Let $V$ and $W$ be $n$-dimensional vector spaces, and let $T:V\rightarrow W$ be a linear transformation. Suppose that $\beta$ is a basis for $V$. Prove that $T$ is an isomorphism if and only if $T(\beta)$ is a basis for $W$.

I know how to prove this question, but there is one part that I didn't quite understand that I need some help.


First suppose that $T$ is an isomorphism and let $\beta$ = {$\beta_1$,$\beta_2$,....., $\beta_n$}

Show that $\beta$ is linearly independent, which means

$$\sum_{i=1}^n a_iT(\beta_i) = 0$$ only when all $a_i=0$, where $a_i$ are scalars.

Note that $\sum_{i=1}^n a_iT(\beta_i)=T(\sum_{i=1}^n a_i\beta_i) = 0$, which implies $\sum_{i=1}^n a_i\beta_i$ is within the null space of $T$

and $\sum_{i=1}^n a_i\beta_i=0$ since $T$ is an isomorphism.

This is where I stuck, someone told me that " $\sum_{i=1}^n a_i\beta_i=0$ since $T$ is an isomorphism." But I don't understand why, can anyone explain to me?


Note that if $T$ is an isomorphism, it is injective (that is, one-to-one). A linear transformation is injective if and only if its nullspace consists only of the $0$ vector.

So, for $\sum_{i=1}^n a_i \beta_i$ to be in the null space of $T$ (or in the null space of any injective linear map), we must have $\sum_{i=1}^n a_i \beta_i=0$.


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