The following exercise is given:
Let $f:\mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation.
Prove, according to your knowledge about Kernel and Image of the Transformation Matrix $M(f)$, the following statements:
- $f$ is surjective $\Rightarrow m \le n$
- $f$ is injective $\Rightarrow m \ge n$
- $f$ is isomorphism $\Rightarrow m = n$
- $m = n \not \Rightarrow f$ is isomorphism
My Solution:
For a matrix $A \in \mathbb{R}^{m \times n}$ the kernel and image are defined:
$ker(A) = \{x \in \mathbb{R}^n : A \cdot x = 0\}$
$im(A) = \{A \cdot x : x \in \mathbb{R}^n\}$
And we know
$dim(ker(A)) + dim(im(A)) = n$
- $f$ is surjective means $im(A)=\mathbb{R}^m$. So $dim(im(A)) = m$ and $dim(ker(A)) \ge 1$ since at least the null vector is in $ker(A)$.
So:
$dim(ker(A)) + m = n$
$\Rightarrow m \le n$
- $f$ is injective means $ker(A)=\{0\}$, so $dim(ker(A))=1$ and $0 \ge dim(im(A)) \lt m$
I'm not sure whether so far my proof(s) are correct, but from here I don't know how to proceed with 2., 3. and 4.
Thank for your help
Best Answer
All these exercises could be derived from the nullity-rank theroem.