Skip to content

Linear Maps

Studying linear maps involves studying mappings between linear spaces.

Linear maps can be represented in matrix form, so many concepts in matrices correspond to concepts in linear maps.

Linear Maps and Linear Transformations

Let \(V\) and \(W\) be two linear spaces over a field \(F\), and let \(T\) be a mapping from \(V\) to \(W\).

If for any vectors \(x\) and \(y\) in \(W\), and any scalars \(k\) and \(l\) in the field \(F\), we have:

\[ T(kx+ly)=kTx+lTy \]

then \(T\) is called a linear map from \(V\) to \(W\). If \(W=V\), then \(T\) is called a linear transformation on \(V\).

For example, the identity transformation \(T_e\) keeps the space unchanged, while the zero transformation \(T_0\) maps the space to the zero space.

We denote \(L(V,W)\) as the set of all linear maps from \(V\) to \(W\). For the set of all linear transformations \(L(V,V)\), we also denote it as \(L(V)\).

Properties

  • A linear map maps the zero vector to the zero vector.
  • A linear map preserves the form of linear operations, i.e., the linear map of a linear combination equals the linear combination of the linear maps.
  • A linear map preserves linear dependence, i.e., if vectors are linearly dependent before mapping, they remain linearly dependent after mapping.

However, a linear map does not preserve linear independence. Vectors that are linearly independent before mapping may not remain linearly independent after mapping.

Matrix Representation of Linear Maps

Let the dimension of \(V\) be \(n\), with a basis \(\alpha_1,\cdots,\alpha_n\) for \(V\). Let the dimension of \(W\) be \(m\), with a basis \(\beta_1,\cdots,\beta_m\) for \(W\). Let \(T\) be a linear map from \(V\) to \(W\).

Express each \(\alpha\) mapped by \(T\) in terms of \(\beta\):

\[ T\alpha_j=a_{1j}\beta_1+\cdots+a_{mj}\beta_m \]

Using matrix notation:

\[ T(\alpha_1,\cdots,\alpha_n)=(T\alpha_1,\cdots,T\alpha_n)=(\beta_1,\cdots,\beta_m)A \]

The matrix \(A\) is called the matrix representation of the linear map \(T\) with respect to these two bases.

Kernel and Image of a Linear Map

The kernel and image discussed here are from the perspective of linear maps. With matrix representation, it can be seen that the kernel and image of a linear map coincide with the kernel and image of its matrix representation.

Let \(T\) be a linear map from space \(V\) to space \(W\), and define:

\[ N(T)=\{x\in V|Tx=0\} \]
\[ R(T)=Im(T)=\{y\in W|y=Tx,Vx\in V\} \]

It can be verified that \(N(T)\) is a subspace of \(V\), and \(R(T)\) is a subspace of \(W\). \(N(T)\) and \(R(T)\) are called the kernel and image of \(V\), respectively. The dimension of \(N(T)\) is called the nullity or defect of \(T\), and the dimension of \(R(T)\) is called the rank of \(T\).

Theorem: Let \(T\) be a linear map from space \(V\) to space \(W\), and let \(V\) be finite-dimensional. Then both \(N(T)\) and \(R(T)\) are finite-dimensional, and we have:

\[ \operatorname{dim} N(T)+\operatorname{dim} R(T)=\operatorname{dim} V \]

That is, the nullity plus the rank of \(T\) equals the dimension of its domain \(V\).

Matrix Representation of Linear Transformations

Let the dimension of \(V\) be \(n\), with a basis \(\alpha_1,\cdots,\alpha_n\) for \(V\). Let \(T\) be a linear transformation on \(V\). Then:

\[ T\alpha_j=a_{1j}\alpha_1+\cdots+a_{nj}\alpha_n \]

Using matrix notation:

\[ T(\alpha_1,\cdots,\alpha_n)=(T\alpha_1,\cdots,T\alpha_n)=(\alpha_1,\cdots,\alpha_n)A \]

The matrix \(A\) is called the matrix representation of the linear transformation \(T\) with respect to this basis.

Due to the structure of the space and the linearity of \(T\), \(T\) is completely determined by \(T\alpha_1,\cdots,T\alpha_n\). Thus, \(T\) uniquely determines a matrix \(A\).

Theorem: Let the dimension of \(V\) be \(n\), and let \(\alpha_1,\cdots,\alpha_n\) be a basis for \(V\). For any \(n \times n\) matrix \(A\), there exists exactly one linear transformation \(T\) from \(V\) to \(V\) such that the matrix of \(T\) is exactly \(A\).

Corollary: There is a one-to-one correspondence between \(L(V,V)\) and the set of all \(n \times n\) matrices.

For example: The zero transformation corresponds to the zero matrix, and the identity transformation corresponds to the identity matrix.

Space of Linear Transformations

Theorem: \(L(V)\) can also form a linear space. Introduce operations in \(L(V)\): For any \(T_1\) and \(T_2\) in \(L(V)\), any \(x\) in \(V\), and any \(k\) in the field \(F\), define:

\[ (T_1+T_2)x=T_1x+T_2x \]
\[ (kT_1)x=k(T_1x) \]

It can be verified that \(L(V)\) is a linear space over \(F\), i.e., the space of linear transformations.

For linear transformations \(T_1\) and \(T_2\) in \(L(V)\), define the product \(T_1T_2\) as:

\[ (T_1T_2)x=T_2(T_1x) \]

It can be verified that \((T_1T_2)\) is also a linear transformation in \(L(V)\). Moreover, the product of linear transformations satisfies associativity but not commutativity, similar to matrix multiplication.

For a linear transformation \(T_1\) in \(L(V)\), if there exists a linear transformation \(T_2\) in \(L(V)\) such that for any vector \(x\) in \(V\), we have:

\[ (T_1T_2)x=T_1(T_2x)=x \]

then \(T_2\) is called the inverse transformation of \(T_1\), denoted as:

\[ T_2=T_1^{-1} \]

and we have:

\[ T_1T_2=T_2T_1=T_e \]

Theorem: Let the dimension of \(V\) be \(n\), and let \(\alpha_1,\cdots,\alpha_n\) be a basis for \(V\). Under this basis, let the matrix of linear transformation \(T_1\) be \(A\), and the matrix of \(T_2\) be \(B\). Then:

  • The matrix of the linear transformation \(T_1+T_2\) is \(A+B\).
  • The matrix of the scalar multiplication \(kT_1\) is \(kA\).
  • The matrix of the product \(T_1T_2\) is \(AB\).
  • If the inverse transformation \(T_1^{-1}\) exists, its matrix is \(A^{-1}\).

Coordinates

Let \(n\) vectors \(x\) be a basis for an \(n\)-dimensional space \(V\). For any vector \(y\) in \(V\), let \(y\) be:

\[ y=a_1x_1+a_2x_2+\cdots+a_nx_n=(x_1,x_2,\cdots,x_n)\begin{pmatrix}a_1\\a_2\\\vdots\\a_n\end{pmatrix} \]

The column vector:

\[ \begin{pmatrix}a_1\\a_2\\\vdots\\a_n\end{pmatrix} \]

is called the coordinates of vector \(y\) with respect to the basis \(x_1,x_2,\cdots,x_n\).

Note that coordinates are column vectors composed of scalars from the field and should be distinguished from vectors in an Abelian group.

Coordinate Transformation Formula

Let the dimension of \(V\) be \(n\), and let \(T\) be a transformation in \(L(V)\). Let the matrix of \(T\) with respect to the basis \(\alpha_1,\cdots,\alpha_n\) be \(A\). Let:

\[ \xi=(\alpha_1,\cdots,\alpha_n)\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix} \]

and also:

\[ T\xi=T(\alpha_1,\cdots,\alpha_n)\begin{pmatrix}y_1\\y_2\\\vdots\\y_n\end{pmatrix} \]

Then we have:

\[ T\xi=T(\alpha_1,\cdots,\alpha_n)\begin{pmatrix}y_1\\y_2\\\vdots\\y_n\end{pmatrix}=(\alpha_1,\cdots,\alpha_n)A\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix} \]

Column vector points in space \(V\) are essentially in the form of "basis times coordinates". A column vector point \(x\) in space \(V\) itself uses the identity matrix \(I\) as the basis, i.e., \(x=Ix\).

Only when the basis is fixed and unchanged, a pure linear transformation \(T\) corresponds to left-multiplying the coordinates by an ordinary matrix.

Think of a linear transformation \(T\) as an observation filter applied to space \(V\). The linear transformation \(T\) acts on space \(V\), distorting it. After applying the filter, the positions of the points themselves do not change.

This theorem also shows that a linear transformation \(T\) on a column vector basis is equivalent to right-multiplying the basis by a transition matrix.

Thus, between different bases, the coordinate relationship is left-multiplication by the inverse of the transition matrix.

Transition Matrix

Let \(n\) vectors \(x\) and \(n\) vectors \(y\) be two bases for space \(V\). For \(1\leq i\leq n\), let the coordinates of each vector \(y_i\) with respect to the basis \(x_1,x_2,\cdots,x_n\) be:

\[ y_i=(x_1,x_2,\cdots,x_n)\begin{pmatrix}a_{1i}\\a_{2i}\\\vdots\\a_{ni}\end{pmatrix} \]

Then the \(n\) vectors \(y\) form the matrix on the left side of the equation, and the \(n\) coordinate columns form the matrix \(A\) on the right side:

\[ (y_1,y_2,\cdots,y_n)=(x_1,x_2,\cdots,x_n)A \]

The matrix \(A\) is called the transition matrix (or transformation matrix) from the basis \(x_1,x_2\cdots,x_n\) to the basis \(y_1,y_2\cdots,y_n\).

Clearly, the transition matrix is invertible. For the above equation, the transition matrix from the basis \(y_1,y_2\cdots,y_n\) to the basis \(x_1,x_2\cdots,x_n\) is \(A^{-1}\).

Note that the transition matrix is a matrix composed of scalars from the field, not a matrix formed by vectors from an Abelian group, and should be distinguished.

Let \(n\) vectors \(x\) and \(n\) vectors \(y\) be two bases for space \(V\). For the same vector \(z\) in space \(V\), we have:

\[ z=(x_1,x_2,\cdots,x_n)\begin{pmatrix}\xi_1\\\xi_2\\\vdots\\\xi_n\end{pmatrix}=(y_1,y_2\cdots,y_n)\begin{pmatrix}\eta_1\\\eta_2\\\vdots\\\eta_n\end{pmatrix} \]

Substituting the earlier equation:

\[ (y_1,y_2\cdots,y_n)=(x_1,x_2\cdots,x_n)A \]

By uniqueness, we obtain:

\[ \begin{pmatrix}\xi_1\\\xi_2\\\vdots\\\xi_n\end{pmatrix}=A\begin{pmatrix}\eta_1\\\eta_2\\\vdots\\\eta_n\end{pmatrix} \]

or

\[ \begin{pmatrix}\eta_1\\\eta_2\\\vdots\\\eta_n\end{pmatrix}=A^{-1}\begin{pmatrix}\xi_1\\\xi_2\\\vdots\\\xi_n\end{pmatrix} \]

This is a pure transformation between coordinates, and the coordinate transformation formulas are within the scalar field. Due to the earlier distinction, vectors in linear spaces and Abelian groups are "abstract vectors", while coordinates and elements of transition matrices are within the scalar field and are considered "concrete vectors". These two types of vectors should be regarded as "different things".

A matrix can transform the entire space, i.e., all coordinates. The column vector \(x\), as coordinates, spans the entire space.

The identity matrix \(I\) is composed of unit vectors. Matrix \(A\) transforms the identity matrix \(I\) to each column vector of \(A\), i.e., transforms unit vectors to each column vector of \(A\). Therefore, left-multiplying by matrix \(A\) can also be viewed as transforming the space in this way.

Left-multiplying a vector by a matrix can also be seen as left-multiplying coordinates by a vector group. From the coordinate perspective:

\[ Iy=Xa \]

The same column vector \(y\), in the "normal" space represented by the identity matrix \(I\), has coordinates \(y\). In the transformed new space, its coordinates will be denoted as \(a\). Thus, matrix \(X\) is not only a basis in the normal space but also the transition matrix from the vector group \(I\) to the vector group \(X\).

A linear transformation \(T\) maps one basis to another, so coordinates are also mapped to other coordinates.

If the transition matrix corresponding to the linear transformation \(T\) that maps basis \(\alpha\) to \(\beta\) is \(A\), then the basis matrices satisfy \(\beta=\alpha A\).

Thus, the coordinate relationship is reversed. Suppose the coordinates after mapping by linear transformation \(T\) are \(b\), i.e., after applying the filter, the observed coordinates are \(b\). Then the point's representation in \(V\) is \(\beta b\). The way to restore it is to use the transition matrix, writing the point's representation in \(V\) as \(\alpha Ab\). Hence, the view that coordinate transformation is left-multiplication by the inverse of the transition matrix becomes clear.

Linear Transformations and Matrix Similarity

Consider a linear transformation \(T\) in space \(V\) with respect to a basis \(\alpha\) of \(V\):

The linear transformation \(T\) acts on the basis \(\alpha\), mapping \(\alpha\) to \(T(\alpha)\), which is equivalent to right-multiplying \(\alpha\) by \(A\), i.e., \(T(\alpha)=\alpha A\).

Matrix similarity considers the problem: The same linear transformation \(T\), described as matrix \(B\) in the space \(V\) with basis \(\beta\), and described as matrix \(A\) in the space \(V\) with basis \(\alpha\).

If the transition matrix is \(C\), i.e., \(\beta=\alpha C\), what is the relationship between the two descriptions \(B\) and \(A\)?

Since it is the same transformation \(T\), we observe that the transition matrix relationship holds before and after transformation:

\[ T(\beta)=T(\alpha)C=\alpha AC \]

From the perspective of basis \(\beta\), the linear transformation \(T\) still corresponds to right-multiplication. Transforming basis \(\beta\) to basis \(\alpha\) and then right-multiplying by \(C\), the transition matrix \(C\) relationship is preserved before and after transformation:

\[ T(\beta)=\beta B=\alpha CB \]

Thus, the problem is solved:

\[ B=C^{-1}AC \]

Theorem: Let \(T\) be a transformation in \(L(V)\). Then the matrices of \(T\) under different bases are similar.

For square matrices \(A\) and \(B\), if there exists an invertible matrix \(C\) such that \(B=C^{-1}AC\), then \(A\) and \(B\) are similar.

Matrix similarity preserves rank, so matrix similarity implies matrix equivalence. However, two equivalent matrices are not necessarily similar.

Since matrix similarity is closely related to shape, there is no relationship between matrix similarity and vector group equivalence or system of equations having the same solution.

Looking back, the explanation of matrix similarity boils down to 4 equations: \(\beta=\alpha C\), \(T(\alpha)=\alpha A\), \(T(\beta)=\beta B\), and \(T(\beta)=T(\alpha)C\).

References

本页面最近更新：，更新历史
发现错误？想一起完善？ 在 GitHub 上编辑此页！
本页面贡献者：OI-wiki
本页面的全部内容在 CC BY-SA 4.0 和 SATA 协议之条款下提供，附加条款亦可能应用