Images and Spans
A linear transformation sends vectors to vectors while respecting addition and scalar multiplication. When the inputs are restricted to a subspace generated by a few vectors, it can be inefficient to test every vector in that subspace separately. Instead, transform the generators. Linearity guarantees that their images contain all the information needed to describe the image subspace.
Let \(T:\mathbb{R}^n\to\mathbb{R}^m\) be a linear transformation. This means that \(T(\mathbf{x}+\mathbf{y})=T(\mathbf{x})+T(\mathbf{y})\) and \(T(c\mathbf{x})=cT(\mathbf{x})\) for all \(\mathbf{x},\mathbf{y}\in\mathbb{R}^n\) and \(c\in\mathbb{R}\). For a subset \(S\) of the domain, its image is the set \(T(S)=\{T(\mathbf{x}):\mathbf{x}\in S\}\). The image of a subspace \(S\) is therefore a set of outputs, not necessarily all of \(\mathbb{R}^m\).
The key fact is that applying a linear transformation to every vector in a span has the same result as spanning the transformed generators. The generators need not be independent, and \(T\) need not be one-to-one or onto.
Proof. First suppose \(k\geq 1\). To prove equality of the two sets, prove that each is contained in the other. Take any vector in the left-hand set. It has the form \(T(\mathbf{x})\) for some \(\mathbf{x}\in\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_k\}\). By the definition of span, \(\mathbf{x}=c_1\mathbf{v}_1+\cdots+c_k\mathbf{v}_k\) for scalars \(c_1,\dots,c_k\). By linearity, applying \(T\) gives \(T(\mathbf{x})=c_1T(\mathbf{v}_1)+\cdots+c_kT(\mathbf{v}_k)\), which belongs to the right-hand set. Conversely, take any vector in the right-hand set. It has the form \(c_1T(\mathbf{v}_1)+\cdots+c_kT(\mathbf{v}_k)\). Linearity makes this equal to \(T(c_1\mathbf{v}_1+\cdots+c_k\mathbf{v}_k)\). The input \(c_1\mathbf{v}_1+\cdots+c_k\mathbf{v}_k\) belongs to the original span, so this output belongs to the left-hand set. Thus both containments hold. If \(k=0\), the input span is \(\{\mathbf{0}\}\), and its image is \(\{T(\mathbf{0})\}=\{\mathbf{0}\}\), since linearity gives \(T(\mathbf{0})=\mathbf{0}\). The span of the empty list of images is also \(\{\mathbf{0}\}\). The equality holds in this case as well. \(\square\)
Finding an Image by Transforming Generators
If a transformation is given by a matrix \(\mathbf{A}\in\mathbb{R}^{m\times n}\), then \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\). To find the image of a span, multiply the matrix by each listed generator and take the span of the resulting vectors. The theorem establishes that this procedure accounts for every linear combination of the original generators, rather than just the generators themselves.
Worked Example: The Image of a Line Under a Shear
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) have standard matrix \(\mathbf{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\). Find the image of the line \(L=\operatorname{span}\{\mathbf{v}\}\), where \(\mathbf{v}=\begin{bmatrix}1\\2\end{bmatrix}\).
First transform the generator:
By the image-of-a-span theorem, \(T(L)=\operatorname{span}\left\{\begin{bmatrix}3\\2\end{bmatrix}\right\}\). To see the set directly, a general vector of \(L\) is \(c\mathbf{v}=\begin{bmatrix}c\\2c\end{bmatrix}\). Its image is
So every image lies in the stated span, and every scalar multiple of \(\begin{bmatrix}3\\2\end{bmatrix}\) is obtained by using that scalar as \(c\). The calculation checks the result for an arbitrary input in the line.
For a line, there is only one image generator. If its image is nonzero, the image is a line; if its image is zero, the image is the subspace \(\{\mathbf{0}\}\). This follows directly from the theorem, since a span of one nonzero vector consists of its scalar multiples, whereas a span of the zero vector contains only the zero vector.
Images of Planes
A plane through the origin in \(\mathbb{R}^3\) can be described as the span of two independent vectors. Its image is the span of the two transformed vectors. Those two images might be independent, might both be zero, or might be dependent with at least one nonzero. Accordingly, the image can be a plane, the zero subspace, or a line. The original plane being two-dimensional does not by itself force its image to be a plane.
Worked Example: A Plane Projected onto the \(xy\)-Plane
Define \(T:\mathbb{R}^3\to\mathbb{R}^3\) by \(T(x,y,z)=(x,y,0)\), and let \(P=\operatorname{span}\{\mathbf{v}_1,\mathbf{v}_2\}\), where \(\mathbf{v}_1=\begin{bmatrix}1\\0\\1\end{bmatrix}\) and \(\mathbf{v}_2=\begin{bmatrix}0\\1\\1\end{bmatrix}\). Find \(T(P)\).
The matrix of \(T\) is \(\mathbf{A}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}\). Transform each generator:
Therefore \(T(P)=\operatorname{span}\left\{\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix}\right\}\), the \(xy\)-plane. Check this for an arbitrary input: \(s\mathbf{v}_1+t\mathbf{v}_2=\begin{bmatrix}s\\t\\s+t\end{bmatrix}\), and
Every pair \(s,t\in\mathbb{R}\) is allowed, so these outputs give exactly the stated plane.
Worked Example: A Plane Collapses to a Line
Let \(S=\operatorname{span}\{\mathbf{w}_1,\mathbf{w}_2\}\subseteq\mathbb{R}^3\), with \(\mathbf{w}_1=\begin{bmatrix}1\\0\\0\end{bmatrix}\) and \(\mathbf{w}_2=\begin{bmatrix}1\\1\\0\end{bmatrix}\). Let \(U:\mathbb{R}^3\to\mathbb{R}^3\) be projection onto the \(x\)-axis, with matrix \(\mathbf{B}=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}\). Find \(U(S)\).
Multiplying gives \(U(\mathbf{w}_1)=\begin{bmatrix}1\\0\\0\end{bmatrix}\) and \(U(\mathbf{w}_2)=\begin{bmatrix}1\\0\\0\end{bmatrix}\). Thus the theorem gives
For a direct check, an input in \(S\) has the form \(s\mathbf{w}_1+t\mathbf{w}_2=\begin{bmatrix}s+t\\t\\0\end{bmatrix}\), so its image is \(\begin{bmatrix}s+t\\0\\0\end{bmatrix}\). This is always a multiple of \(\begin{bmatrix}1\\0\\0\end{bmatrix}\), and every multiple \(a\begin{bmatrix}1\\0\\0\end{bmatrix}\) occurs by choosing \(s=a\) and \(t=0\). The calculation verifies that the image is exactly a line.
Subspaces Generated by a Basis
A basis of a subspace is an independent list that spans that subspace. When a subspace is specified by a basis, the image-of-a-span theorem says to transform the basis vectors and span their outputs. The transformed list need not remain independent: a transformation can send distinct basis vectors to the same vector or to zero.
Worked Example: The Image of a Basis-Generated Subspace
Let \(T:\mathbb{R}^3\to\mathbb{R}^2\) have matrix \(\mathbf{C}=\begin{bmatrix}1&2&0\\0&1&1\end{bmatrix}\), and let \(V=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}=\mathbb{R}^3\). Find \(T(V)\).
The images of the standard basis vectors are the columns of \(\mathbf{C}\):
Hence \(T(V)=\operatorname{span}\left\{\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}2\\1\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\right\}\). The first and third vectors are the standard basis of \(\mathbb{R}^2\), so this span is \(\mathbb{R}^2\). To check that every output can be obtained, take any \(\begin{bmatrix}a\\b\end{bmatrix}\in\mathbb{R}^2\) and choose \(\mathbf{x}=\begin{bmatrix}a\\0\\b\end{bmatrix}\). Then
Thus the image is all of \(\mathbb{R}^2\), as the span calculation states.
Common Pitfall: Assuming the Image Keeps the Same Shape
A common mistake is to assume that a plane must map to a plane because it began as a plane. The projection example above gives a concrete counterexample: two independent vectors \(\mathbf{w}_1=\begin{bmatrix}1\\0\\0\end{bmatrix}\) and \(\mathbf{w}_2=\begin{bmatrix}1\\1\\0\end{bmatrix}\) generate a plane, but their images under \(U\) are equal. A wrong calculation would keep both original vectors and conclude \(U(S)=\operatorname{span}\{\mathbf{w}_1,\mathbf{w}_2\}=S\). That computes the original subspace, not its image.
The correction is to apply \(U\) to the generators before taking their span: \(U(\mathbf{w}_1)=U(\mathbf{w}_2)=\begin{bmatrix}1\\0\\0\end{bmatrix}\), so \(U(S)=\operatorname{span}\left\{\begin{bmatrix}1\\0\\0\end{bmatrix}\right\}\). The general rule is to transform every generator, including any generator whose image may be zero, and then take the span of the resulting list. The theorem requires linearity; for a transformation that is not linear, the formula is not guaranteed.
This result is useful because it reduces an entire subspace to a finite calculation. If a subspace is supplied by generators, its image is determined by a finite list of matrix-vector products. The same method also helps distinguish the image of a particular subspace \(S\) from the image of the whole domain: \(T(S)\) uses only inputs in \(S\), while \(T(\mathbb{R}^n)\) allows every input in the domain.
Check Your Understanding
Use the image-of-a-span theorem, and check your conclusions by applying the transformation to a general linear combination.
- Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) have matrix \(\begin{bmatrix}1&0\\2&0\end{bmatrix}\). Find the image of \(\operatorname{span}\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}\).
- For \(P=\operatorname{span}\left\{\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\1\end{bmatrix}\right\}\), find its image under the map \(T(x,y,z)=(x+y,y+z)\).
- Explain why the image of \(\operatorname{span}\{\mathbf{v}\}\) is \(\{\mathbf{0}\}\) when \(T(\mathbf{v})=\mathbf{0}\).
- Give a matrix transformation from \(\mathbb{R}^3\) to \(\mathbb{R}^2\) whose image of \(\mathbb{R}^3\) is a line, and verify the image using the three standard basis vectors.
- In the image-of-a-span proof, identify the step that uses linearity in each direction of the set equality.
