Linear Transformations, Kernels, and Lost Information

The big idea

A linear transformation is a rule that moves vectors while preserving addition and scaling. Its kernel is the set of all input vectors that the rule sends to the zero vector. In symbols, \(\operatorname{ker}(T)=\{v:T(v)=0\}\). Finding a kernel therefore means solving a homogeneous system: set every output coordinate equal to zero and solve all the resulting equations together.

The kernel records lost information. If the only vector sent to zero is the zero vector itself, the transformation is one-to-one: two different inputs cannot produce the same output. If a nonzero vector lies in the kernel, adding that vector to an input does not change its output, so collisions occur.

Worked example

Let \(T(x,y)=(x+y,2x+2y)\). To find its kernel, solve

\[ x+y=0, \qquad 2x+2y=0. \]

The second equation repeats the first, so let \(x=t\) and write \(y=-t\). The complete solution set is

\[ \operatorname{ker}(T)=\{(t,-t):t\in\mathbb R\}. \]

Geometrically, this is a line through the origin. For example, both \((0,0)\) and \((1,-1)\) produce the output \((0,0)\), so this transformation is not one-to-one.

How children may show it

A strong solution connects three representations:

• the scalar equations obtained from \(T(v)=0\);
• the full solution set, often written with a parameter; and
• the geometric subspace, such as the origin or a line through it.

For a proof, the learner may use linearity to write \(T(p-q)=T(p)-T(q)\). This turns a question about two equal outputs into a question about one vector in the kernel.

Common mix-up

A learner may find one vector sent to zero and call that vector “the kernel.” The kernel is the entire set of such vectors. Another mix-up is treating the kernel as an empty set when no nonzero solution appears. The zero vector is always in the kernel of a linear transformation, so the smallest possible kernel is \(\{0\}\), called the trivial kernel.

It is also important not to confuse an inverse image with an inverse function. A fiber \(T^{-1}(\{s\})\) is simply the set of all inputs that produce \(s\); the notation makes sense even when \(T\) is not invertible.

Try it together

Choose a simple matrix and multiply it by \((x,y)\). Set the output equal to \((0,0)\), solve the two equations, and sketch the solution set. Then pick one input \(p\) and add a kernel vector \(k\). Compare \(T(p+k)\) with \(T(p)\). Ask which step uses linearity and what the comparison says about information the matrix keeps or loses.