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Let $V$ be a finite dimensional vector space over $F$, and let $$T:V\to V$$ be a linear map.

Suppose that given any two bases $B$ and $C$ for $V$, we have that the matrix of $T$ with respect $B$ is equal to that with respect to $C$.

How can we show that this implies that there exists some $\lambda\in F$ such that $T(v)=\lambda v$, $\forall v\in V$?

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Let $A$ be the matrix of $T$. Suppose $Tv_1=v_2$ for some vectors $v_1,v_2$ that are not linearly dependent. Extend $\{v_1,v_2\}$ to a basis $\{v_1,v_2,\ldots,v_n\}$ of $V$. By assumption, the matrices of $T$ w.r.t. the ordered bases $\{v_1,v_2,v_3,\ldots,v_n\}$ and $\{v_2,v_1,v_3,\ldots,v_n\}$ must both be equal to $A$. Therefore $Tv_2=v_1$. But then $T(v_1+v_2)=v_1+v_2\not=v_1-v_2$. Hence the matrix of $T$ w.r.t. the ordered basis $\{v_1+v_2,\ v_1-v_2,\ v_3,\ldots, v_n\}$ would be different from $A$, which is a contradiction.

Therefore, for every vector $v\in V$, $Tv$ and $v$ must be linearly dependent, i.e. $Tv=\lambda_vv$ for some scalar $\lambda_v$. Let $v$ be the first vector in an ordered basis of $V$, we see that $\lambda_v$ is equal to the $(1,1)$-th entry of $A$, which is a constant. Hence the result follows.

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  • $\begingroup$ Thanks. By "not linearly independent" I guess you mean "not linearly dependent", right? The second sentence. $\endgroup$ – Spenser Feb 18 '13 at 12:25
  • $\begingroup$ @Spenser Thanks for catching the typo. It's now corrected. $\endgroup$ – user1551 Feb 18 '13 at 13:59
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Hint: This is equivalent to say that any nonzero vector is an eigenvector of $T$

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  • $\begingroup$ Would you care to elaborate? I don't see how it is equivalent. $\endgroup$ – EuYu Feb 18 '13 at 15:28

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