When we say that a transformation is linear, we are saying that we can “pull” constants out before applying the transformation and break the transformation up over addition and subtraction. Mathematically, this means that the following two rules hold for any vectors \(\vec{u}\) and \(\vec{v}\) in the domain and all scalars, \(c\) and \(d\).

- \(T(c\vec{v}) = cT(\vec{v})\)
- \(T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})\)

These two rules can be combined into the following, equivalent rule.

\(T(c\vec{u} + d\vec{v}) = cT(\vec{u}) + dT(\vec{v})\)

## Using this rule to prove a transformation is linear

Let’s use an example to see how you would use this definition to prove a given transformation is linear.

### Example

Show that \(T\left(\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}\right) = \begin{bmatrix} x \\ 5y \\ x + z \\ \end{bmatrix}\) is a linear transformation, using the definition.

### Solution

Looking at the rule, this transformation takes vectors in \(R^3\) to vectors in \(R^3\), as the input and output vectors both have 3 entries. We must show that the definition above holds for ANY vectors in the domain \(R^3\) and any scalars, \(c\) and \(d\).

Anytime you start a proof like this, make sure you define any variables you use. Notice that this is essentially the first line of the proof below.

Overall, since our goal is to show that \(T(c\vec{u} + d\vec{v}) = cT(\vec{u}) + dT(\vec{v})\), we will calculate one side of this equation and then the other, finally showing that they are equal.

### Proof

Let \(\vec{u} = \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ \end{bmatrix}\) and \(\vec{v} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ \end{bmatrix}\) be vectors in \(R^3\) and \(c\) and \(d\) be scalars.

Then:

\(T(c\vec{u} + d\vec{v}) =\)

Idea: combine the vectors and then apply the rule for T.

\(\begin{align}T\left(c\begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ \end{bmatrix} + d\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ \end{bmatrix}\right) &= T\left(\begin{bmatrix} cu_1 + dv_1 \\ cu_2 + dv_2 \\ cu_3 + dv_3\\\end{bmatrix}\right)\\ &= \begin{bmatrix} cu_1 + dv_1 \\ 5(cu_2 + dv_2) \\ (cu_1 + dv_1) + (cu_3 + dv_3)\\\end{bmatrix}\end{align}\)

And:

\(cT(\vec{u}) + dT(\vec{v}) =\)

Idea: apply the rule for T first and then combine the vectors.

\(\begin{align} cT\left(\begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ \end{bmatrix}\right) + dT\left(\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ \end{bmatrix}\right) &= c\begin{bmatrix} u_1 \\ 5u_2 \\ u_1 + u_3 \\ \end{bmatrix} + d\begin{bmatrix} v_1 \\ 5v_2 \\ v_1 + v_3 \\ \end{bmatrix}\\ &= \begin{bmatrix} cu_1 + dv_1\\ 5cu_2 + 5dv_2 \\ c(u_1 + u_3) + d(v_1 + v_3) \\ \end{bmatrix}\end{align}\)

Idea: to conclude the proof, explain how this shows the definition holds and show the two calculations are equal.

Since:

\(\begin{bmatrix} cu_1 + dv_1 \\ 5(cu_2 + dv_2) \\ (cu_1 + dv_1) + (cu_3 + dv_3)\\\end{bmatrix} = \begin{bmatrix} cu_1 + dv_1 \\ 5cu_2 + 5dv_2 \\ (cu_1 + cu_3) + (dv_3 + dv_3)\\\end{bmatrix} = \begin{bmatrix} cu_1 + dv_1\\ 5cu_2 + 5dv_2 \\ c(u_1 + u_3) + d(v_1 + v_3) \\ \end{bmatrix}\)

we have shown that \(T(c\vec{u} + d\vec{v}) = cT(\vec{u}) + dT(\vec{v})\). Thus, by definition, the transformation is linear. \(\blacksquare\)

## Important

Notice that the proof above did not use any specific vectors like \(\begin{bmatrix} 1 \\ 5 \\ 2 \\ \end{bmatrix}\) or \(\begin{bmatrix} 6 \\ 9 \\ 0 \\ \end{bmatrix}\). Showing the rule works for specific vectors or scalars only shows that it works for those specific values! The idea is that we must show it works for ANY GENERAL vectors and scalars.

## Other methods

When a transformation maps vectors from \(R^n\) to \(R^m\) for some n and m (like the one above, for instance), then we have other methods that we can apply to show that it is linear. For example, we can show that T is a matrix transformation, since every matrix transformation is a linear transformation. To learn how to find such a matrix, check out this article: Finding the standard matrix for a linear transformation.

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