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 $ker(A)$.


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

$\Rightarrow m \le n$

  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

  • 1
    $\begingroup$ For the first 3 problems, just use $\dim(ker(f))+\dim(im(f))=n$ along with the fact that $\dim(ker(f))\ge0$ and if $ker(f)=\{0\}$, then $\dim(ker(f))=0$. Note that dimension of the zero space is 0 and not 1. For the last problem, just take $m=n$ and take $f$ to be the zero map. $\endgroup$ – VENKITESH Dec 17 '17 at 21:31
  • $\begingroup$ 1. $dim(ker(f)) \ge 0$ and $im(f) = \mathbb{R}^m \Rightarrow dim(im(f)) = m \Rightarrow m \le n$ Is it correct? $\endgroup$ – John Dec 17 '17 at 22:45
  • $\begingroup$ 2. $ker(f) = \{0\} \Rightarrow dim(ker(f)) = 0 \Rightarrow 0 + dim(im(f)) = n$ $dim(im(f)) \le m$ But this doesn't prove $m \ge n$ $\endgroup$ – John Dec 17 '17 at 22:45
  • $\begingroup$ 3. $dim(ker(f)) \ge 0$ and $im(f) = \mathbb{R}^m \Rightarrow dim(im(f)) = m \Rightarrow m = n$ This doesn't seem correct. Since isomorphism means that there is a bijective transformation between $\mathbb{R}^n$ and $\mathbb{R}^m$, isn't the following correct? $dim(ker(f)) = 0$ and $im(f) = \mathbb{R}^m \Rightarrow dim(im(f)) = m \Rightarrow m = n$ $\endgroup$ – John Dec 17 '17 at 22:46
  • $\begingroup$ 4. Let $M(f) \in \mathbb{R}^{n \times n}$ be the zero matrix. Then it is true, $f$ is not injective, since all vectors from the domain of $f$ get mapped to the zero vector in the codomain. Therefore $f$ is not bijective and hence $f$ is not isomorphism. Is it correct? $\endgroup$ – John Dec 17 '17 at 22:46

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.
  1. $f$ is surjective $\Rightarrow m \le n$:

$\color{green}{\text{True: at least m columns are needed.}}$

  1. $f$ is injective $\Rightarrow m \ge n$:

$\color{green}{\text{True: indeed if f is injective M can't have more than m columns otherwise ker(M)\neq 0}}$

  1. $f$ is isomorphism $\Rightarrow m = n$

$\color{green}{\text{True: indeed the matrix M must be full rank}}$

  1. $m = n \not \Rightarrow f$ is isomorphism

$\color{green}{\text{True: indeed implication holds only if M is full rank.}}$

  • 1
    $\begingroup$ Please correct this answer: for 2, OP has now swapped $m\le n$ with $m\ge n$; for 4, you claim it true, not false, considering there’s $\not\Rightarrow$ and not $\Rightarrow$. $\endgroup$ – arseniiv Dec 17 '17 at 23:19
  • $\begingroup$ @arseniiv Thanks! $\endgroup$ – user Dec 17 '17 at 23:23
  • $\begingroup$ Thanks for your Answer. Could you please also comment on my new solutions above? $\endgroup$ – John Dec 18 '17 at 2:22
  • $\begingroup$ I guess we should prove all four statements with the nullity-rank theorem. Thanks for your help! $\endgroup$ – John Dec 18 '17 at 22:48

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