# Demonstration: If all vectors of $V$ are eigenvectors of $T$, then there is one $\lambda$ such that $T(v) = \lambda v$ for all $v \in V$.

Let $T: V \rightarrow V$ be a linear operator.

I need to demonstrate that if all nonzero vectors of $V$ are eigenvectors of $T$, then there is one specific $\lambda \in K$ such that $T(v) = \lambda v$, for all $v \in V$.

I understand that, if all nonzero vectors of $V$ are eigenvectors of $T$, then $T$ must be a scaling transformation. It just stretch or shrinks vectors, but doesn't change their directions.

So, the statement says that if it happens, then, there is a single $\lambda$ such that $T(v) = \lambda v$. In other words, if there is such transformation, then it scales all vectors by the same scalar $\lambda$.

Applying the transformation to our standard basis vectors, we have: $$T(e_1) = \lambda_1 e_1 \\ T(e_2) = \lambda_2 e_2 \\ \vdots \\ T(e_n) = \lambda_n e_n$$

I understand I need to prove that $\lambda_1 = \lambda_2 = \dots = \lambda_n$, but I can't see how!

EDIT

$$v = c_1e_1 + c_2e_2 + \dots + c_ne_n \\ T(v) = \mu v = \lambda_1c_1e_1 + \lambda_2c_2e_2 + \dots + \lambda_nc_ne_n \\$$

Since what's multiplying $v$ coordinates is $\lambda_i$, then all of them must be $\mu$. I'm not sure how to 'mathematize' this. Is this idea correct?

EDIT 2 Extending the left hand side of EDIT 1, we have: $$\mu v = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ \mu(c_1e_1 + \dots + c_ne_n) = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ \mu c_1e_1 + \dots + \mu c_ne_n = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\$$

And since $e_i$ are linearly independent, $\mu = \lambda_1 = \lambda_2 = \dots = \lambda_n$. Is this proof correct?

• Hint: consider $e_1+e_2$. May 17, 2012 at 18:23
• I believe I've come to something. If ALL vectors of V are eigenvectors, then $T(v)$ is also an eigenvector, since T maps $V \rightarrow V$. Then it can be written as $\mu v$, where $\mu$ is another eigenvalue. Than $\mu = \lambda_1 = \lambda_2 = \dots = \lambda_n$. Is it correct? May 17, 2012 at 18:28
• Your edit shows that $\mu = \lambda_i c_i$, $\forall i$, since the $e_i$ are linearly independent. May 17, 2012 at 18:40
• But $\mu$ is a scalar multiplying $v$, and $c_i$ are $v$ coordinates on standard basis, i.e., $c_ie_i$ are $v$ coordinates. If $\mu$ is multiplying $v$, why does $c_i$ comes to $\mu$ equation? May 17, 2012 at 18:44
• You have now asked THREE different questions in the same post. What happens to the answers to your FIRST question, posted while this was the only one?
– Did
May 17, 2012 at 19:13

Since all $v\in V$ are eigenvectors, we can choose $e_i$, the $i$th unit vector. Then by assumption we have $T e_i = \lambda_i e_i$ for some $\lambda_i$. It follows that $T$ is diagonal, with elements $\lambda_1,...,\lambda_n$ on the diagonal.

Now choose $v=e_1+...+e_n$, again for some $\lambda$, we have $Tv=\lambda v$, so we have $$T v = T(e_1+...+e_n) = \lambda_1 e_n +... + \lambda_n e_n = \lambda (e_1+...+e_n).$$ Since the $e_i$ are linearly independent, it follows that $\lambda = \lambda_1 = ... = \lambda_n$. Hence $Tx = \lambda x$, $\forall x$.

• When you write $v = e_1 + \dots + e_n$, you're considering it for unit vectors. Shouldn't you consider $v = c_1e_1 + \dots + c_ne_n$ with $c_i \in K$, in order to extend it to all vectors of $V$? May 17, 2012 at 18:51
• What needs to be extended? If you know $T$ on a basis, you know $T$ everywhere, by linearity. First I show $T$ is diagonal, since $T(\sum x_i e_i) = \sum \lambda_i x_i e_i$, then I show that all diagonal elements are equal. May 17, 2012 at 19:02
• Sorry, I got confused mixing it with my try of demonstration. I'm still trying to assimilate your answer ;). May 17, 2012 at 19:11
• This is a very nice proof. +1 May 17, 2012 at 19:43
• @copper.hat $v$ is just a specific vector that sets $v = e_1+...+e_n$, you shouldn't show that it works for all $u \in V$? Sep 25, 2017 at 7:46

Assume for a contradiction that $v,w$ are eigenvectors for $\lambda\neq\mu$, respectively, (in particular they are nonzero) and that $T(v+w)=\nu(v+w)$ for some $\nu\in K$. Since any nonzero scalar multiple of an eigenvector is an eigenvector for the same eigenvalue, $v$ and $w$ cannot be linearly dependent. Then by this linear independence $\nu v+\nu w=T(v+w)=T(v)+T(w)=\lambda v+\mu w$ implies $(\nu,\nu)=(\lambda,\mu)$, which contradicts $\lambda\neq\mu$.

So if all nonzero vectors are eigenvectors, then all of them must be so for the same eigenvalue$~\lambda$, and one has $T=\lambda I$. (Pedantically, if $\dim V=0$ there are no eigenvectors at all, and one is free to choose$~\lambda$; "specific" in the question is not justified in this case.)

Hint: Assume that $Tu=\lambda u$ and $Tv=\mu v$ for some nonzero vectors $u$ and $v$ and some $\lambda$ and $\mu$.

• Show that $\{u,v\}$ is a linearly independent family.
• Show that $\{u+v,au+bv\}$ is a linearly independent family, for every $a\ne b$.
• Show that $\{T(u+v),u+v\}$ is not linearly independent.
• Conclude that $\lambda=\mu$.
• What is a linearly free family? May 17, 2012 at 18:38
• @copper.hat: A Gallicism.
– Did
May 17, 2012 at 18:44
• A straight couple from the 60's maybe? May 17, 2012 at 18:49
• @copper.hat: Nice... :-)
– Did
May 17, 2012 at 18:58