Linear Mapping

Studying linear mappings means studying mappings between linear spaces.

A linear mapping can be represented in matrix form, so many concepts from matrices have corresponding interpretations in the study of linear mappings.

Linear Mappings and Linear Transformations

Let $V$ and $W$ be two linear spaces over the 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 $F$:

$$T(kx+ly)=kTx+lTy$$

then $T$ is called a linear mapping from $V$ to $W$. If $W=V$, then $T$ is called a linear transformation on $V$.

For example, the identity transformation $T_e$ leaves the space unchanged, while the zero transformation $T_0$ maps the space to the zero space.

Let $L(V,W)$ denote the set of all linear mappings from $V$ to $W$. The set of all linear transformations $L(V,V)$ is also denoted by $L(V)$.

Properties

• A linear mapping maps the zero vector to the zero vector.
• A linear mapping preserves the form of linear operations: the linear mapping of a linear operation equals the linear operation of the linear mappings.
• A linear mapping preserves linear dependence: if vectors are linearly dependent before mapping, they remain linearly dependent after mapping.

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

Matrix Representation of a Linear Mapping

Let the dimension of $V$ be $n$, let $\alpha_1,\cdots,\alpha_n$ be a basis of $V$, let the dimension of $W$ be $m$, let $\beta_1,\cdots,\beta_m$ be a basis of $W$, and let $T$ be a linear mapping from $V$ to $W$.

Represent the image under $T$ of each $\alpha$ using $\beta$:

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

In 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 mapping $T$ under these two bases.

Null Space and Image Space of a Linear Mapping

The null space and image space here are described from the perspective of linear mappings. Through the matrix representation, we can see that the null space and image space of a linear mapping agree with the null space and image space of its matrix.

Let $T$ be a linear mapping from $V$ to $W$, and define:

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

It is easy to verify that $N(T)$ is a subspace of $V$ and $R(T)$ is a subspace of $W$. We call $N(T)$ and $R(T)$ the null space and image space of $V$, respectively. The dimension of $N(T)$ is called the nullity or deficiency of $T$, and the dimension of $R(T)$ is called the rank of $T$.

Theorem: Let $T$ be a linear mapping from $V$ to $W$. If $V$ is finite-dimensional, then both $N(T)$ and $R(T)$ are finite-dimensional, and:

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

That is, the nullity plus the rank of $T$ equals the dimension of its domain $V$.

Matrix Representation of a Linear Transformation

Let the dimension of $V$ be $n$, let $\alpha_1,\cdots,\alpha_n$ be a basis of $V$, and let $T$ be a linear transformation on $V$. Then:

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

In 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 $T$ under this basis.

By the structure of the space and the linearity of $T$, $T$ is completely determined by $T\alpha_1,\cdots,T\alpha_n$, so $T$ uniquely determines a matrix $A$.

Theorem: Let the dimension of $V$ be $n$, let $\alpha_1,\cdots,\alpha_n$ be a basis of $V$, and take any square matrix of order $n$, $A$. There is exactly one linear transformation from $V$ to $V$, namely $T$, such that the matrix of $T$ is exactly $A$.

Corollary: There is a one-to-one correspondence between $L(V,V)$ and all square matrices of order $n$.

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

The Space of Linear Transformations

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

$$(T_1+T_2)x=T_1x+T_2x$$ $$(kT_1)x=k(T_1x)$$

It is easy to verify that $L(V)$ is a linear space over $F$, namely, the space of linear transformations.

For linear transformations $T_1$ and $T_2$ in $L(V)$, define the product $T_1T_2$ of $T_1$ and $T_2$ by:

$$(T_1T_2)x=T_2(T_1x)$$

It can be verified that $(T_1T_2)$ is also a linear transformation in $L(V)$, and that the product of linear transformations is associative but not commutative, just like matrix multiplication.

For a linear transformation $T_1$ in $L(V)$, if there is a linear transformation $T_2$ in $L(V)$ such that for every vector $x$ in $V$:

$$(T_1T_2)x=T_1(T_2x)=x$$

then $T_2$ is called the inverse transformation of $T_1$, written:

$$T_2=T_1^{-1}$$

and:

$$T_1T_2=T_2T_1=T_e$$

Theorem: Let the dimension of $V$ be $n$, let $\alpha_1,\cdots,\alpha_n$ be a basis of $V$, and let the matrices of the linear transformations $T_1$ and $T_2$ under this basis be $A$ and $B$, respectively. Then:

• The matrix of the sum $T_1+T_2$ of linear transformations is $A+B$
• The matrix of the scalar multiple $kT_1$ of a linear transformation is $kA$
• The matrix of the product $T_1T_2$ of linear transformations is $AB$
• If the inverse transformation of $T_1$ exists, its matrix is $A^{-1}$

Coordinates

Let $n$ vectors $x$ form a basis of the $n$-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 coordinate of vector $y$ under the basis $x_1,x_2,\cdots,x_n$.

Clearly, coordinates are column vectors composed of scalars from the field and must be distinguished from vectors in an Abelian group.

Coordinate Transformation Formula

Let the dimension of $V$ be $n$, let $T$ be a transformation in $L(V)$, and let the matrix of $T$ under 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 suppose that:

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

Then:

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

Column-vector points in the space $V$ are essentially of the form “basis times coordinates”. The column-vector point $x$ in $V$ itself uses the identity matrix $I$ as its basis, namely $x=Ix$.

Only when the basis is fixed, and it is the same basis, is a pure linear transformation $T$ simply an ordinary matrix multiplying coordinates on the left.

Think of the linear transformation $T$ as an observation filter for the space $V$. The object acted on by $T$ is the space $V$, and the space $V$ is distorted. After the filter is applied, the point itself has not moved.

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

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 of the space $V$. For $1\leq i\leq n$, let the coordinates of each vector $y_i$ under the basis $x_1,x_2,\cdots,x_n$ be:

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

Thus, the $n$ vectors $y$ form the matrix on the left-hand side of the equation, and the $n$ coordinate vectors form the matrix $A$ on the right-hand side:

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

Matrix $A$ is called the transition matrix from basis $x_1,x_2\cdots,x_n$ to basis $y_1,y_2\cdots,y_n$, and is also called a change-of-basis matrix.

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

Thus, the transition matrix is a matrix composed of scalars from the field, not a matrix whose columns are vectors from an Abelian group; the distinction should be kept in mind.

Let $n$ vectors $x$ and $n$ vectors $y$ be two bases of the space $V$. For the same vector $z$ in the space $V$:

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

Substitute the result above:

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

By uniqueness:

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

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 transformation purely between coordinates; the coordinate transformation formulas are all in the scalar field. As distinguished above, vectors in linear spaces and Abelian groups are abstract vectors, while the coordinates and the entries of transition matrices are in the scalar field and are regarded as concrete vectors. The two kinds of vectors should be regarded as different things.

A matrix can transform the entire space, that is, all coordinates, with the column vector $x$ ranging over the entire space.

The identity matrix $I$ consists of unit vectors. Matrix $A$ transforms the identity matrix $I$ into each column vector of matrix $A$, that is, it transforms unit vectors into the column vectors of matrix $A$. Therefore, left multiplication by matrix $A$ can also be viewed as applying this transformation to the space.

Multiplying a vector by a matrix on the left can also be viewed as multiplying a vector group by coordinates on the left. From the coordinate perspective:

$$Iy=Xa$$

For the same column vector $y$, its coordinates in the “ordinary” space represented by the identity matrix $I$ are $y$; in the new space after the transformation, its coordinates are denoted by $a$. Thus, matrix $X$ is not only a basis in the ordinary space, but also the transition matrix from vector group $I$ to vector group $X$.

The linear transformation $T$ maps one basis to another, and thus maps one set of coordinates to another.

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

Thus, the relationship between coordinates is exactly reversed. Suppose the coordinates after the linear transformation $T$ are $b$, that is, the coordinates observed after applying the filter are $b$. Then the representation of the point in $V$ is $\beta b$. To restore it, use the transition matrix to write the representation of the point in $V$ as $\alpha Ab$. This makes clear why coordinate transformation is viewed as left multiplication by the inverse of the transition matrix.

Linear Transformations and Matrix Similarity

Consider a linear transformation $T$ in the space $V$ and its relationship to the basis $\alpha$ of $V$:

The linear transformation $T$ acts on the basis $\alpha$ and maps basis $\alpha$ to $T(\alpha)$, which is equivalent to multiplying the basis $\alpha$ on the right by $A$, namely $T(\alpha)=\alpha A$.

Matrix similarity asks: the same linear transformation $T$ is represented by matrix $B$ in space $V$ with basis $\beta$, and by matrix $A$ in space $V$ with basis $\alpha$.

If the transition matrix is $C$, that is, $\beta=\alpha C$, what is the relationship between the two representations $B$ and $A$?

Because it is the same transformation $T$, one fact is clear: the transition-matrix relationship before and after the transformation always holds, namely:

$$T(\beta)=T(\alpha)C=\alpha AC$$

From the perspective of basis $\beta$, the linear transformation $T$ is still right multiplication. Converting basis $\beta$ to basis $\alpha$ and then multiplying on the right by $C$ preserves the transition-matrix relationship $C$ before and after the 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 is 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 need not be similar.

Because matrix similarity is closely related to shape, it has no relationship with vector-group equivalence or systems having the same solutions.

In retrospect, the explanation of matrix similarity is the four equations: $\beta=\alpha C$, $T(\alpha)=\alpha A$, $T(\beta)=\beta B$, and $T(\beta)=T(\alpha)C$.

References

• 【Official bilingual/collection】The Essence of Linear Algebra - Series Collection P13 09 - Change of Basis