Answer by xbh for proof linear transformation injective, surjective, isomorphism
All these exercises could be derived from the nullity-rank theroem. You've got it right.Since $f$ is injective, we have $\mathrm {Ker}(f) = 0 \implies \dim (\mathrm {Ker}(f)) = 0 \implies \dim (\mathrm...
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$f$ is surjective $\Rightarrow m \le n$: $\color{green}{\text{True: at least m columns are needed.}}$$f$ is injective $\Rightarrow m \ge n$:$\color{green}{\text{True: indeed if f is injective M can't...
View Articleproof linear transformation injective, surjective, isomorphism
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...
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