Linear transformations · Definition

Rank–nullity theorem

linear algebralinear transformations

Before you start you need to know

This theorem splits the dimension of the domain between what the map destroys and what it keeps, so one of the two quantities can be found from the other.

Definition

For a linear transformation \(T\) with finite-dimensional domain \(V\), the dimension of the kernel plus that of the image equals the dimension of \(V\).

\[\dim V=\dim\ker T+\dim\operatorname{Im}T\]

The dimension of the kernel is called the nullity and that of the image, the rank. For an \(m\times n\) matrix \(A\), viewed as \(T(x)=Ax\), the domain is \(\mathbb{R}^n\) and the theorem reads \(\text{rank}(A)+\text{nullity}(A)=n\): the number of columns is split between the pivot columns (which count the rank) and the free variables (which count the nullity).

The idea of the proof is to complete a basis. Take a basis \(u_1,\dots,u_k\) of the kernel and extend it to a basis \(u_1,\dots,u_k,w_1,\dots,w_r\) of the domain. The vectors \(T(w_1),\dots,T(w_r)\) form a basis of the image: they span because the \(u_i\) are sent to zero, and they are independent because a relation among them would give a combination of the \(w_j\) that \(T\) sends to zero, that is, a kernel vector that is not a combination of the \(u_i\). Hence \(\dim V=k+r\).

Example: for \(T(x,y,z)=(x+y,\,y+z)\) the kernel is the line spanned by \((1,-1,1)\), of dimension \(1\), and the image is \(\mathbb{R}^2\), of dimension \(2\); indeed \(1+2=3\), the dimension of \(\mathbb{R}^3\).

Immediate consequences: if \(\dim V>\dim W\), no map \(V\to W\) can be injective, since the kernel is nontrivial; if \(\dim V<\dim W\), none can be onto. When \(\dim V=\dim W\), injective and onto are equivalent. A common mistake is to apply the theorem to the codomain: the sum gives the dimension of the domain, not that of \(W\).

Think about it

A map \(T:\mathbb{R}^5\to\mathbb{R}^5\) is injective. Is it onto?

Yes: the kernel is \(\{0\}\), so the image has dimension \(5\), which is the whole codomain.

Why does a homogeneous system with more unknowns than equations always have nontrivial solutions?

Because the rank is at most the number of equations, so the nullity \(n-\text{rank}\) is positive.

Exercises 10

  1. A \(3\times5\) matrix has rank \(2\). What is the dimension of its kernel?
  2. A linear map \(T:\mathbb{R}^4\to\mathbb{R}^3\) is onto. What is the dimension of its kernel?
  3. A linear map defined on a space of dimension \(6\) has a kernel of dimension \(2\). What is the dimension of its image?
  4. Let \(A=\begin{pmatrix}1&2&3&4\\2&4&6&8\\1&0&1&0\end{pmatrix}\). What is the dimension of its kernel?
  5. What is the dimension of the image of an injective linear map \(T:\mathbb{R}^3\to\mathbb{R}^5\)?
  6. Let \(T(x,y,z)=(x-y,\;y-z,\;x-z)\). What is the dimension of its image?
  7. Can there be an injective linear map \(T:\mathbb{R}^3\to\mathbb{R}^2\)?
  8. The matrix \(C=\begin{pmatrix}1&0&2&-1\\0&1&1&3\\1&1&3&2\\2&1&5&1\end{pmatrix}\) defines \(T(x)=Cx\) from \(\mathbb{R}^4\) to \(\mathbb{R}^4\). What do the dimensions of its kernel and image add up to?
  9. If \(A\) is a \(3\times6\) matrix, what is the smallest possible dimension of its kernel?
  10. A \(5\times7\) matrix \(A\) has a kernel of dimension \(3\). What is the dimension of the space spanned by its columns?

References

  • Sheldon Axler, Linear Algebra Done Right, 4th edition, Springer, 2024. §3.B (teorema fundamental de las transformaciones lineales)
  • Gilbert Strang, Introduction to Linear Algebra, 6th edition, Wellesley-Cambridge Press, 2023. §3.4
  • David C. Lay, Steven R. Lay, Judi J. McDonald, Linear Algebra and Its Applications, 6th edition, Pearson, 2021. §4.6