Definition and Linear-Combination Preservation

What It Means to Preserve Linear Structure

A function between vector spaces can send vectors to quite different vectors, but a linear transformation respects two basic operations: adding vectors and multiplying them by scalars. These requirements let us calculate the image of a complicated combination from the images of its component vectors. They also impose consequences that are not separate assumptions, such as sending the zero vector to the zero vector.

Throughout this tutorial, scalars are real. Let \(n,m\in\mathbb{N}\), and let \(T:\mathbb{R}^n\to\mathbb{R}^m\) be a function. The inputs and outputs are column vectors. The addition on the input side takes place in \(\mathbb{R}^n\), while the addition on the output side takes place in \(\mathbb{R}^m\); the scalar multiplication in each space is the usual coordinatewise operation.

Both conditions matter. Preservation of addition concerns the image of a sum, and preservation of scalar multiplication concerns the image of a scaled vector. A linear transformation is required to satisfy them for every permitted input and scalar, not just for selected examples. The definition itself is the starting point; the rules below follow from it.

Consequences for Zero, Negatives and Differences

Write \(\mathbf{0}_n\) for the zero vector in the domain \(\mathbb{R}^n\), and \(\mathbf{0}_m\) for the zero vector in the codomain \(\mathbb{R}^m\). They may have different numbers of entries, so keeping track of which space contains a zero vector is useful.

Proof. Since \(\mathbf{0}_n+\mathbf{0}_n=\mathbf{0}_n\), preservation of addition gives \(T(\mathbf{0}_n)=T(\mathbf{0}_n)+T(\mathbf{0}_n)\). Subtracting \(T(\mathbf{0}_n)\) from both sides in \(\mathbb{R}^m\) yields \(T(\mathbf{0}_n)=\mathbf{0}_m\). Next, \(\mathbf{u}+(-\mathbf{u})=\mathbf{0}_n\), so \(T(\mathbf{u})+T(-\mathbf{u})=T(\mathbf{0}_n)=\mathbf{0}_m\). Thus \(T(-\mathbf{u})=-T(\mathbf{u})\). Finally, \(\mathbf{u}-\mathbf{v}=\mathbf{u}+(-\mathbf{v})\), and preservation of addition together with the negative-vector identity gives \(T(\mathbf{u}-\mathbf{v})=T(\mathbf{u})+T(-\mathbf{v})=T(\mathbf{u})-T(\mathbf{v})\). \(\square\)

These identities are consequences, not extra conditions to add to the definition. In particular, the zero-vector result follows from preservation of addition; it is not necessary to assume separately that \(T\) fixes zero. The theorem also explains why subtraction can be handled by subtracting the corresponding images.

Preservation of Finite Linear Combinations

A linear combination of a list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) in \(\mathbb{R}^n\) is a vector \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where each coefficient \(c_i\) is real. The two defining properties can be applied repeatedly, so the image of this whole combination is determined by the images of the individual vectors.

Proof. We prove the statement by induction on \(p\). For \(p=1\), the formula is precisely preservation of scalar multiplication: \(T(c_1\mathbf{v}_1)=c_1T(\mathbf{v}_1)\). Now suppose the formula holds for a positive integer \(p\), and consider \(p+1\) vectors. By preservation of addition, then the induction hypothesis, and then preservation of scalar multiplication,

Thus the formula holds for \(p+1\). By induction, it holds for every positive integer \(p\). \(\square\)

For a single term, the theorem is the scalar multiplication rule; for two terms, it uses both parts of the definition. The induction establishes that no matter how many terms appear, each coefficient can be taken outside \(T\), and the image of the sum is the sum of the images. The theorem is stated for \(p\geq 1\); the case of an empty sum, if needed, is exactly the already-proved fact \(T(\mathbf{0}_n)=\mathbf{0}_m\).

Relations and Consistent Representations

The linear-combination theorem has a useful consequence for any relation that equals the zero vector. It also ensures that if one input has two different expressions as a linear combination, those expressions cannot lead to different outputs under a linear transformation.

Proof. For the first assertion, apply linear-combination preservation to the left side of the assumed relation: \(c_1T(\mathbf{v}_1)+\cdots+c_pT(\mathbf{v}_p)=T(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p)=T(\mathbf{0}_n)=\mathbf{0}_m\). For the second assertion, subtract the right-hand input combination from the left-hand input combination. The result is zero by the assumed equality. The first assertion applied to the combined list \(\mathbf{v}_1,\dots,\mathbf{v}_p,\mathbf{w}_1,\dots,\mathbf{w}_q\), with coefficients \(c_1,\dots,c_p,-d_1,\dots,-d_q\), gives \(\sum_{i=1}^p c_iT(\mathbf{v}_i)-\sum_{j=1}^q d_jT(\mathbf{w}_j)=\mathbf{0}_m\). Rearranging proves the claimed equality of outputs. \(\square\)

Common Pitfall: Applying a Coefficient to the Wrong Vector

When evaluating \(T(c\mathbf{u})\), the coefficient \(c\) multiplies \(T(\mathbf{u})\), not the original input \(\mathbf{u}\) in place of its image. For a concrete illustration, suppose \(T:\mathbb{R}^2\to\mathbb{R}^2\) is linear, \(\mathbf{u}=(1,0)^\top\), \(\mathbf{v}=(0,1)^\top\), \(T(\mathbf{u})=(2,1)^\top\), and \(T(\mathbf{v})=(-1,3)^\top\). A wrong calculation is \(T(2\mathbf{u}-\mathbf{v})=2T(\mathbf{u})-\mathbf{v}=(4,2)^\top-(0,1)^\top=(4,1)^\top\). The term being subtracted is not \(\mathbf{v}\); it is \(T(\mathbf{v})\). The correct calculation is

The distinction is especially important when domain and codomain differ: \(\mathbf{v}\) belongs to the input space, while \(T(\mathbf{v})\) belongs to the output space. Even when both spaces are \(\mathbb{R}^2\), as in this example, those are still different vectors in general.

Why These Rules Are Useful

Once the images of some vectors are known, linear-combination preservation turns many other image calculations into ordinary arithmetic with vectors. Relations among inputs also impose relations among their outputs. This gives a reliable way to simplify calculations, check whether proposed image values are consistent with a known relation, and compare two expressions for the same input. The key discipline is to apply \(T\) to the entire input expression first, then use the defining rules to expand it.