Space of matrices with bounded row space
Let $V$ be a vector space of $m \times n$ matrices over a field $\mathbb{F}$ with the property that, for all $M \in V, \ \operatorname{rowsp}(M) \subseteq W$, where $W \subseteq \mathbb{F}^n$ is a subspace which is a row space of some $M \in V$. Prove that $\operatorname{dim}_\mathbb{F}(V) = \operatorname{max}\{n,m\} \operatorname{dim}_\mathbb{F}(W)$.
Ribs
114
Answer
Answers can only be viewed under the following conditions:
- The questioner was satisfied with and accepted the answer, or
- The answer was evaluated as being 100% correct by the judge.
Martin
1.7K
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 1270 views
- $10.00
Related Questions
- [Linear Algebra] Diagonalizable operator and Spectrum
- Find eigenvalues and eigenvectors of $\begin{pmatrix} 1 & 6 & 0 \\ 0& 2 & 1 \\ 0 & 1 & 2 \end{pmatrix} $
- Diagonalization of linear transformations
- Show that eigenvectors of a symmetric matrix are orthogonal
- Space of all matrices with given column space
- Find $a,b,c$ so that $\begin{bmatrix} 0 & 1& 0 \\ 0 & 0 & 1\\ a & b & c \end{bmatrix} $ has the characteristic polynomial $-\lambda^3+4\lambda^2+5\lambda+6=0$
- Certain isometry overfinite ring is product of isometries over each local factor
- Linearly independent vector subsets.
If you want equality for the dimension you need to also assume that W is itself the row space of some M in V. Otherwise V=0 is a counterexample.
You're right, edited, thanks