## Definition

Let $$V$$ and $$W$$ be vector spaces over the field $$\mathbb{F}$$ having the same finite dimension. Let $$T:V\rightarrow W$$ be a linear transformation.

$$T$$ is said to be invertible if there is a linear transformation $$S:W\rightarrow V$$ such that $$S(T(x)) = x$$ for all $$x\in V$$. $$S$$ is called the inverse of $$T$$. In casual terms, $$S$$ undoes whatever $$T$$ does to an input $$x$$.

In fact, under the assumptions at the beginning, $$T$$ is invertible if and only if $$T$$ is bijective. Here, we give a proof that bijectivity implies invertibility. The other direction is left as an exercise.

Suppose that $$T$$ is bijective. So for each $$x \in V$$, there is a unique vector in $$W$$, call it $$y_x$$ such that $$T(x) = y_x$$. Define $$S:W \rightarrow V$$ as follows: for each $$y \in W$$, let $$x \in V$$ be such that $$y = y_x$$. Note that such an $$x$$ exists as $$T$$ is surjective and the choice is unique as $$T$$ is injective. Hence, $$S$$ is a well-defined function from $$W$$ to $$V$$ and it satisfies $$S(T(x)) = x$$. It remains to show that $$S$$ is a linear transformation.

Let $$y_1, y_2 \in W$$ and $$\alpha \in \mathbb{F}$$. We show that $$S$$ satisfies $$S(y_1 + y_2) = S(y_1) + S(y_2)$$ and $$S(\alpha y_1) = \alpha S(y_1)$$.

Let $$x_1, x_1 \in V$$ be such that $$T(x_i) = y_i$$ for $$i = 1,2$$. Then $$S(y_i) = x_i$$ for $$i = 1,2$$. Using that $$T$$ is a linear transformation, we have \begin{eqnarray*} S(y_1 + y_2) &=& S(T(x_1)+T(x_2))\\ & =& S(T(x_1 + x_2)) \\ & =& x_1 + x_2 \\ & =& S(y_1) + S(y_2) \end{eqnarray*} and \begin{eqnarray*} S(\alpha y_1) & = & S(\alpha T(x_1)) \\ & = & S(T(\alpha x_1)) \\ & = & \alpha x_1 \\ & = & \alpha S(y_1). \end{eqnarray*} This shows that $$S$$ is a linear transformation.

### Examples

1. We have seen a while back that linear transformations corresponding to elementary row operations are invertible.

2. Let $$T:\mathbb{R}^3\rightarrow \mathbb{R}^3$$ be given by $$T\left(\begin{bmatrix} x\\y\\z\end{bmatrix}\right) =\begin{bmatrix} x\\y\\0\end{bmatrix}.$$ We argue that $$T$$ is not invertible. If you plot $$p=\begin{bmatrix} x\\y\\z\end{bmatrix}$$ in the $$x$$-$$y$$-$$z$$ space with the $$x$$-$$y$$ plane as the horizontal plane, then $$T(p)$$ is the intersection of the vertical line through $$p$$ with the $$x$$-$$y$$ plane. Call this point of intersection $$p'$$. Now, every point on this vertical line gets mapped to $$p'$$ by $$T$$. Hence, given $$p'$$, there is no unique element in $$\mathbb{R}^3$$ that gets mapped to $$p'$$, implying that no inverse of $$T$$ exists.

We mention in passing that $$T$$ is a called a projection because it “projects” points in 3-space onto the $$x$$-$$y$$ plane.

## Testing invertibility

If $$T$$ is given by $$T(x) = Ax$$ where $$A$$ is a square matrix with entries from a field, then $$T$$ is invertible if and only if $$A$$ is invertible.

## Exercises

For each of the following linear transformations, determine if it is invertible.

1. $$T:\mathbb{R}^2\rightarrow \mathbb{R}^2$$ given by $$T\left(\begin{bmatrix} x\\y\end{bmatrix}\right) = \begin{bmatrix} x+y \\ 2x-y \end{bmatrix}$$.

2. $$T:\mathbb{C}^2\rightarrow \mathbb{C}^2$$ given by $$T(z) = Az$$ where $$A = \begin{bmatrix} i & 2 \\ 1 & -2i \end{bmatrix}$$.