From One Solution to Every Solution
For a linear map, solving \(T(\mathbf{v})=\mathbf{w}\) has two distinct parts. First, determine whether the target \(\mathbf{w}\) is reached by the map at all. If it is reached, find one vector that maps to it. Then the kernel describes exactly how that solution can be changed without changing its image. This gives a complete description of the solution set, not just a way to find one answer.
Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), and let \(T:V\to W\) be linear. Linearity means that \(T(a\mathbf{u}+b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v})\) for all \(\mathbf{u},\mathbf{v}\in V\) and scalars \(a,b\in\mathbb{F}\). No finite-dimensional assumption is needed for the results in this tutorial.
The image answers the existence question: a solution exists precisely when the target belongs to \(\operatorname{im}T\). Once a solution \(\mathbf{v}_0\) has been found, the kernel accounts for all other solutions. The next result states both facts and proves the full description.
Proof. By the definition of image, \(\mathbf{w}\in\operatorname{im}T\) means that there is some \(\mathbf{v}_0\in V\) with \(T(\mathbf{v}_0)=\mathbf{w}\). This is exactly the condition that the equation has a solution. Now suppose such a \(\mathbf{v}_0\) exists. If \(\mathbf{v}\in S_{\mathbf{w}}\), then \(T(\mathbf{v})=\mathbf{w}=T(\mathbf{v}_0)\). By linearity, \(T(\mathbf{v}-\mathbf{v}_0)=T(\mathbf{v})-T(\mathbf{v}_0)=\mathbf{0}\), so \(\mathbf{v}-\mathbf{v}_0\in\ker T\). Thus \(\mathbf{v}=\mathbf{v}_0+\mathbf{u}\) for some \(\mathbf{u}\in\ker T\), and \(S_{\mathbf{w}}\subseteq\mathbf{v}_0+\ker T\). Conversely, if \(\mathbf{u}\in\ker T\), then \(T(\mathbf{v}_0+\mathbf{u})=T(\mathbf{v}_0)+T(\mathbf{u})=\mathbf{w}+\mathbf{0}=\mathbf{w}\). Therefore \(\mathbf{v}_0+\mathbf{u}\in S_{\mathbf{w}}\), proving the reverse inclusion. If there is no solution, the set of solutions has no elements, so it is empty. \(\square\)
The two directions in the set equality are important. Subtracting a known solution from any solution shows that the difference must lie in the kernel. Adding any kernel vector to a known solution gives another solution. Together, these statements rule out both missing solutions and extra vectors in the proposed description.
Finding a Particular Solution and the Kernel
In practice, separate the work into two tasks. Find one particular solution \(\mathbf{v}_0\), then solve the homogeneous equation \(T(\mathbf{u})=\mathbf{0}\) to find \(\ker T\). The theorem combines these answers. If the target is not in the image, there is no particular solution to start from, and the fiber is empty; a kernel calculation cannot change that fact.
Worked Example: A Map from R³ to R²
Define \(T:\mathbb{R}^3\to\mathbb{R}^2\) by \(T\left(\begin{bmatrix}x\\y\\z\end{bmatrix}\right)=\begin{bmatrix}x+y\\y+z\end{bmatrix}\). Find every solution of \(T(\mathbf{v})=\begin{bmatrix}2\\1\end{bmatrix}\).
First choose \(y=0\). The two target equations then require \(x=2\) and \(z=1\), so one particular solution is \(\mathbf{v}_0=\begin{bmatrix}2\\0\\1\end{bmatrix}\). Check it directly: \(T(\mathbf{v}_0)=\begin{bmatrix}2+0\\0+1\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}\).
Next find the kernel by solving \(x+y=0\) and \(y+z=0\). Set \(y=t\), where \(t\in\mathbb{R}\). Then \(x=-t\) and \(z=-t\), so \(\ker T=\left\{\begin{bmatrix}-t\\t\\-t\end{bmatrix}:t\in\mathbb{R}\right\}\). The theorem gives every solution as \(\begin{bmatrix}2\\0\\1\end{bmatrix}+\begin{bmatrix}-t\\t\\-t\end{bmatrix} =\begin{bmatrix}2-t\\t\\1-t\end{bmatrix}\), for \(t\in\mathbb{R}\).
Check: Applying \(T\) to the claimed general solution gives \(T\left(\begin{bmatrix}2-t\\t\\1-t\end{bmatrix}\right) =\begin{bmatrix}(2-t)+t\\t+(1-t)\end{bmatrix} =\begin{bmatrix}2\\1\end{bmatrix}\). Thus every parameter value gives a solution, and the theorem ensures there are no others.
A particular solution need not be unique. The next proposition explains why choosing a different one cannot change the final solution set.
Proof. Since both vectors map to \(\mathbf{w}\), linearity gives \(T(\mathbf{v}_1-\mathbf{v}_0)=T(\mathbf{v}_1)-T(\mathbf{v}_0)=\mathbf{w}-\mathbf{w}=\mathbf{0}\). Hence \(\mathbf{k}=\mathbf{v}_1-\mathbf{v}_0\in\ker T\), so \(\mathbf{v}_1=\mathbf{v}_0+\mathbf{k}\). If \(\mathbf{u}\in\ker T\), then \(\mathbf{v}_1+\mathbf{u}=\mathbf{v}_0+(\mathbf{k}+\mathbf{u})\), and \(\mathbf{k}+\mathbf{u}\in\ker T\), because the kernel is a subspace. This proves \(\mathbf{v}_1+\ker T\subseteq\mathbf{v}_0+\ker T\). Also, \(-\mathbf{k}\in\ker T\); applying the same reasoning with \(\mathbf{v}_0=\mathbf{v}_1-\mathbf{k}\) gives the reverse inclusion. Therefore the two translates are equal. \(\square\)
Examples Across Different Vector Spaces
Worked Example: Differentiation of Polynomials
Let \(T:P_2\to P_1\) be differentiation over \(\mathbb{R}\), where \(P_2\) is the space of real polynomials of degree at most \(2\), and \(P_1\) is the space of real polynomials of degree at most \(1\). Find all \(p\in P_2\) such that \(T(p)=2x+3\).
A particular solution is \(p_0=x^2+3x\), because \(T(p_0)=2x+3\). To compute the kernel, write a general polynomial as \(p(x)=ax^2+bx+c\). Then \(T(p)=2ax+b\), which is the zero polynomial exactly when \(2a=0\) and \(b=0\). Thus \(a=b=0\), while \(c\) is unrestricted, and \(\ker T\) consists of the constant polynomials.
All solutions are therefore \(p(x)=x^2+3x+c\), where \(c\in\mathbb{R}\). Check: differentiating gives \(2x+3+0=2x+3\) for every \(c\). Every solution is included because the difference between any two solutions has derivative zero and is therefore constant.
Worked Example: A Trace Target
Define \(T:\mathbb{R}^{2\times2}\to\mathbb{R}\) by \(T(\mathbf{A})=\operatorname{tr}\mathbf{A}\), the sum of the diagonal entries. Find every matrix \(\mathbf{A}\) for which \(T(\mathbf{A})=5\).
One solution is \(\mathbf{A}_0=\begin{bmatrix}5&0\\0&0\end{bmatrix}\), since its trace is \(5+0=5\). A general matrix in the kernel has trace zero. Writing its entries and setting the trace equal to zero gives \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\) with \(a+d=0\), so \(d=-a\). Hence \(\ker T=\left\{\begin{bmatrix}a&b\\c&-a\end{bmatrix}:a,b,c\in\mathbb{R}\right\}\).
Adding a kernel matrix to \(\mathbf{A}_0\) yields all solutions: \(\mathbf{A}=\begin{bmatrix}5+a&b\\c&-a\end{bmatrix}\), where \(a,b,c\in\mathbb{R}\). Check: the trace is \((5+a)+(-a)=5\). Conversely, any \(2\times2\) matrix of trace \(5\) has diagonal entries whose sum is \(5\); setting \(a\) equal to its first diagonal entry minus \(5\), and taking \(b,c\) to be its off-diagonal entries, gives exactly the displayed form.
Worked Example: A Target with No Preimage
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be given by \(T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}x\\0\end{bmatrix}\). Determine whether \(T(\mathbf{v})=\begin{bmatrix}1\\2\end{bmatrix}\) has a solution.
For every input \(\begin{bmatrix}x\\y\end{bmatrix}\), the second coordinate of its image is \(0\). The target has second coordinate \(2\), so no input can map to it. Equivalently, \(\operatorname{im}T=\left\{\begin{bmatrix}x\\0\end{bmatrix}:x\in\mathbb{R}\right\}\), and \(\begin{bmatrix}1\\2\end{bmatrix}\notin\operatorname{im}T\). Thus the solution set is empty. The kernel here is \(\left\{\begin{bmatrix}0\\y\end{bmatrix}:y\in\mathbb{R}\right\}\), but adding its vectors cannot produce a solution when no particular preimage exists.
Common Pitfall: Stopping at One Solution
Finding one vector that maps to the target does not usually solve the equation completely. For example, let \(T:\mathbb{R}^2\to\mathbb{R}\) be \(T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=x+y\), and consider the target \(3\). The vector \(\begin{bmatrix}3\\0\end{bmatrix}\) is a particular solution, but it is wrong to conclude that it is the only one: \(T\left(\begin{bmatrix}2\\1\end{bmatrix}\right)=2+1=3\) as well.
The kernel consists of vectors satisfying \(x+y=0\), so it is \(\left\{\begin{bmatrix}-t\\t\end{bmatrix}:t\in\mathbb{R}\right\}\). The full set of solutions is \(\begin{bmatrix}3\\0\end{bmatrix}+\begin{bmatrix}-t\\t\end{bmatrix} =\begin{bmatrix}3-t\\t\end{bmatrix}\). Substitution confirms \(T\left(\begin{bmatrix}3-t\\t\end{bmatrix}\right)=(3-t)+t=3\). In particular, \(t=1\) recovers the second solution \(\begin{bmatrix}2\\1\end{bmatrix}\). The correction is to report one preimage together with the entire kernel, not to stop after finding the preimage.
There is also a useful special case. If \(\ker T=\{\mathbf{0}\}\) and a solution \(\mathbf{v}_0\) exists, the theorem gives \(S_{\mathbf{w}}=\mathbf{v}_0+\{\mathbf{0}\}=\{\mathbf{v}_0\}\), so the solution is unique. If the kernel contains a nonzero vector \(\mathbf{u}\), then \(\mathbf{v}_0+\mathbf{u}\ne\mathbf{v}_0\), and both vectors solve the equation; in that case a nonempty solution set has more than one element.
The procedure is therefore precise: test whether the target lies in the image; if it does, find one preimage; then add every vector in the kernel. This method applies to maps on coordinate spaces, polynomial spaces, and matrix spaces alike. It also keeps existence separate from the number and form of solutions: the image determines whether solutions exist, while the kernel determines how solutions vary.
Check Your Understanding
For each problem, distinguish the existence question from the description of all solutions.
- For \(T:\mathbb{R}^2\to\mathbb{R}\), \(T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=2x-y\), find all solutions to \(T(\mathbf{v})=4\) by finding one preimage and the kernel.
- For \(D:P_2\to P_1\), \(D(p)=p'\), determine whether \(D(p)=x^2+1\) has a solution. Explain using the codomain and the form of derivatives of polynomials in \(P_2\).
- Suppose \(T(\mathbf{v}_0)=\mathbf{w}\) and \(\mathbf{u}\in\ker T\). Prove directly from linearity that \(\mathbf{v}_0+\mathbf{u}\) is also a solution.
- If two vectors \(\mathbf{v}_0,\mathbf{v}_1\) both map to the same target, what must be true of \(\mathbf{v}_1-\mathbf{v}_0\)? Justify the answer.
- For the trace map on \(\mathbb{R}^{2\times2}\), give a matrix with trace \(0\) and use it to produce a second solution from \(\begin{bmatrix}5&0\\0&0\end{bmatrix}\) to the target \(5\).
