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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.


  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.

Quick Quiz


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 \(A = \begin{bmatrix} i & 2 \\ 1 & -2i \end{bmatrix}\).