From a Formula to a Kernel and an Image
A formula for a linear map gives a direct way to ask two different questions. Which inputs produce the zero vector? Which outputs can the map produce at all? The first question asks for the kernel; the second asks for the image. To answer them, we translate the formula into equations, solve those equations, and describe the resulting sets precisely.
Throughout, the spaces in the examples are finite-dimensional real vector spaces. For a linear map \(T:V\to W\), the kernel and image are the sets
To compute the kernel, set every output coordinate equal to zero and solve for the input coordinates. To compute the image, write the output coordinates in terms of the input coordinates, then describe exactly which output vectors arise. Giving only a collection of sample outputs is not enough: the description must show that every stated output is attainable and that no other outputs are attainable.
When the formula is already given in standard coordinates, we can read off this matrix directly. If \(\mathbf{A}\) is the matrix of \(T\), then the output coordinate column is \(\mathbf{A}\mathbf{c}\). This turns the kernel question into the homogeneous system \(\mathbf{A}\mathbf{c}=\mathbf{0}\). For the image, each output is a linear combination of the columns of \(\mathbf{A}\), with the input coordinates as its coefficients.
Why the Coordinate Method Works
Proof. Write \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\), so \(\mathbf{c}=[\mathbf{v}]_{\mathcal{B}}\). By linearity, \(T(\mathbf{v})=c_1T(\mathbf{b}_1)+\cdots+c_nT(\mathbf{b}_n)\). The columns of \(\mathbf{A}\) are the \(\mathcal{C}\)-coordinates of these vectors, so the \(\mathcal{C}\)-coordinate vector of \(T(\mathbf{v})\) is \(\mathbf{A}\mathbf{c}\). A vector is zero exactly when its coordinate vector is zero. Thus \(T(\mathbf{v})=\mathbf{0}\) exactly when \(\mathbf{A}\mathbf{c}=\mathbf{0}\). For the image, every output has the form \(T(\mathbf{v})=c_1T(\mathbf{b}_1)+\cdots+c_nT(\mathbf{b}_n)\), so it belongs to the stated span. Conversely, any linear combination \(c_1T(\mathbf{b}_1)+\cdots+c_nT(\mathbf{b}_n)\) equals \(T(c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n)\), so it is an output. Both inclusions hold, proving the image description. \(\square\)
The proposition gives two complementary procedures. For the kernel, solve the homogeneous system and convert each solution coordinate column back into a vector of \(V\). For the image, use the outputs of a basis of the domain as generators. These procedures do not require the map to be one-to-one or onto, and neither conclusion should be assumed before doing the calculation.
Solving for the Kernel
Worked Example: A Kernel in \(\mathbb{R}^3\)
Let \(T:\mathbb{R}^3\to\mathbb{R}^2\) be given by \(T(x,y,z)=(x+2y-z,\;2x+4y-2z)\). Find \(\ker T\).
An input belongs to the kernel exactly when both output coordinates are zero. The equations are \(x+2y-z=0\) and \(2x+4y-2z=0\). The second equation is twice the first, as the row reduction confirms:
Thus the only constraint is \(x+2y-z=0\). Choose \(y=s\) and \(z=t\) freely. Then \(x=-2s+t\), and every kernel vector has the form
Therefore \(\ker T=\operatorname{span}\{(-2,1,0),(1,0,1)\}\). Check: for \((-2,1,0)\), the output coordinates are \(-2+2-0=0\) and \(-4+4-0=0\). For \((1,0,1)\), they are \(1+0-1=0\) and \(2+0-2=0\). Since the equation \(x+2y-z=0\) was solved with arbitrary \(s\) and \(t\), the parameterization includes every kernel vector, not only the two checked generators.
Free parameters are not extra restrictions: they record the choices that remain after solving the equations. To establish a kernel description, it is useful to verify both directions. First, substitute a proposed general vector into the formula and confirm its image is zero. Then explain why solving the equations forces every kernel vector to have that form.
Describing Every Attainable Output
Proof. If \(\mathbf{w}\in\operatorname{im}T\), then \(\mathbf{w}=T(\mathbf{v})\) for some \(\mathbf{v}\in V\). Since \(\mathcal{B}\) is a basis, \(\mathbf{v}=\sum_{j=1}^n c_j\mathbf{b}_j\) for scalars \(c_j\). Linearity gives \(\mathbf{w}=\sum_{j=1}^n c_jT(\mathbf{b}_j)\), so \(\mathbf{w}\) belongs to the span. In the other direction, take any vector \(\sum_{j=1}^n c_jT(\mathbf{b}_j)\) in that span. Linearity gives \(\sum_{j=1}^n c_jT(\mathbf{b}_j)=T(\sum_{j=1}^n c_j\mathbf{b}_j)\), which is an output of \(T\). Therefore the two sets are equal. \(\square\)
Sometimes the span description is already the clearest answer. In other cases, it is helpful to eliminate the input parameters and give equations that outputs must satisfy. The next example does both, and checks that the equations describe exactly the attainable set.
Worked Example: An Image as a Plane in \(\mathbb{R}^3\)
Define \(S:\mathbb{R}^2\to\mathbb{R}^3\) by \(S(s,t)=(s+t,\;2s-t,\;3s)\). Describe \(\operatorname{im}S\).
Write a general output as \((a,b,c)\). The formula says \(a=s+t\), \(b=2s-t\), and \(c=3s\). Adding the first two equations eliminates \(t\): \(a+b=3s=c\). Thus every output satisfies \(a+b=c\).
To check that this condition is sufficient, suppose \(a+b=c\). Set \(s=c/3\) and \(t=a-c/3\). Then the first output coordinate is \(s+t=c/3+a-c/3=a\). The second is \(2s-t=2c/3-(a-c/3)=c-a=b\), using \(a+b=c\). The third is \(3s=c\). Hence \((a,b,c)\) is attained, and
Equivalently, evaluating \(S\) on the standard basis vectors gives \(S(1,0)=(1,2,3)\) and \(S(0,1)=(1,-1,0)\), so \(\operatorname{im}S=\operatorname{span}\{(1,2,3),(1,-1,0)\}\). Check: the general combination \(s(1,2,3)+t(1,-1,0)\) is \((s+t,2s-t,3s)\), whose first coordinate plus second is \(3s\), its third coordinate. This agrees with the equation description.
A Map Defined on Matrices
Worked Example: Kernel and Image of a Matrix Formula
Let \(U:\mathbb{R}^{2\times2}\to\mathbb{R}^2\) be defined by \(U\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)=(a+d,\;b+c)\). Find its kernel and image.
For the kernel, set both output coordinates equal to zero. The equations are \(a+d=0\) and \(b+c=0\), so \(d=-a\) and \(c=-b\). Therefore
For the image, let \((u,v)\in\mathbb{R}^2\) be any proposed output. Choose the input matrix \(\begin{bmatrix}u&v\\0&0\end{bmatrix}\). Its output is \((u+0,v+0)=(u,v)\), so every vector in \(\mathbb{R}^2\) is attained and \(\operatorname{im}U=\mathbb{R}^2\).
Checks: applying \(U\) to \(\begin{bmatrix}a&b\\-b&-a\end{bmatrix}\) gives \((a-a,b-b)=(0,0)\). For the proposed preimage of \((u,v)\), direct substitution gives \(U\left(\begin{bmatrix}u&v\\0&0\end{bmatrix}\right)=(u+0,v+0)=(u,v)\). Thus both descriptions follow from the defining formula.
Common Pitfall: Row Reduction Does Not Preserve the Image
When finding a kernel, row reduction is appropriate because reversible row operations preserve the solutions of a homogeneous system. But row operations generally change the columns, so the columns of the reduced matrix should not be used as the original map's output vectors.
For a concrete example, consider the map with matrix \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\), from \(\mathbb{R}^2\) to \(\mathbb{R}^3\). The columns of \(\mathbf{A}\) are \((1,0,1)\) and \((0,1,1)\), so its image is their span. A mistaken calculation might row-reduce and then take the nonzero rows, \((1,0)\) and \((0,1)\), as generators for the image. Those rows are vectors in \(\mathbb{R}^2\), not \(\mathbb{R}^3\), so they cannot describe outputs of this map.
The correction is to use the original columns for the image, or derive the output formula directly. For input \((x,y)\), the output is \((x,y,x+y)\). Hence the image is \(\{(a,b,c)\in\mathbb{R}^3:c=a+b\}\). Every such vector is attained by choosing input \((a,b)\), and every output has the required equation. Row reduction may help solve for the kernel, but the image belongs to the codomain and must be described there.
Check Your Understanding
For each problem, state the full set you obtain and verify it from the defining formula.
- For \(T(x,y)=(x-y,\;2x-2y)\), find \(\ker T\).
- For \(S(s,t)=(s+2t,\;3s+t)\), describe the image as a span of vectors obtained from the standard basis.
- For \(P(x,y,z)=(x+y,\;y+z)\), describe every attainable output in \(\mathbb{R}^2\).
- For \(Q\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)=(a-b,\;c+d)\), find the kernel.
- Explain why row-reducing a matrix is valid for solving its homogeneous system, but its reduced columns need not describe the original map's image.
