$V$ is a subspace of $\mathbb{R}^{n\times n},\dim(V)\geq kn+1.$ How to show that there exist $ A\in V$ and $r(A)\geq k+1?$

Asked Oct 16 '19 at 11:54

Active Oct 16 '19 at 12:54

Viewed 50 times

I want to find the maximal dimension of a subspace of $\mathbb{R}^{n\times n}$ such that all members of it are not invertible. I think the problem above may help,but I don't know how to prove it.Any help will be thanked.

edited Oct 16 '19 at 12:54

Bernard

179,256

asked Oct 16 '19 at 11:54

Tree23

1,198

See this post – Ben Grossmann Oct 16 '19 at 12:14
What is $r(A)$? – Bernard Oct 16 '19 at 12:55
@Bernard Presumably the rank. Note that the question was tagged as being related to rank. – Ben Grossmann Oct 16 '19 at 14:13
The rank of a vector??? – Bernard Oct 16 '19 at 14:34
@Bernard $V$ is a subspace of $\Bbb R^{n \times n}$, and so is a set of matrices. Of course every element of $V$ has a rank. – Ben Grossmann Oct 16 '19 at 22:11
@Omnomnomnom: Of course. I hadn't realised that $\mathbf R^{n\times n}$ might denote here the space of square matrices of order $n$ (which I usually denote $M_n(\mathbf R)$). – Bernard Oct 16 '19 at 22:18

$V$ is a subspace of $\mathbb{R}^{n\times n},\dim(V)\geq kn+1.$ How to show that there exist $ A\in V$ and $r(A)\geq k+1?$

0 Answers0

Linked