Is The Composition of Two Linear Transformations Invertible Assume we have two linear compositions $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ and $S: \mathbb{R}^n \rightarrow \mathbb{R}^n$. How would I go about proving that $S \circ T$ ( Composition of $S$ and $T$) is an invertible linear transformation
T(x1,x2)=(x1-x2,3x1-2x2)
S(x1,x2)=(2x1+3x2,-x1+x2)
 A: Note that a linear map $T$ is invertible iff $\det T \neq 0$.
Thus, we need to check precisely when $\det(S\circ T) \neq 0$.
However, note that
$$\det(S\circ T) = \det S\cdot\det T.$$
Thus, we get that
$$\begin{align}\det(S\circ T) \neq 0 &\iff \det S\cdot\det T \neq 0\\
& \iff \det S \neq 0 \text{ and } \det T \neq 0
\end{align}$$
Thus, we get that $S \circ T$ is invertible iff $S$ and $T$ are invertible.

Remark. We crucially used that $S$ and $T$ were linear maps between spaces of the same dimension. (When we wrote $\det(S\circ T) = \det S \cdot \det T$.)
If we were in the situation $S:\Bbb R^m \to \Bbb R^n$ and $T:\Bbb R^n \to \Bbb R^m$, the same would not be true.
A: Please check these links:
https://math.mit.edu/classes/18.013A/MathML/chapter04/section02.xhtml
Linear Algebra, meaning of 0 determinant in linear transformations
If the linear transform is not invertible, it will transform a set of basis vectors into a zero volume due to the columns of the transformation matrix being lin. dependent
