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Let $T:V\longrightarrow W$ be a linear transformation, where $V(F)$ and $W(F)$ are finite dimensional vector spaces. Show that dim $V$=dim $W$ iff $T$ is non singular.

A textbook has the following solution:

We know that dim $V$=dim $ R(T)$+dim $N(T)$. Therefore, dim $V$=dim $ R(T)$=dim $W$. If and only if dim $N(T)=0$, i.e., if and only if $N(T)=\lbrace 0 \rbrace$, i.e. , if and only if $T$ in non singular.

The above proof is very unsatisfactory. How can we write dim $R(T)$= dim $W$? Converse part is also not clear.

gete
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  • $T$ nonsingular implies that $T$ is surjective and injective. Hence $\dim R(T)=\dim W$. – Dietrich Burde Nov 03 '19 at 14:58
  • It seems that there is a stop that it should not be. Does "Therefore, dim $V$=dim $ R(T)$=dim $W$, if and only if dim $N(T)=0$, i.e., if and only if $N(T)=\lbrace 0 \rbrace$, i.e. , if and only if $T$ in non singular." make more sense to you? – ajotatxe Nov 03 '19 at 15:09
  • @ajotatxe You have pointed right. That full stop in the textbook made me more confused. – gete Nov 03 '19 at 15:14
  • @Dietrich Burde Thanks, i am not very much familiar with the notion of non singular transformations. Does $T$ being non singular imply that $T$ is invertible? – gete Nov 03 '19 at 15:18
  • @DietrichBurde T non singular means $kerT=(0)$ as per standard definition,for a linear operator i.e. transformation on the same set,it is equivalent to saying bijective. – Kishalay Sarkar Nov 03 '19 at 15:42

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The question is wrong,perhaps the term non singular that means $kerT=\{0\}$ should be replaced by the term invertible.For example take the inclusion map $(x,y)\to (x,y,0)$ from $2D$ space to $3D$ space which is nonsingular,but $dim(V)\neq dim(W)$.
Addendum $Ker(T)=(0)$ iff $T$ is injective.You can easily check $T$ is injective,but $T$ is not surjective.So $T$ is not an isomorphism.You probably know that $T$ is isomorphic means $T$ is bijective linear transformation(definition).

Kishalay Sarkar
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  • this problem was taken from a textbook entitled Linear Algebra by K. P Gupta, page no. 126. To my knowledge, the same problem are also available in some other literatures though i couldn't recall the exact title of the literatures. – gete Nov 03 '19 at 15:57
  • @gete You must see the definition of non singular,I have used standard definition. – Kishalay Sarkar Nov 03 '19 at 16:00
  • Yes that is true and you seem to have pointed right. However, i am not that good at this topic. So, i couldn't understand how did you claim that your counter example is a non singular map? I am bit unfamiliar with such notation of mappings. – gete Nov 03 '19 at 16:10
  • Let me add something to make you understand,see the edit. – Kishalay Sarkar Nov 03 '19 at 16:33
  • Got it, but in the text book which i referred, $T$ being non singular seems to imply that $T$ is isomorphism also. – gete Nov 03 '19 at 17:03
  • Then the statement is true,what you wrote. – Kishalay Sarkar Nov 03 '19 at 17:16
  • @gete See https://math.stackexchange.com/questions/679558/proving-v-is-isomorphic-to-w-iff-dim-v-dim-w – Kishalay Sarkar Nov 03 '19 at 17:17