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6 Linear Transformations

A progressive guide to linear transformations, their kernels and images, matrix representations, composition, change of basis, and the rank-nullity theorem.

Linear transformations: definition and basis determination

A is a function between vector spaces that preserves linear combinations. For vectors u,v\mathbf{u},\mathbf{v} and scalars a,ba,b,

T(au+bv)=aT(u)+bT(v).T(a\mathbf{u}+b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v}).

This implies preservation of addition and scalar multiplication:

T(u+v)=T(u)+T(v),T(cu)=cT(u).T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v}), \qquad T(c\mathbf{u})=cT(\mathbf{u}).

It also implies T(0)=0T(\mathbf{0})=\mathbf{0}. Examples include matrix multiplication, differentiation on a polynomial space, integration with a fixed lower limit, and rotations, reflections, or projections through the origin.

A is completely determined by its values on a basis. If B={v1,…,vn}\mathcal{B}=\{\mathbf{v}_1,\ldots,\mathbf{v}_n\} and

v=c1v1+⋯+cnvn,\mathbf{v}=c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n,

then

T(v)=c1T(v1)+⋯+cnT(vn).T(\mathbf{v})=c_1T(\mathbf{v}_1)+\cdots+c_nT(\mathbf{v}_n).

Takeaway: Understanding the images of the basis vectors determines the transformation everywhere.

and

The records the input vectors that disappear under the transformation:

ker⁡T={v∈V:T(v)=0}.\ker T=\{\mathbf{v}\in V:T(\mathbf{v})=\mathbf{0}\}.

The records every output that can be produced:

im⁡T={T(v):v∈V}.\operatorname{im}T=\{T(\mathbf{v}):\mathbf{v}\in V\}.

Both sets are subspaces. For a matrix transformation TA(x)=AxT_A(\mathbf{x})=A\mathbf{x}, the is the solution space of Ax=0A\mathbf{x}=\mathbf{0}, and the is the column space of AA.

Consider

T:R3→R2,T(x,y,z)=(x+y,y+z).T:\mathbb{R}^3\to\mathbb{R}^2, \qquad T(x,y,z)=(x+y,y+z).

Solving T(x,y,z)=(0,0)T(x,y,z)=(0,0) gives x+y=0x+y=0 and y+z=0y+z=0, so

(x,y,z)=t(−1,1,−1)(x,y,z)=t(-1,1,-1)

and

ker⁡T=span⁡{(−1,1,−1)}.\ker T=\operatorname{span}\{(-1,1,-1)\}.

The matrix columns are (1,0)(1,0) and (1,1)(1,1), which are linearly independent and span R2\mathbb{R}^2. Therefore,

im⁡T=R2.\operatorname{im}T=\mathbb{R}^2.

Takeaway: The describes lost input directions, while the describes attainable outputs.

and

means that distinct inputs have distinct outputs. For a , this is equivalent to having only the zero vector in the :

T is injective  ⟺  ker⁡T={0}.T\text{ is injective}\iff \ker T=\{\mathbf{0}\}.

means that every vector in the codomain is reached:

T is surjective  ⟺  im⁡T=W.T\text{ is surjective}\iff \operatorname{im}T=W.

For the example T:R3→R2T:\mathbb{R}^3\to\mathbb{R}^2, the contains the nonzero vector (−1,1,−1)(-1,1,-1), so the transformation is not injective. Its is all of R2\mathbb{R}^2, so it is surjective.

Dimension gives useful tests. If dim⁡V>dim⁡W\dim V>\dim W, no from VV to WW can be injective. If dim⁡V<dim⁡W\dim V<\dim W, no from VV to WW can be surjective. When the domain and codomain have equal finite dimensions, and are equivalent.

Takeaway: Test through the and through the .

Matrix representations

A depends on ordered bases. If B={v1,…,vn}\mathcal{B}=\{\mathbf{v}_1,\ldots,\mathbf{v}_n\} is an input basis and C\mathcal{C} is an output basis, then

[T]C←B=[[T(v1)]C⋯[T(vn)]C].[T]_{\mathcal{C}\leftarrow\mathcal{B}}= \begin{bmatrix} [T(\mathbf{v}_1)]_{\mathcal{C}}&\cdots&[T(\mathbf{v}_n)]_{\mathcal{C}} \end{bmatrix}.

The matrix equation is

[T(v)]C=[T]C←B[v]B.[T(\mathbf{v})]_{\mathcal{C}}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}.

Thus, there is one column for each input basis vector and one row for each output coordinate.

For example, if T:R2→R2T:\mathbb{R}^2\to\mathbb{R}^2 satisfies

T(1,0)=(2,1),T(0,1)=(−1,3),T(1,0)=(2,1), \qquad T(0,1)=(-1,3),

then, using standard bases,

[T]=[2−113].[T]=\begin{bmatrix}2&-1\\1&3\end{bmatrix}.

For an arbitrary input, this gives

T(x,y)=(2x−y,x+3y).T(x,y)=(2x-y,x+3y).

Takeaway: Build a transformation matrix by placing the images of the input-basis vectors into its columns.

and matrix multiplication

The of two transformations applies one transformation first and the next transformation second. If U:U0→VU:U_0\to V and T:V→WT:V\to W, then

(T∘U)(x)=T(U(x)).(T\circ U)(\mathbf{x})=T(U(\mathbf{x})).

If AA represents UU and BB represents TT, then

[T∘U]=BA.[T\circ U]=BA.

The rightmost matrix acts first, so the order of multiplication follows the order of application. In general,

BA≠AB.BA\ne AB.

The and also satisfy useful inclusions:

ker⁡U⊆ker⁡(T∘U)\ker U\subseteq\ker(T\circ U)

because a vector sent to zero by UU remains zero after applying TT, and

im⁡(T∘U)⊆im⁡T\operatorname{im}(T\circ U)\subseteq\operatorname{im}T

because every output of the is an output of TT.

Takeaway: For compositions, track the domains and the order, then multiply matrices from right to left.

and similarity

A alters coordinate descriptions without changing the underlying vector or transformation. If B\mathcal{B} and B′\mathcal{B}' are bases for the same space, then

[v]B=PB←B′[v]B′.[\mathbf{v}]_{\mathcal{B}}=P_{\mathcal{B}\leftarrow\mathcal{B}'}[\mathbf{v}]_{\mathcal{B}'}.

The columns of PB←B′P_{\mathcal{B}\leftarrow\mathcal{B}'} are the vectors of B′\mathcal{B}' written in the coordinates of B\mathcal{B}.

For a linear operator on one vector space, let AA be its matrix in the old basis and A′A' its matrix in the new basis. If P=PB←B′P=P_{\mathcal{B}\leftarrow\mathcal{B}'}, then

A′=P−1AP.A'=P^{-1}AP.

This is a similarity relation. The entries of the matrix may change, but rank, nullity, determinant, and eigenvalues remain unchanged.

For example, with the standard basis E\mathcal{E} and basis B′={(1,1),(1,−1)}\mathcal{B}'=\{(1,1),(1,-1)\},

PE←B′=[111−1].P_{\mathcal{E}\leftarrow\mathcal{B}'}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}.

Thus, if [v]B′=(2,3)T[\mathbf{v}]_{\mathcal{B}'}=(2,3)^T, then

[v]E=[111−1][23]=[5−1].[\mathbf{v}]_{\mathcal{E}}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{bmatrix}2\\3\end{bmatrix}=\begin{bmatrix}5\\-1\end{bmatrix}.

Takeaway: Coordinate changes modify the matrix description, not the underlying linear operator.

and problem-solving procedure

The connects the dimensions of the domain, , and :

dim⁡V=rank⁡(T)+nullity⁡(T).\dim V=\operatorname{rank}(T)+\operatorname{nullity}(T).

Here,

rank⁡(T)=dim⁡(im⁡T)\operatorname{rank}(T)=\dim(\operatorname{im}T)

and

nullity⁡(T)=dim⁡(ker⁡T).\operatorname{nullity}(T)=\dim(\ker T).

For an m×nm\times n matrix viewed as a transformation from Rn\mathbb{R}^n to Rm\mathbb{R}^m, the formula becomes

rank⁡(A)+nullity⁡(A)=n.\operatorname{rank}(A)+\operatorname{nullity}(A)=n.

The number on the right is the number of columns because it is the dimension of the domain. For the transformation T(x,y,z)=(x+y,y+z)T(x,y,z)=(x+y,y+z), the has dimension 11 and the has dimension 22. Therefore,

nullity⁡(T)+rank⁡(T)=1+2=3=dim⁡R3.\operatorname{nullity}(T)+\operatorname{rank}(T)=1+2=3=\dim\mathbb{R}^3.

This explains why one independent input direction is lost while two independent output directions remain.

Practical procedure

  1. Solve Ax=0A\mathbf{x}=\mathbf{0} to find the .

  2. Count free variables to find the nullity.

  3. Row-reduce AA and count pivots to find the rank.

  4. Use the original pivot columns as a basis for the .

  5. Verify rank⁡(A)+nullity⁡(A)=n\operatorname{rank}(A)+\operatorname{nullity}(A)=n.

  6. For compositions, multiply matrices in application order from right to left.

  7. For new bases, use the appropriate change-of-coordinate matrices on the left and right.

Takeaway: Rank-nullity provides both a dimension theorem and a reliable check on computations.