Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to recall a question from a past exam to review for an upcoming exam; I think it went like this:

Suppose a finite-dimensional linear operator $T:V \to V$ has the same matrix representation in every basis. Show that $T$ must be a scalar multiple of the identity transformation.

First, does it sound like my recollection of the problem is correct? Second, any suggestions on how to approach a proof?

share|improve this question

3 Answers 3

up vote 6 down vote accepted

A proof sketch could be:

1. Every (nonzero) vector is an eigenvector. Let $v\ne 0$ and suppose $Tv$ is not a multiple of $v$. Then $v$ and $Tv$ are linearly independent; extend $\langle v,Tv\rangle$ to a basis $\langle v, Tv, v_3,v_4,\ldots,v_n\rangle$. By assumption $T$ has the same matrix representation $M$ in this basis and in the basis $\langle v,v+Tv,v_3,v_4,\ldots,v_n\rangle$. But that means that the first column of $M$ is simultaneously $(0,1,0,\ldots,0)^{\mathsf t}$ and $(-1,1,0,\ldots,0)^{\mathsf t}$, which is absurd.

2. All eigenvalues are the same. Since every vector is an eigenvector, there exists an eigenbasis. Therefore $M$ is diagonal. It can only be invariant under permutations of the basis vectors if all of the diagonal entries are equal..

Therefore $T$ must be scalar multiplication by the common eigenvalue.

share|improve this answer
    
If all eigenvalues are the same then $T$ could be a Jordan-normal form matrix with diagonal entries equal and 1's on the upper diagonal so argument 2 is incomplete. –  user2566092 Mar 27 at 20:15
    
Nevermind, you're giving a two-step proof not two different proof sketches –  user2566092 Mar 27 at 20:16
    
The system just tried to test with this as a "possible spam". LOL. math.stackexchange.com/review/low-quality-posts/196488 :-P –  Asaf Karagila Apr 18 at 18:27

If you know that change of basis is realised by conjugating by an appropriate invertible matrix, then you can reason in terms of matrices as follows. $E_{i,j}$ is the matrix with unique nonzero entry $1$ at position $i,j$.

  • The (unique) matrix $M$ of $T$ can have no nonzero off-diagonal entries: if $a_{i,j}$ were such an entry, then conjugating by $I+E_{j,i}$ adds $a_{i,j}$ to the diagonal entries at $(j,j)$, and subtracts it from the entry at $(i,i)$, while it was supposed to leave all entries unchanged.

  • Being diagonal, $M$ must have all diagonal entries equal, since conjugating by a permutation matrix permutes the diagonal entries.

share|improve this answer

If you don't know what an eigenvalue is, and if you're not worried about elegance, then here is a more direct approach (assuming you're working over a field not of characteristic two).

There exist scalars $\lambda_{ij}$ for $1\leq i,j\leq n$ such that for every basis $\{v_1,\dotsc,v_n\}$ of $V$ we have $$ \begin{array}{ccccccc} T(v_1) & = & \lambda_{11}v_1 & + & \dotsb & + & \lambda_{n1}v_n \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ T(v_n) & = & \lambda_{1n}v_1 & + & \dotsb & + & \lambda_{nn}v_n \end{array}\tag{1} $$ Now, fix $v\in V$ and note that there exists a basis $\{v_1,\dotsc,v_i,\dotsc,v_n\}$ of $V$ such that $v_i=v$. Equation $(1)$ then implies $$ T(v)=\lambda_{1i}v_1+\dotsb+\lambda_{ii}v_i+\dotsb+\lambda_{ni}v_n\tag{2} $$ Next, since $$\{-v_1,\dotsc,-v_{i-1},v_i,-v_{i+1},\dotsc,-v_n\}$$ is also a basis for $V$, equation $(1)$ also implies $$ T(v)=-\lambda_{1i}v_1-\dotsb-\lambda_{i-1,i}\cdot v_{i-1}+\lambda_{ii}v_i-\lambda_{i+1,i}\cdot v_{i+1}-\dotsb-\lambda_{ni}v_n\tag{3} $$ Subtracting equation $(3)$ from equation $(2)$ gives $$ \mathbf{0}=2\lambda_{1i}v_1+\dotsb+2\lambda_{i-1,i}\cdot v_{i-1}+2\lambda_{i+1,i}\cdot v_{i+1}+\dotsb+2\lambda_{ni}v_{n}\tag{4} $$ Since $\{v_1,\dotsc,v_n\}$ are linearly independent, $(4)$ implies $$ \lambda_{1i}=\dotsb=\lambda_{i-1,i}=\lambda_{i+1,i}=\dotsb=\lambda_{ni}=0\tag{5} $$ Since our choice of $i$ was arbitrary, equation $(5)$ implies $$ \lambda_{kl}=0\tag{6} $$ whenever $k\neq l$. Moreover, equations $(2)$ and $(6)$ imply that $T(v)=\lambda_{kk}v=\lambda_{ll}v$ for all $k$ and $l$ so that $$ \lambda_{kk}=\lambda_{ll} $$ for all $k$ and $l$. That is, there exists a scalar $\lambda$ such that $$ \lambda_{kl}= \begin{cases} 0 & k\neq l \\ \lambda & k=l \end{cases}\tag{7} $$

Finally, we wish to show that there exists a scalar $\lambda$ such that $T(v)=\lambda v$ for every $v\in V$. To do so, let $v\in V$ and note that $(2)$ and $(7)$ imply $T(v)=\lambda_{ii} v=\lambda v$.

share|improve this answer
1  
Wow. This was way nicer in my head before I wrote this down. –  Brian Fitzpatrick Mar 27 at 20:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.