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I was doing some linear algebra exercises and came across the following tough problem :

Let $M_{n\times n}(\mathbf{R})$ denote the set of all the matrices whose entries are real numbers. Suppose $\phi:M_{n\times n}(\mathbf{R})\to M_{n\times n}(\mathbf{R})$ is a nonzero linear transform (i.e. there is a matrix $A$ such that $\phi(A)\neq 0$) such that for all $A,B\in M_{n\times n}(\mathbf{R})$ $$\phi(AB)=\phi(A)\phi(B).$$ Prove that there exists a invertible matrix $T\in M_{n\times n}(\mathbf{R})$ such that $$\phi(A)=TAT^{-1}$$ for all $A\in M_{n\times n}(\mathbf{R})$.

This is an exercise from my textbook and I am all thumbs when I attempted to solve it .

Can someone tell me as to how should I , at least , start the problem ?

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    $\begingroup$ One direction should be rather trivial (Given $T\in GL(n)$ show that $A\mapsto T^{-1}AT$ satisfies the properties). A useful start might be to see that $\phi(I) = I$. $\endgroup$
    – AlexR
    Apr 22, 2015 at 14:48
  • $\begingroup$ It true that $\phi(I_n)=I_n$, but how to prove this? $\phi$ is nonzero may be an important hint. $\endgroup$
    – Xiang Yu
    Apr 22, 2015 at 14:50
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    $\begingroup$ @Brenton That's not enough. Instead see that $$\phi(I) \phi(B) = \phi(IB) = \phi(B)$$ For all $B\in\mathbb R^{n\times n}$, wich is the definition of multiplicative identity. $\endgroup$
    – AlexR
    Apr 22, 2015 at 14:53
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    $\begingroup$ @AlexR how do we know that $\phi(B)$ is invertible for any $B$? $\endgroup$ Apr 22, 2015 at 14:54
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    $\begingroup$ If I well understand we have $\phi(A^n)=[\phi(A)]^n$, so it seems that $A$ and $\phi(A)$ have the same minimal polynomial. $\endgroup$ Apr 22, 2015 at 15:37

3 Answers 3

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This kind of problems are known as linear preserver problems in the literature. The following is a sketch of proof that immediately comes to my mind. Certainly there are simpler ways to solve the problem (especially if one makes use of existing results on linear preserver problems), but anyway, let $\{e_1,\ldots,e_n\}$ be the standard basis of $\mathbb R^n$ and $E_{ij}=e_ie_j^T$.

  1. Prove that $\phi$ is injective. Hint. Suppose the contrary that $\phi(X)=0$ for some matrix $X$ whose $(r,s)$-th entry is nonzero. Now consider $\phi(E_{ir}XE_{sj})$ for every $(i,j)$.

  2. Prove that

    • $\phi$ preserves non-invertibility (hint: if $X$ is singular, then $XY=0$ for some nonzero $Y$),
    • $\phi$ preserves invertibility (hint: if $\phi(P)$ is singular for some invertible $P$, then $\phi(P)Y=0$ for some nonzero matrix $Y$; since $\phi$ is an injective linear operator over a finite dimensional vector space, $Y=\phi(B)$ for some nonzero $B$, but then ...),
    • $\phi(I)=I$.
  3. This is the only interesting step in the whole proof: show that every $\phi(E_{ii})$ is a rank-1 idempotent matrix. Hint: the rank of an idempotent matrix is equal to its trace.

  4. Argue that without loss of generality, we may assume that $\phi(E_{11})=E_{11}$.

  5. Show that whenever $i,j\ne1$, the first column and the first row of $\phi(E_{ij})$ are zero (hint: $E_{ij}E_{11}=0=E_{11}E_{ij}$). By mathematical induction/recursion, show that we may further assume that $\phi(E_{ii})=E_{ii}$ for every $i$.

  6. For any off-diagonal coordinate pair $(i,j)$, show that $\phi(E_{ij})$ is a scalar multiple of $E_{ij}$ (hint: we have $E_{kk}E_{ij}=0$ for every $k\ne i$ and $E_{ij}E_{kk}=0$ for every $k\ne j$).

  7. Hence prove that in addition to all the previous assumptions (i.e. $\phi(E_{ii})=E_{ii}$ and $\phi(E_{ij})$ is a scalar multiple of $E_{ij}$ for every $i,j$), we may further assume that $\phi(E_{\color{red}{1}j})=E_{\color{red}{1}j}$ for every $j$.

  8. Since $\phi$ preserves invertibility and non-invertibility, prove that $\phi(E_{ij})=E_{ij}$ for every $(i,j)$.

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  • $\begingroup$ You seem to have all the answers! +1 $\endgroup$ Apr 22, 2015 at 19:21
  • $\begingroup$ A key here seems to be to find a basis such that $B_{ij} B_{mn} = \delta_{jm} B_{in}$. $\endgroup$
    – copper.hat
    Apr 22, 2015 at 23:48
  • $\begingroup$ May I ask a qustion?How can we show that $\mathrm{rank}(\phi(E_{ii})=1$?sine we only have $\phi(E_{ii})=\phi(E_{ii}^2)=\phi(E_{ii})^2$, which implies $\phi(E_{ii})$ is a idempotent matrix. $\endgroup$
    – Xiang Yu
    Apr 23, 2015 at 1:09
  • $\begingroup$ @XiangYu Well, $\phi$ is injective and linear and $\phi(E_{ii})$ is idempotent. Therefore the rank of $\phi(E_{ii})$ is at least one. Now the sum of traces of all $\phi(E_{ii})$ is equal to $n$ because $\phi(I)=I$. So ... $\endgroup$
    – user1551
    Apr 23, 2015 at 1:43
  • $\begingroup$ @copper.hat Yes, sort of. In principle this problem is not that different from those more familiar linear algebra exercises that can be solved by manipulating those $E_{ij}$s (such as finding the center of $M_n(\mathbb R)$). This one just needs a few more steps. $\endgroup$
    – user1551
    Apr 23, 2015 at 2:00
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Here is an alternative proof that is much less computational. It works over any field. Let $E_{ij}$ be the matrix with a $1$ at the $(i,j)$-th entry and zeros elsewhere.

  1. $\phi$ is injective and hence an automorphism. Suppose the contrary that $\phi(A)=0$ for some nonzero $A$. Since $A$ is nonzero, every square matrix $B$ can be written as a finite sum of the form $\sum_k X_kAY_k$ (that's simply because Kronecker products of the form $Y_k^T\otimes X_k$ span the set of all $n^2\times n^2$ matrices). But then we would have $\phi(B)=\sum_k\phi(X_k)\phi(A)\phi(Y_k)=0$ for every $B$, which is a contradiction because $\phi\ne0$.

  2. $\phi$ preserves the rank/nullity of a matrix. As $\phi$ is an automorphism, $\phi$ and $\phi^{-1}$ preserve the linear dependence/independence of any set of matrices. In turn, $$ \frac1n\dim\{X\in M_n(\mathbb C):\,AX=0\}=\frac1n\dim\{Y\in M_n(\mathbb C):\,\phi(A)Y=0\}. $$ As the two sides are the respective nullities of $A$ and $\phi(A)$, the conclusion follows. In particular, $\phi$ preserves singularity/invertibility of matrices and $\phi(I)=I$.

  3. As $\phi$ is an automorphism, a polynomial $p$ annihilates a matrix $A$ if and only if it annihilates $\phi(A)$. It follows that

    • $A$ and $\phi(A)$ always have identical minimal polynomials.
    • So, if $A$ is diagonalisable, $\phi(A)$ must be diagonalisable too and the two matrices must share the same set of eigenvalues. The multiplicities of the eigenvalues must be identical too, because $A-\lambda I$ and $\phi(A)-\lambda I$ have the same rank for every scalar $\lambda$. In other words, $\phi(A)$ must be similar to $A$.
    • As $\phi$ also preserve the linear independence of any set of matrices as well as matrix rank, it must map the set of all diagonal matrices onto an $n$-dimensional subspace of commuting diagonalisable matrices. Since commuting diagonalisable matrices are simultaneously diagonalisable, we may assume that $\phi$ maps diagonal matrices to diagonal matrices.
    • So, each $\phi(E_{ii})$ is a diagonal matrix that is similar to $E_{ii}$. Hence $\phi(E_{ii})=E_{jj}$ for some $j$. As $\phi$ is injective, we may assume without loss of generality that $\phi(E_{ii})=E_{ii}$ for each $i$. Hence $\phi(D)=D$ for every diagonal matrix $D$.
  4. Here comes the computational part. Let $C$ be the circulant permutation matrix whose super-diagonal as well as the bottom left entry are filled with ones. By comparing the coefficients on both sides of $\phi(D_1CD_2)=D_1\phi(C)D_2$ for arbitrary diagonal matrices $D_1$ and $D_2$, we see that $\phi(C)=DC$ for some diagonal matrix $D$. Note that $\det D=1$ because $I=\phi(C^n)=(DC)^n=\det(D)I$. Write $D=\operatorname{diag}(d_1,\ldots,d_n)$. Then $DC=\tilde{D}C\tilde{D}^{-1}$ where $$ \tilde{D}=\operatorname{diag}\left(\prod_{j=1}^nd_j,\ \prod_{j=2}^nd_j,\ \prod_{j=3}^nd_j,\ \ldots,\ d_n\right). $$ Since every matrix $A$ can be written as $A=\sum_{k=0}^{n-1}D_kC^k$ for some diagonal matrices $D_0,D_1,\ldots,D_{n-1}$, we obtain \begin{align} \phi(A) &=\sum_{k=0}^{n-1}\phi(D_k)\,\phi(C)^k\\ &=\sum_{k=0}^{n-1}D_k\left(\tilde{D}C\tilde{D}^{-1}\right)^k\\ &=\sum_{k=0}^{n-1}D_k\tilde{D}C^k\tilde{D}^{-1}\\ &=\sum_{k=0}^{n-1}\tilde{D}D_kC^k\tilde{D}^{-1}\\ &=\tilde{D}A\tilde{D}^{-1}. \end{align}

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Since $\varphi$ is linear, it's only natural to investigate how it acts on the canonical basis of $M_n(k)$. Let $E_{ij}$ denote the matrix with a single one for the entry $(i,j)$ and $0$'s elsewhere. It is standard that $E_{ij}E_{kl} = \delta_{jk}E_{il}$, hence $\varphi(E_{ij})\varphi(E_{kl}) = \delta_{jk}\varphi(E_{il})$.

Let $e_k$ denote the $k$-th vector of the canonical basis of $k^n$.Note that if $\varphi(E_{ij}) = CE_{ij}C^{-1}$, then $\varphi(E_{ij})Ce_k = \delta_{jk}Ce_i$. With $j=k$, $$\varphi(E_{ij})Ce_j=Ce_i$$ which yields a construction of $C$ (fix $j$ and let $i$ vary), as follows.

Let $I$ denote the identity matrix. Since $\varphi \neq 0$, there is some $A$ such that $\varphi(A)\neq 0$, hence $\varphi(A)\varphi(I)\neq 0$, hence $\varphi(I)\neq 0$. Since $I=\sum_{k=1}^n E_{kk}$, there exists some $k$ such that $\varphi(E_{kk})\neq 0$. WLOG suppose $k=1$.
Since $\varphi(E_{11})\neq 0$, there exists $e_1'\in \operatorname{im}\varphi(E_{11})\setminus\{0\}$. Define $e_i':=\varphi(E_{i1})(e_1')$. It's easy to check that every $e_i'$ is non-zero and that $(e_1',\ldots, e_n')$ is linearly independent, hence a basis. Define $C$ the matrix with columns $(e_1',\ldots, e_n')$. $C$ is invertible by construction. It is easily checked that for fixed $(i,j)$, the following holds: $$\forall k, \varphi(E_{ij})e_k' = CE_{ij}C^{-1}e_k'$$ thus $$\varphi(E_{ij}) = CE_{ij}C^{-1}$$

Hence by linearity, $\varphi(A) = CAC^{-1}$ for all $A$.

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  • $\begingroup$ Duplicate answers are automatically spotted by a script and they should be avoided. If two different questions can be answered by exactly the same lines, one of such questions is probably a duplicate and it is the case to mark it as such. $\endgroup$ Apr 29, 2018 at 17:43
  • $\begingroup$ How to prove $(e_1',\ldots, e_n')$ is linearly independent? $\endgroup$
    – cjdx
    Apr 10 at 4:05

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