Suppose that $V$ be a finite dimensional vector space, and $f:V \longrightarrow V$ be a non zero linear map. If the matrix of $f$ with respect to any basis of $V$ gives a diagonal matrix, why is $f= \lambda Id. $ where Id is an identity map ?


I am trying to show by contradiction. Let $f \neq \lambda Id.$ This implies that for any $v \in V$, $fv \neq \lambda Id v $ which is same as $fv \neq \lambda v. $ Thus, for all $v \in V$, $v$ and $fv$ are linearly independent. I don't know where this leads now.

  • $\begingroup$ It would be better to add $f$ is non zero linear map.... $\endgroup$ – user87543 Oct 13 '13 at 19:10
  • $\begingroup$ do you know how to write matrix corresponding to linear map? $\endgroup$ – user87543 Oct 13 '13 at 19:13

You are on right path :

Suppose $f\neq \lambda Id$ for any $\lambda \in F$.

Fix some basis $\{v_1,v_2,\dots v_n\}$ for $V$ over $F$.

As $f$ is non zero linear map, I have for some $v_i \in \{v_1,v_2,\dots v_n\}$ : $f(v_i)\neq 0$ and $f(v_i)\neq \lambda v_i$.

So, by writing corresponding matrix with $\{v_1,v_2,\dots v_n\}$ as basis, you get some non zero element in $i^{th}$ column different from $a_{ii}$..

Does this say anything about matrix being diagonal?

EDIT :how do you write matrix corresponding to linear map????

take a basis vector $v_i$, and write $f(v_i)$ as linear combination of $\{v_1,v_2,\dots v_n\}$

and write coefficients as corresponding column to $v_i$....

you have $f(v_i)\neq \lambda v_i$ and $f(v_i)\neq 0$.

so, there does exists some non zero coefficient to some basis vector $v_j$ for $i\neq j$ which gives some non zero element which is out of diagonal i mean which is not in the diagonal.

i.e., you have some non zero element in a non diagonal entry...

Thus matrix is not diagonal....

  • $\begingroup$ sorry I don't see why that says matrix is not diagonal.. $\endgroup$ – ugstudent1243 Oct 13 '13 at 19:40
  • $\begingroup$ check the edit please.. $\endgroup$ – user87543 Oct 13 '13 at 20:16
  • $\begingroup$ This is still completely false. The line "there does exists some non zero coefficient" does not follows from what you wrote before. For instance, I can use your "proof" to show that the matrix (1 0)(0 2) is not diagonal, which is doubtful. $\endgroup$ – D. Thomine Oct 14 '13 at 11:26
  • $\begingroup$ I said, there exists some non zero element which is not in the diagonal... so your example $(1 0)(0 2)$ does not fit into this... $\endgroup$ – user87543 Oct 14 '13 at 11:29

Let $v$ and $u$ be two vectors from a basis so $f (u)=\lambda_u u$ and $f(v)=\lambda_v v$ and $$f(u+v)=\lambda_{u+v}(u+v)=\lambda_u u+ \lambda_v v\tag{1}$$ It's easy to prove from $(1)$ that $\lambda_u=\lambda_{u+v}=\lambda_v$ so for any vectors $w$ of the basis we have $f(w)=\lambda w$ and then $f$ is a homothetie.

  • $\begingroup$ Why two vectors only? $\endgroup$ – Pedro Tamaroff Oct 13 '13 at 19:28
  • $\begingroup$ They are two arbitrary vectors so my proof is valid for all the vectors of the basis taken two by two. $\endgroup$ – user63181 Oct 13 '13 at 19:30
  • $\begingroup$ Sorry, I accidentally ready "form" instead of "from"! I added the word "be", if you don't mind. $\endgroup$ – Pedro Tamaroff Oct 13 '13 at 19:31
  • $\begingroup$ You're welcome @PedroTamaroff :) $\endgroup$ – user63181 Oct 13 '13 at 19:34

Let $\{a_1,a_2,\ldots\}$ be a basis, $f(a_i)=\lambda_ia_i$. Suppose $\lambda_1\ne\lambda_2$. Consider the new basis $\{b_1,b_2,\ldots\}= \{a_1+a_2,a_2,\ldots\}$. Then $f(b_1)=\lambda_1b_1+(\lambda_2-\lambda_1)b_2$, hence in the new basis the matrix of $f$ is not diagonal if $\lambda_1\ne\lambda_2$.


$f$ is diagonal on any basis. So $f(v_{i})=a_{i}v_{i}$ where $a_{i}$ is some scalar dependent on which vector $v_{i}$ you use. If $v_{1}$ and $v_{2}$ are two linearly independent vectors then $$f(v_{1}+v_{2})=a_{1+2}(v_{1}+v_{2})=a_{1+2}v_{1}+a_{1+2}v_{2}$$ and $$f(v_{1}+v_{2})=f(v_{1})+f(v_{2})=a_{1}v_{1}+a_{2}v_{2}$$

So $a_{1}=a_{1+2}=a_{2}$. This must now be true for all vectors in $V$. Hence $f=aI$.


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