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 $V$ be a finite-dimensional vector space over $\mathbb F$ ($\text{char}\ \mathbb F =0$) and $T$ is a linear transformation such that for any basis the matrix of linear transformation has at least one zero.

I want to show that there exists $c\in \mathbb F$ $\,$such that $T(x)=cx$ for any $x \in V.$

Any suggestion?

share|improve this question

1 Answer 1

up vote 6 down vote accepted

It certainly won't work in general. Over the field $GF(2)$, the only matrix with no zero entries is all $1$'s, and there are lots of $n \times n$ matrices that are not similar to this besides $0$ and $I$.

EDIT: with the added condition $\text{char } {\mathbb F} = 0$, it works. Let $u$ be a vector such that $u$ and $Tu$ are linearly independent. There is a basis $b_1, b_2, \ldots, b_n$ with $b_1 = u$ and $b_2 = Tu$. Let $B$ be the matrix with these basis vectors as columns, so $(B^{-1} T B)_{2,1} \ne 0$. Consider $C = \sum_{P} c_P B P$ where the sum is over all $n \times n$ permutation matrices and $c_P$ are indeterminates. $\det(C)$ is a polynomial in the $c_P$, and not identically $0$ (note that $\det(C) = \pm \det(B) \ne 0$ if any one of the $c_P$ is $1$ and the others are $0$). Similarly, any matrix element of $(\text{Adj}(C) T C)_{ij}$ (where $\text{Adj}$ is the adjugate or classical adjoint matrix) is a polynomial in the $c_P$, and not identically $0$: if we take one particular $c_P$ for permutation $\pi$ to be $1$ and the others $0$, $(\text{Adj}(C) T C)_{ij} = \det(BP) (P^{-1} B^{-1} T B P)_{ij}$, and if $\pi(j) = 1$ and $\pi(i)=2$ this is $\det(BP)$. There will therefore be some set of rational values of the $c_P$ for which all those polynomials are nonzero, and the columns of the corresponding $C$ are a basis in which all matrix elements of $T$ are nonzero.

share|improve this answer
    
yes, I have forgotten $char F=0$ –  Babak Miraftab Jul 3 '12 at 19:07
5  
Not that such things are very important, but I don't understand the downvote on this answer. It deals with the original question very well. –  Dylan Moreland Jul 3 '12 at 19:17

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.