# problem about the dimensions of two subspaces

Let $\{v_1,\ldots,v_k\}$ be vectors of $\mathbb R^n$ with $k\le n$ and consider the following subspaces:

$M=\{A \in \mathbb R^{n\times n} : Av_i=0, i=1, \ldots ,k \}$

$N=\operatorname{span}\{v_1,\ldots, v_k\}$.

what is the relationship between the $\dim(M)$ and $\dim(N)$?

please see if there is any error in this solution, and correct. Thank you

sol.

let $\quad \beta =\{A_1,A_2,\ldots,A_p\}$ a base of $M$ $\quad (p\le n \times n)$

and $\quad \gamma =\{v_1,\ldots,v_q\}$ a base of $N$ $\quad (q\le k).$

let's consider the following matrices

$$W=\begin{bmatrix} A_1 \\ A_2 \\ \vdots \\ A_p \end{bmatrix} \quad and \quad B=\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_q \end{bmatrix}$$

Note that: $W \in \mathbb R^{(p.n)\times n},\quad$$B \in \mathbb R^{n\times q}\quad$ and $WB=O$ then $B\in Nu(W)$ $\Rightarrow$ $q\le Nu(W) \quad$ (as the $posto(B)=q$)

by the Rank-Nullity Theorem we have to

$n=dim(Nu(W)+posto(W)$ $\quad \Rightarrow\quad$ $posto (W)\le n-q \quad$

As the

$posto(W)=dim(M), \quad q=posto(B)=dim(N)$ then

$dim(M)+dim(N)\le n$

please see if there is any error in this solution, and correct. Thank you

• This is the "Rank-Nullity Theorem" – Doug M Dec 7 '17 at 1:38
• It's not clear to me that by the definition you gave M is a matrix space whereas N is a vector space. Do you confirm? – gimusi Dec 7 '17 at 13:08
• M and N satisfy the definition of subspace, M y N son subespacios diferentes, pero si creo que haya una relacion entre sus dimenciones – ros Dec 7 '17 at 13:53
• Consider the matrix $B$ whose columns are $v_1,\dots,v_k$. Each row of $A$ is in the (left) nullspace of $B$. The dimension of the span of $\{\,v_1,\dots,v_k\,\}$ is the rank of $B$. – Gerry Myerson Dec 8 '17 at 4:58
• Any thoughts, ros? – Gerry Myerson Dec 9 '17 at 11:55

You mean $q\le k-i+1{}{}{}{}{}{}{}{}{}{}$?
• I already corregi was $\gamma=\{v_1, . . .,v_q\}$ – ros Dec 9 '17 at 3:20