Closed razeghi71 closed 8 months ago
I believe the statement is correct as it is: If dim(V) = dim(W) then it holds that
injective <=> surjective <=> bijective
Assume dim(V) = 1, dim(Im(\Phi)) = 1, dim (W) = 2
We can have an injective mapping, but that wouldn't be surjective or bijective.
I may miss something, though.
That's true, what I wrote instead was also wrong, but lets say $V = W = R^2$ and $\Phi$ is a function that maps all vectors in $V$ say $[x, y]^T$ to $[x, 0]^T$ in $W$, here $Im(\Phi) \subseteq W$ but $\Phi$ is not surjective, injective.
OK. Then I'll close this issue
Describe the mistake The last consequence that has been listed at the end of page 60 is that: if $dim(v) = dim(w)$ then $\Phi$ is injective and surjective and bijective. but I think the right version should be: If $dim(v) = dim(Im(\Phi))$ then $\Phi$ is injective and surjective and bijective.
Location Please provide the
Proposed solution Change $dim(v) = dim(w)$ to $dim(v) = dim(Im(\Phi))$