[Math] proof linear transformation injective, surjective, isomorphism

linear algebralinear-transformations

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:

  1. $f$ is surjective $\Rightarrow m \le n$
  2. $f$ is injective $\Rightarrow m \ge n$
  3. $f$ is isomorphism $\Rightarrow m = n$
  4. $m = n \not \Rightarrow f$ 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

$dim(ker(A)) + dim(im(A)) = n$

  1. $f$ 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:

$dim(ker(A)) + m = n$

$\Rightarrow m \le n$

  1. $f$ 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.

  1. You've got it right.
  2. Since $f$ is injective, we have $\mathrm {Ker}(f) = 0 \implies \dim (\mathrm {Ker}(f)) = 0 \implies \dim (\mathrm {Im}(f)) = n - \dim (\mathrm {Ker}(f)) = n.$ Also notice that for $f \in \mathcal L(\mathbb R^n, \mathbb R^m)$, it always holds that $\mathrm {Im}(f)$ is a subspace of $\mathbb R^m$. Therefore, $$ n = \dim(\mathrm {Im}(f)) \leqslant \dim (\mathbb R^m) = m. $$
  3. The combination of 1, 2. A bijection is a mapping that is both injective and surjective. An isomorphism is a bijective linear mapping.
  4. The zero map is surely one counterexample.
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