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?


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.

  • $\begingroup$ yes, I have forgotten $char F=0$ $\endgroup$ – Babak Miraftab Jul 3 '12 at 19:07
  • 5
    $\begingroup$ 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. $\endgroup$ – Dylan Moreland Jul 3 '12 at 19:17

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