What Linearity Preserves
A linear map respects the two operations used to build vectors: addition and scalar multiplication. As a result, whenever an input is assembled from several vectors using those operations, its image can be assembled from the images using the same coefficients. This rule is useful when a vector is easier to describe as a combination of vectors whose images are already known.
Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), and let \(T:V\to W\). We use the definition that \(T\) is linear when, for all \(\mathbf{u},\mathbf{v}\in V\) and \(c\in\mathbb{F}\), it satisfies additivity and homogeneity:
A finite linear combination of \(\mathbf{v}_1,\dots,\mathbf{v}_p\in V\) is an expression \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where each \(c_i\in\mathbb{F}\). The coefficients can be zero, negative, or—over \(\mathbb{C}\)—complex. The preservation rule says that applying \(T\) to the whole expression gives the same coefficients on the images.
Proof. First, homogeneity with scalar \(0\) gives \(T(\mathbf{0})=T(0\mathbf{v})=0T(\mathbf{v})=\mathbf{0}\), for any \(\mathbf{v}\in V\). This proves the assertion about the empty combination. We now prove the formula for positive \(p\) by induction. For \(p=1\), the formula is exactly homogeneity: \(T(c_1\mathbf{v}_1)=c_1T(\mathbf{v}_1)\). Suppose the formula holds for a list of \(p\) vectors. For \(p+1\) vectors, additivity and then the induction hypothesis and homogeneity give $$ \begin{aligned} T\left(\sum_{i=1}^{p+1}c_i\mathbf{v}_i\right) &=T\left(\sum_{i=1}^{p}c_i\mathbf{v}_i+c_{p+1}\mathbf{v}_{p+1}\right)\\ &=T\left(\sum_{i=1}^{p}c_i\mathbf{v}_i\right)+T(c_{p+1}\mathbf{v}_{p+1})\\ &=\sum_{i=1}^{p}c_iT(\mathbf{v}_i)+c_{p+1}T(\mathbf{v}_{p+1}). \end{aligned} $$ Thus the formula holds for \(p+1\). By induction, it holds for every positive integer \(p\). \(\square\)
The theorem applies to a list of inputs, not just to distinct vectors: some \(\mathbf{v}_i\) may be equal, and some coefficients may be zero. Those cases require no extra conditions. The rule also does not require that the vectors form a basis or be linearly independent.
Computing an Image from Known Images
Worked Example: A Vector in \(\mathbb{R}^2\)
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be \(T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}2x-y\\x+3y\end{bmatrix}\). Compute the image of \(\begin{bmatrix}5\\1\end{bmatrix}\) by writing the input as a combination of vectors with known images.
Choose \(\mathbf{u}=\begin{bmatrix}1\\1\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}1\\-1\end{bmatrix}\). Solving \(\begin{bmatrix}5\\1\end{bmatrix}=a\mathbf{u}+b\mathbf{v}\) gives \(a+b=5\) and \(a-b=1\). Adding the equations gives \(2a=6\), so \(a=3\); then \(b=2\). Therefore $$ \begin{bmatrix}5\\1\end{bmatrix} =3\begin{bmatrix}1\\1\end{bmatrix} +2\begin{bmatrix}1\\-1\end{bmatrix}. $$ The map is linear: its formula is a linear combination of the input coordinates in each output coordinate, so it respects addition and scalar multiplication. Its images on the two chosen vectors are $$ T(\mathbf{u})=\begin{bmatrix}2(1)-1\\1+3(1)\end{bmatrix} =\begin{bmatrix}1\\4\end{bmatrix}, \qquad T(\mathbf{v})=\begin{bmatrix}2(1)-(-1)\\1+3(-1)\end{bmatrix} =\begin{bmatrix}3\\-2\end{bmatrix}. $$ By preservation of linear combinations, $$ T\left(\begin{bmatrix}5\\1\end{bmatrix}\right) =3\begin{bmatrix}1\\4\end{bmatrix} +2\begin{bmatrix}3\\-2\end{bmatrix} =\begin{bmatrix}3+6\\12-4\end{bmatrix} =\begin{bmatrix}9\\8\end{bmatrix}. $$ Check: using the defining formula directly gives \(\begin{bmatrix}2(5)-1\\5+3(1)\end{bmatrix}=\begin{bmatrix}9\\8\end{bmatrix}\), the same result.
The same argument works in vector spaces whose elements are not coordinate columns. For example, polynomials can be combined by adding them and multiplying them by scalars. Their coefficients can serve as the inputs to a linear map.
Worked Example: A Map on Polynomials
Let \(P_2(\mathbb{R})\) be the vector space of real polynomials of degree at most \(2\), and define \(T:P_2(\mathbb{R})\to\mathbb{R}^2\) by \(T(a+bx+cx^2)=\begin{bmatrix}a+2b-c\\3a-b+2c\end{bmatrix}\). Compute \(T(2-3x+4x^2)\).
This map is linear: the coefficients of a sum of polynomials are the sums of the corresponding coefficients, and scaling a polynomial scales each coefficient. Each output coordinate is a fixed linear combination of those coefficients, so both additivity and homogeneity hold. The input decomposes as \(2\cdot 1+(-3)x+4x^2\). The images of the three monomials are $$ T(1)=\begin{bmatrix}1\\3\end{bmatrix},\qquad T(x)=\begin{bmatrix}2\\-1\end{bmatrix},\qquad T(x^2)=\begin{bmatrix}-1\\2\end{bmatrix}. $$ Therefore, $$ \begin{aligned} T(2-3x+4x^2) &=2T(1)-3T(x)+4T(x^2)\\ &=2\begin{bmatrix}1\\3\end{bmatrix} -3\begin{bmatrix}2\\-1\end{bmatrix} +4\begin{bmatrix}-1\\2\end{bmatrix}\\ &=\begin{bmatrix}2\\6\end{bmatrix} +\begin{bmatrix}-6\\3\end{bmatrix} +\begin{bmatrix}-4\\8\end{bmatrix} =\begin{bmatrix}-8\\17\end{bmatrix}. \end{aligned} $$ Check: substituting \(a=2\), \(b=-3\), \(c=4\) into the formula gives \(\begin{bmatrix}2+2(-3)-4\\3(2)-(-3)+2(4)\end{bmatrix}=\begin{bmatrix}-8\\17\end{bmatrix}\).
Images of Matrix Combinations
A matrix space is also a vector space: matrices of the same size can be added and multiplied by scalars entry by entry. A matrix in that space can be written as a combination of matrices that have one entry equal to \(1\) and all other entries equal to \(0\). Preservation then reduces a computation to the images of those elementary matrices.
Worked Example: A Map from \(2\times2\) Matrices
For \(\mathbf{A}=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\), define \(T(\mathbf{A})=\begin{bmatrix}a_{11}+a_{12}-2a_{21}\\3a_{11}-a_{22}\end{bmatrix}\). Compute the image of \(\begin{bmatrix}2&-1\\3&4\end{bmatrix}\) using its entries.
The formula is linear in the four entries: adding matrices adds each entry, and scaling a matrix scales each entry. Thus \(T\) is linear. Let \(\mathbf{E}_{ij}\) denote the \(2\times2\) matrix whose entry in row \(i\), column \(j\) is \(1\) and whose other entries are \(0\). The input decomposes into its entry contributions: $$ \begin{bmatrix}2&-1\\3&4\end{bmatrix} =2\mathbf{E}_{11}-\mathbf{E}_{12}+3\mathbf{E}_{21}+4\mathbf{E}_{22}. $$ Applying the formula to each elementary matrix gives $$ T(\mathbf{E}_{11})=\begin{bmatrix}1\\3\end{bmatrix},\quad T(\mathbf{E}_{12})=\begin{bmatrix}1\\0\end{bmatrix},\quad T(\mathbf{E}_{21})=\begin{bmatrix}-2\\0\end{bmatrix},\quad T(\mathbf{E}_{22})=\begin{bmatrix}0\\-1\end{bmatrix}. $$ Preservation of the combination now yields $$ \begin{aligned} T\left(\begin{bmatrix}2&-1\\3&4\end{bmatrix}\right) &=2\begin{bmatrix}1\\3\end{bmatrix} -\begin{bmatrix}1\\0\end{bmatrix} +3\begin{bmatrix}-2\\0\end{bmatrix} +4\begin{bmatrix}0\\-1\end{bmatrix}\\ &=\begin{bmatrix}2\\6\end{bmatrix} +\begin{bmatrix}-1\\0\end{bmatrix} +\begin{bmatrix}-6\\0\end{bmatrix} +\begin{bmatrix}0\\-4\end{bmatrix} =\begin{bmatrix}-5\\2\end{bmatrix}. \end{aligned} $$ Check: direct substitution gives \(\begin{bmatrix}2+(-1)-2(3)\\3(2)-4\end{bmatrix}=\begin{bmatrix}-5\\2\end{bmatrix}\).
Why Values on a Spanning Set Are Enough
A set \(S\subseteq V\) spans \(V\) when every vector in \(V\) is a finite linear combination of elements of \(S\). A basis is an ordered list that is both linearly independent and spanning. For determining a linear map, the spanning property is the key: the preservation rule allows the image of any combination to be calculated from the images of its ingredients.
Proof. Fix \(\mathbf{v}\in V\). Because \(S\) spans \(V\), there are a nonnegative integer \(p\), vectors \(\mathbf{s}_1,\dots,\mathbf{s}_p\in S\), and scalars \(c_1,\dots,c_p\) such that \(\mathbf{v}=\sum_{i=1}^{p}c_i\mathbf{s}_i\). If \(p=0\), then \(\mathbf{v}=\mathbf{0}\), and linearity gives \(T(\mathbf{v})=\mathbf{0}=U(\mathbf{v})\). If \(p>0\), the preservation theorem and the assumption that the maps agree on \(S\) give $$ T(\mathbf{v})=\sum_{i=1}^{p}c_iT(\mathbf{s}_i) =\sum_{i=1}^{p}c_iU(\mathbf{s}_i) =U(\mathbf{v}). $$ Since \(\mathbf{v}\) was arbitrary, the maps agree on all of \(V\). A basis spans \(V\), so the final statement follows. \(\square\)
This result explains why a finite-dimensional linear map can be described by its values on a basis: every input is a combination of basis vectors, and its image must use the same coefficients on their images. The theorem asserts uniqueness from those values; it does not say that arbitrary assigned values automatically define a linear map. A proposed definition must still be consistent with the linear-map requirements.
Common Mistake: Dropping or Changing Coefficients
When applying the preservation rule, each coefficient stays attached to the image of the vector it originally multiplied. For instance, suppose \(T:\mathbb{R}^2\to\mathbb{R}^2\) is \(T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}x+y\\x-y\end{bmatrix}\), \(\mathbf{u}=\begin{bmatrix}1\\0\end{bmatrix}\), and \(\mathbf{v}=\begin{bmatrix}0\\1\end{bmatrix}\). Then \(T(\mathbf{u})=\begin{bmatrix}1\\1\end{bmatrix}\) and \(T(\mathbf{v})=\begin{bmatrix}1\\-1\end{bmatrix}\).
A wrong calculation might claim $$ T(2\mathbf{u}-3\mathbf{v})=T(\mathbf{u})-T(\mathbf{v}) =\begin{bmatrix}0\\2\end{bmatrix}. $$ This drops the coefficients \(2\) and \(-3\). The correct calculation is $$ \begin{aligned} T(2\mathbf{u}-3\mathbf{v}) &=2T(\mathbf{u})-3T(\mathbf{v})\\ &=2\begin{bmatrix}1\\1\end{bmatrix} -3\begin{bmatrix}1\\-1\end{bmatrix} =\begin{bmatrix}2\\2\end{bmatrix} -\begin{bmatrix}3\\-3\end{bmatrix} =\begin{bmatrix}-1\\5\end{bmatrix}. \end{aligned} $$ Check: \(2\mathbf{u}-3\mathbf{v}=\begin{bmatrix}2\\-3\end{bmatrix}\), and the defining formula gives \(T\left(\begin{bmatrix}2\\-3\end{bmatrix}\right)=\begin{bmatrix}2-3\\2-(-3)\end{bmatrix}=\begin{bmatrix}-1\\5\end{bmatrix}\). The coefficients are preserved, including their signs.
Check Your Understanding
Use the preservation rule explicitly in each calculation.
- If \(T\) is linear, \(\mathbf{v}=2\mathbf{u}-\mathbf{w}\), \(T(\mathbf{u})=\begin{bmatrix}1\\3\end{bmatrix}\), and \(T(\mathbf{w})=\begin{bmatrix}4\\-2\end{bmatrix}\), find \(T(\mathbf{v})\).
- Suppose a linear map \(T\) satisfies \(T(1)=\begin{bmatrix}1\\0\end{bmatrix}\), \(T(x)=\begin{bmatrix}0\\1\end{bmatrix}\), and \(T(x^2)=\begin{bmatrix}1\\1\end{bmatrix}\). Find \(T(3-2x+5x^2)\).
- Write \(\begin{bmatrix}1&2\\-1&3\end{bmatrix}\) as a combination of the matrices \(\mathbf{E}_{ij}\). How does a linear map act on that combination?
- Prove that two linear maps agreeing on a spanning set agree at the zero vector, including when the spanning set is empty.
- Identify and correct the coefficient error in the claim \(T(4\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\).
