# Linear operator with same representation in any basis

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?

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## 3 Answers

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.

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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.

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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 $T(v)=\lambda_{ii} v=\lambda v$.

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