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Let $S: \mathbb R^3 \to \mathbb R^4$ and $T: \mathbb R^4 \to \mathbb R^3$ be linear transformation such that $T\circ S$ is the identity map of $\mathbb R^3$. Then

  1. $S\circ T$ is the identity map of $\mathbb R^4$.

  2. $S \circ T$ is one-one, but not onto.

  3. $S \circ T$ is not one-one,but onto.

  4. $S \circ T$ is neither one-one nor onto.

Since $T\circ S$ is the identity map of $\mathbb R^3$ then it must be one-one and the dimension of domain ($\mathbb R^3$) and co-domain ($\mathbb R^3$) are same then it is a bijective map. Then we also say that $(T\circ S)^{-1}$ also exist. Then how to show which one is correct from the above options.

user642796
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Sayan rakshit
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1 Answers1

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Since $T\circ S$ is the identity map (so bijective) then it's a classic result that $T$ is onto and $S$ is one to one. Hence by the rank nullity theorem $\dim\ker T=1$ and $\dim{\rm im}S=3$ and then $S\circ T$ is neither one to one nor onto.

Added We can see the result using contradiction. If one of the options $1,2$ or $3$ holds then $T$ or $S$ or both become bijective which's a contradiction since $\Bbb R^3$ isn't isomorphic to $\Bbb R^4$.