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Let $V$ be a finite dimensional vector space, and let $T\in\mathscr{L}(V)$ (where $\mathscr{L}(V)$ is the set of linear maps $V\to V$). Show that $T$ is the identity multiplied by a scalar iff $TS = ST$ for all $S\in\mathscr{L}(V)$.

This problem is taken from Linear Algebra Done Right, by Sheldon Axler (Ch 3), and is given before the concepts of eigenvalues, eigenvectors, operators and inner products are introduced. Up to this point only vector spaces, subspaces, bases (spanning and linear independence), linear maps, matrices and the correspondence between them are introduced.

I have no idea how to approach this problem, and looking for hints that may help me solve it. Any help will be appreciated.

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3 Answers 3

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Only the interesting part: Let $TS = ST$ for all $S \in \mathcal{L}(V)$. This means that for all $x \in V$ it holds $TSx=STx$. You can then find $S$ and $x$ such that $x \in \ker(S) \neq \{0\}$ and --provided that $T \neq \mathrm{Id}$-- such that the preimage of $Tx$ meets $S^{-1}Tx\neq \ker(S)$: Simply take an $x$, that is mapped by $T$ outside its own 1-dimensional subspace and and take a $S$ that maps this $x$ to $0$. The latter can be done by modifying the identity mapping.

This proof works also with infinite dimensional vector spaces or if you are not yet allowed to use the isomorphism between finite dimensional vector spaces and matrices.

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  • $\begingroup$ Could you perhaps elaborate on the construction of $S$ and $x$? $\endgroup$ Mar 22, 2014 at 10:54
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Just use the "unit matrices" $E_{ij}$ which are $1$ on the $(i,j)$ coordinate and zero elsewhere for $S$, and you'll quickly discover that the off-diagonal elements have to be zero, and that the on-diagonal elements have to coincide.

This shows that being a constant diagonal matrix is equivalent to this commutativity condition. The fact that the action of these matrices coincide with the action of scalar multiplication is readily verified.

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$\Longrightarrow$ Given $aI\in\mathscr{L}(V)$, for some $a\in \mathbb{F}$, one can easily verify for all $S\in\mathscr{L}(V)$: $$(aI)S=a(IS)=aS=a(SI)=S(aI)$$ $\Longleftarrow$ Let $T\in\mathscr{L}(V)$ be such that for all $S\in\mathscr{L}(V)$: $ST=TS$. Let $(v_1,...,v_n)$ be a basis of $V$, and define: $$ S_iv_i=v_i, \,\,\,\,\, S_iv_j=0\,\,\,(j\neq i) $$ Also, let us write: $$ Tv_i = a_{1,i}v_1\,+\,...\,+\,a_{n,i}v_n $$ We can now see that: $$ a_{i,i}v_i=S_i(a_{1,i}v_1\,+\,...\,+\,a_{n,i}v_n)=S_iTv_i=TSv_i=Tv_i $$ If $\,\dim V = 1$ then all maps, including $T$, are scalar multiplications, So assume that $\dim V > 1$. We can select any two indices $i\neq j$, and define: $$Sv_i = v_j,\,\,\,\,Sv_j=v_i,\,\,\,\, Sv_k = 0\,\,(j\neq k\neq i)$$ and so: $$ a_{i,i}v_j=S(a_{i,i}v_i)=STv_i=TSv_i = Tv_j = a_{j,j}v_j $$ So we must have $a=a_{1,1}=\cdots =a_{n,n}$. Let $v\in V$ be arbitrary, and write $v=b_1v1\,+\,...\,+\,b_nv_n$, then we must have: $$ Tv = T(b_1v_1\,+\,...\,+\,b_nv_n)=b_1Tv_1\,+\,...\,+\,b_nTv_n=b_1(av_1)\,+\,...\,+\,b_n(av_n)=a(b_1v_1\,+\,...\,+\,b_nv_n)=av $$ or in other words: $T = aI$.

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