Let $A\in M_{n\times n}(\mathbb{R})$ be of rank $m.$ Then the map $\mathbb{R}^{n}\rightarrow \mathbb{R}^{m}$ given by $v\rightarrow Av$ is not injective.
checking of injectiveness of the function
-1
$\begingroup$
real-analysis
linear-algebra
matrices
matrix-rank
-
0presumably $m
– 2017-01-28
1 Answers
2
The matrix $A$ should be in $M_{m\times n}(\mathbb{R})$ in order for $v\mapsto Av$ to be a linear map $\mathbb{R}^n\to\mathbb{R}^m$.
If $n>m$, then the map cannot be injective because of the rank-nullity theorem.
If $n=m$, the map is injective.
It is not possible that $n
-
0yes. got it. thanks – 2017-01-28