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.

Let $K$ be a field, suppose that $D\colon M_{n\times n}(K) \to K$ is a function such that $D(AB)=D(A)\cdot D(B)$ and $D(I) \neq D(0)$, where $0$ is the zero matrix. Show that if $\operatorname{rank}(A) < n$, then $D(A)=0$.

My consideration is that: first by $D(0)=D(0)D(I)$ and $D(0)\neq D(I)$, I can show $D(0)=0$ and $D(I)=1$. then I want to show $D(I_k)=0$, where $I_k $ is $n\times n$ diagonal matrix with k diagonal entries equal to $1$ and others $0$. Then $D(A)=D(P^{-1}I_kP)=0$. However, I fail to prove $D(I_k)=0$.

Any suggestions?

Thanks a lot

share|improve this question
    
consider the column vectors, rank is the maxi number of these vectors s.t. they are linearly independent –  David Apr 27 '12 at 21:21
    
Writing $D(A)=D(P^{-1}I_kP^)$ suggests that you write $A=P^{-1}I_kP$, hence in particular $A$ is diagonalizable, which doesn't need to be true (for example, $A=\pmatrix{1&1\\\ 0&1}$). Maybe you used an other argument that you should specify. –  Davide Giraudo Apr 28 '12 at 10:22

3 Answers 3

Note that since $I_k^2=I_k$, then $D(I_k)$ is either $1$ or $0$. Note also that it is enough to show that $D(I_{n-1})=0$, as $D(I_k)=D(I_kI_{n-1})=D(I_k)D(I_{n-1})$.

So suppose that $D(I_{n-1})=1$. For convenience I will write $I_{n-1}=\sum_{j=1}^{n-1}E_{jj}$, where $\{E_{kj}\}$ are the canonical matrix units. Now we can obtain $\sum_{j=2}^nE_{jj}$ by permutations, so $D(\sum_{j=2}^nE_{jj})=1$. But then $$ D(\sum_{j=2}^{n-1}E_{jj})=D(\sum_{j=1}^{n-1}E_{jj}\sum_{j=2}^{n}E_{jj}) =D(\sum_{j=1}^{n-1}E_{jj})D(\sum_{j=2}^{n}E_{jj})=1. $$ As $\sum_{j=2}^{n-1}E_{jj}$ is similar to $I_{n-2}$, we conclude that $D(I_{n-2})=1$. By repeating the argument, we get in the end that $D(E_{jj})=1$ for all $j$. But then $$ 0=D(0)=D(I_{n-1}E_{nn})=D(I_{n-1})D(E_{nn})=1, $$ a contradiction. So $D(I_{n-1})=0$, and we are done.

share|improve this answer

I think there is a theorem (at least for real matrices) which says that for any matrix $A$ there exist invertible matrices $P,Q$ such that $I_k=PAQ$ (where $P,Q$ come from elementary operations on rows and columns)

If a matrix is invertible, then $D(PP^{-1})=D(I)=1$, so $D(P) \neq 0$. So indeed you can reduce the problem to proving that $D(I_k)=0$.

First $D(I_1)=0$ since there exist matrices $P,Q$ invertible such that $PI_1Q$ has only the element on the position $2,2$ equal to $1$ and the rest zero. If $D(I_1)=1$ then $D(PI_1Q)=1$, and $D(0)=D(I_1 \cdot (PI_1Q))=1$. Contradiction.

Suppose now that $D(I_k)\neq 0$. Then $D(I_k)=1$ since $D(I_k)=(D(I_k))^2$. If $k<n$ you can pick $P,Q$ invertible such that $PI_kQ$ has the diagonal elements on positions $2,..,k+1$ equal to $1$ and the rest $0$. Then $D(PI_kQ)=1$ and the product $I_k \cdot PI_kQ$ has only $k-1$ diagonal entries equal to 1 and the value of $D$ on this matrix is $1$. Therefore you can conclude that $D(I_{k-1})=1$. Inductively you reach $D(I_1)=1$ which is a contradiction.

It's not a very pretty solution, but I guess it works.

share|improve this answer

I will use the result mentioned by Beni Bogossel: for a matrix $A$ of rank $r$ we can find invertible matrices $P$ and $Q$ such that $A=PA'Q$, where $A'$ is a diagonal matrix, with $r$ $1$ and $n-r$ zeros. For $S\subset\{1,\ldots,n\}$, we denote $M_S$ the diagonal matrix, with $M_S(i,i)=1$ if $i\in S$ and $0$ otherwise. If $S_1$ and $S_2$ are two subsets of $\{1,\ldots,n\}$ which have the same cardinality, the matrices $M_{S_1}$ and $M_{S_2}$ are similar, and $D(M_{S_1})=D(M_{S_2})$ because $D(M_S)\in\{0,1\}$ for all $S$ and $D(A)\neq 0$ if $A$ is invertible. So if $|S|=1$, then $D(M_S)=0$, because (if we assumed $n\geq 2$) we can find $S'$ disjoint of $S$ with cardinal $1$ and $0=D(M_{S\cap S'})=D(M_S)D(M_{S'})=D(M_S)^2$. Now let $S\subsetneq\{1,\ldots,n\}$. We can find $x\notin S$ so $$D(M_S)=D(M_SM_{\{x\}})=D(M_{S\cap \{x\}})=D(M_S)D(M_{\{x\}})=0$$ and we are done (without induction).

share|improve this answer

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.