_2) = T(v_1) + T(v_2) = \phi(v_1)v_1 + \phi(v_2)v_2; \tag{6}$$

combining (5) and (6):

$$\phi(v_1 + v_2) v_1 + \phi(v_1 + v_2) v_2 = \phi(v_1)v_1 + \phi(v_2)v_2; \tag{7}$$

by the linear independence of $$v_1$$ and $$v_2$$, we see from (7) that

$$\phi(v_1) = \phi(v_1 + v_2) = \phi(v_2); \tag{8}$$

since $$v_1, v_2$$ are an arbitrary pair of linearly independent vectors, this result combined with (3)-(4) shows that $$\phi(v)$$ is a constant; taking $$\alpha = \phi(v)$$ for any $$v \in V$$ then yields the complete solution. QED.

• Please, remember point three here for future posts.
– Pedro
Jan 4, 2015 at 6:16
• @PedroTamaroff: very well, I will try to refrain from ancient and other foreign tongues for the moment. Having said that, was there something you wanted to talk to me about? Jan 5, 2015 at 5:12
• @PedroTamaroff: well, I looked up your cited "point three" and I must say I can't see how it bears on the case at hand. Sep 20, 2021 at 3:24
• The original message refers to the signature that was removed: "Hope this helps. Cheers, And As Ever, Fiat Lux!!!"
– Pedro
Sep 20, 2021 at 10:56