Linear Transformations and Matrix Theory in Vector Spaces

Linear Transformations on $F$-spaces: Definitions and Examples

In the study of vector spaces over a field $F$, a central concept is the mapping between these spaces that preserves their algebraic structure. Let $V$ and $W$ be $F$-spaces. A mapping $T: V \rightarrow W$ is defined as a linear transformation if it satisfies two fundamental properties. First, it must satisfy additivity, meaning $T(u + v) = T(u) + T(v)$ for all vectors $u$ and $v$ in $V$. Second, it must satisfy homogeneity of degree one, meaning $T(\alpha u) = \alpha T(u)$ for all scalars $\alpha \in F$ and for all vectors $u \in V$. Alternatively, these two requirements can be combined into a single condition: $T(\alpha u + v) = \alpha T(u) + T(v)$ for all $u, v \in V$ and all $\alpha \in F$. If the transformation maps a space into itself, such that $T: V \rightarrow V$, it is referred to as a linear operator on $V$.

Consider Example 2.1.2, where a mapping $T: F^{(3)} \rightarrow F^{(2)}$ is defined by $T(x_1, x_2, x_3) = (x_1 + x_2, x_3)$. To verify linearity, let $u = (x_1, x_2, x_3)$ and $v = (y_1, y_2, y_3)$ be elements of $F^{(3)}$, and let $\alpha \in F$. The sum of the vectors is $u + v = (x_1 + y_1, x_2 + y_2, x_3 + y_3)$. Applying the transformation, $T(u + v) = (x_1 + y_1 + x_2 + y_2, x_3 + y_3)$, which can be rearranged as $(x_1 + x_2, x_3) + (y_1 + y_2, y_3) = T(u) + T(v)$. Similarly, for scalar multiplication, $\alpha u = (\alpha x_1, \alpha x_2, \alpha x_3)$, and $T(\alpha u) = (\alpha x_1 + \alpha x_2, \alpha x_3)$. Factoring out the scalar yields $\alpha(x_1 + x_2, x_3) = \alpha T(u)$. Since both conditions hold, $T$ is a linear transformation.

In contrast, Example 2.1.3 presents a mapping $T: F^{(3)} \rightarrow F^{(2)}$ defined by $T(x_1, x_2, x_3) = (x_1 + 1, x_3)$. Using Method 1 to disprove linearity, let $u = (1, 0, 0)$ and $v = (2, 0, 0)$. Then $u + v = (3, 0, 0)$. Calculating the values, $T(u + v) = T(3, 0, 0) = (4, 0)$, while $T(u) = (2, 0)$ and $T(v) = (3, 0)$. Adding the results, $T(u) + T(v) = (5, 0)$. Because $(4, 0) \neq (5, 0)$, the additivity condition fails. Method 2 demonstrates failure of homogeneity: let $u = (1, 0, 0)$ and $\alpha = 2$. Then $\alpha u = (2, 0, 0)$. Here, $T(\alpha u) = (3, 0)$ but $\alpha T(u) = 2(2, 0) = (4, 0)$. Since $(3, 0) \neq (4, 0)$, the mapping is not linear.

Additional examples include matrix-based transformations and special trivial cases. Example 2.1.4 defines $T: M_{m \times n}(F) \rightarrow M_{m \times n}(F)$ by $T(A) = MAN$, where $M$ and $N$ are fixed $m \times m$ and $n \times n$ matrices respectively. For matrices $A$ and $B$, $T(A + B) = M(A + B)N = MAN + MBN = T(A) + T(B)$, and $T(\alpha A) = M(\alpha A)N = \alpha(MAN) = \alpha T(A)$, proving linearity. Example 2.1.5 introduces the identity linear transformation $I: V \rightarrow V$, where $I(u) = u$. It satisfies $I(u + v) = u + v = I(u) + I(v)$ and $I(\alpha u) = \alpha u = \alpha I(u)$. Example 2.1.6 defines the zero linear transformation $Z: V \rightarrow W$ by $Z(u) = 0$. It satisfies $Z(u + v) = 0 = 0 + 0 = Z(u) + Z(v)$ and $Z(\alpha u) = 0 = \alpha \cdot 0 = \alpha Z(u)$.

Certain remarks and lemmas further characterize these transformations. Remark 2.1.7 notes that for any linear transformation $T: V \rightarrow W$, it is always true that $T(0) = 0$. Furthermore, the transformation respects linear combinations: $T(\alpha_1 u_1 + \alpha_2 u_2 + \dots + \alpha_n u_n) = \alpha_1 T(u_1) + \alpha_2 T(u_2) + \dots + \alpha_n T(u_n)$. Lemma 2.1.8 establishes the existence and uniqueness of linear transformations: if $V$ has a basis $B = \{v_1, v_2, \dots, v_n\}$, and $S = \{w_1, w_2, \dots, w_n\}$ is any set of vectors in $W$, there exists a unique linear transformation $T: V \rightarrow W$ such that $T(v_j) = w_j$ for each $j = 1, 2, \dots, n$.

Matrix of Linear Transformation On $F$-spaces

A linear transformation between finite-dimensional vector spaces can be represented as a matrix. Let $V$ and $W$ be $F$-spaces with dimensions $n$ and $m$ respectively. Let $B = \{v_1, v_2, \dots, v_n\}$ be a basis for $V$ and $S = \{w_1, w_2, \dots, w_m\}$ be a basis for $W$. For a linear transformation $T: V \rightarrow W$, each image vector $T(v_j)$ can be expressed as a unique linear combination of the basis vectors in $S$: $T(v_j) = \sum_{i=1}^m a_{ij} w_i$. The matrix $A = (a_{ij}) \in M_{m \times n}(F)$ formed by these coordinates is called the matrix of $T$ relative to the bases $B$ and $S$, denoted as $A = [T]_{B,S}$.

Example 2.2.2 illustrates this process. Let $T: F^{(3)} \rightarrow F^{(2)}$ be defined by $T(x_1, x_2, x_3) = (x_1 + x_2, 2x_1 + x_3)$. To find the matrix $A$ relative to standard bases $B = \{e_1, e_2, e_3\}$ and $S = \{f_1, f_2\}$, we calculate: $T(e_1) = T(1, 0, 0) = (1, 2) = 1f_1 + 2f_2$, $T(e_2) = T(0, 1, 0) = (1, 0) = 1f_1 + 0f_2$, and $T(e_3) = T(0, 0, 1) = (-1, 1) = -1f_1 + 1f_2$. The resulting matrix is: $A = \begin{pmatrix} 1 & 1 & -1 \\ 2 & 0 & 1 \end{pmatrix}$

In part (2) of the same example, we find a different matrix $B$ relative to different bases $B' = \{v_1=(1, 0, -1), v_2=(1, 1, 1), v_3=(1, 0, 0)\}$ and $S' = \{w_1=(0, 1), w_2=(1, 0)\}$. We compute: $T(v_1) = T(1, 0, -1) = (1, 1) = 1w_1 + 1w_2$, $T(v_2) = T(1, 1, 1) = (2, 3) = 3w_1 + 2w_2$, and $T(v_3) = T(1, 0, 0) = (1, 2) = 2w_1 + 1w_2$. The resulting matrix is: $B = \begin{pmatrix} 1 & 3 & 2 \\ 1 & 2 & 1 \end{pmatrix}$ Note: The coordinate vectors for the images are placed as columns in the matrix $B$.

The Effect of Change of Bases

When the bases of the vector spaces are changed, the matrix representing the linear transformation also changes. If $V$ has bases $B$ and $B'$, and $W$ has bases $S$ and $S'$, the relationship between the matrix $A = [T]_{B,S}$ and $B = [T]_{B',S'}$ involves change-of-basis matrices. Lemma 2.1.8 implies the existence of linear transformations $P: V \rightarrow V$ and $Q: W \rightarrow W$ related to these coordinate shifts. Let $C$ be the matrix of $P$ relative to $B$ and $B'$, and $D$ be the matrix of $Q$ relative to $S$ and $S'$. In the general case, $B = D A C^{-1}$. For a linear operator $T: V \rightarrow V$, where the same basis transition applies to both the domain and codomain, $D = C$, leading to the relation $B = C^{-1} A C$.

Example 2.3.2 demonstrates this with a linear transformation $T: V \rightarrow V$ where $\dim(V) = 3$ and the original matrix relative to basis $B = \{u, v, w\}$ is: $A = \begin{pmatrix} 1 & 1 & 0 \\ -1 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ We wish to find matrix $B$ relative to a new basis $B' = \{u+v, u-2v+w, v-w\}$. The change-of-basis matrix $C$ is constructed from the coordinates of the new basis vectors relative to the old basis: $C = \begin{pmatrix} 1 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -1 \end{pmatrix}$

To find $C^{-1}$, we follow several steps. First, calculate the determinant $\det(C) = 1(2-1) - 1(-1-0) + 0 = 1 + 1 = 2$. Next, determine the matrix of cofactors: $c_{11}=1, c_{12}=1, c_{13}=1, c_{21}=1, c_{22}=-1, c_{23}=-1, c_{31}=1, c_{32}=-1, c_{33}=-3$. The adjoint matrix is the transpose of the cofactor matrix: $\text{Adj}(C) = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & -1 & -3 \end{pmatrix}$ The inverse is $C^{-1} = \frac{1}{2} \text{Adj}(C)$. Finally, the new matrix is obtained via $B = C^{-1} A C$: $B = \frac{1}{2} \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & -1 & -3 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \\ -1 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -1 \end{pmatrix} = \dots = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 1 & 0 \end{pmatrix}$

The Kernel and Image of a Linear Transformation

For a linear transformation $T: V \rightarrow W$, two critical subspaces are defined: the kernel and the image. The kernel of $T$, denoted $\text{ker}(T)$, consists of all vectors $u \in V$ such that $T(u) = 0$. The image of $T$, denoted $\text{im}(T)$, consists of all vectors $w \in W$ such that $w = T(u)$ for some $u \in V$. Lemma 2.4.2 provides the proof that $\text{ker}(T)$ is a subspace of $V$ and $\text{im}(T)$ is a subspace of $W$. For the kernel, the zero vector is included because $T(0) = 0$. If $u, v \in \text{ker}(T)$, then $T(u+v) = T(u) + T(v) = 0 + 0 = 0$, proving closure under addition. Similarly, $T(\alpha u) = \alpha T(u) = \alpha(0) = 0$, proving closure under scalar multiplication. Similar logic applies to the image subspace.

Definition 2.4.3 introduces dimensionality terms: the nullity of $T$ is $\dim(\text{ker} T)$, and the rank of $T$ is $\dim(\text{im} T)$. Theorem 2.4.5, known as the Rank-Nullity Theorem, states that for a transformation $T: V \rightarrow W$, $\text{nullity}(T) + \text{rank}(T) = \dim(V)$.

In Example 2.4.6, $T: F^{(3)} \rightarrow F^{(3)}$ is defined by $T(x, y, z) = (x+2z, 2x+y, -2y+2z)$. To find a basis for $\text{im}(T)$, we look at the images of standard basis vectors: $v_1 = (1, 2, 0)$, $v_2 = (0, 1, -2)$, and $v_3 = (2, 0, 2)$. These generate the image. Checking for independence via the determinant: $\det(\dots) = 0$, indicating dependence. Testing the subset $S = \{(1, 2, 0), (0, 1, -2)\}$, we find they are linearly independent. Thus, $\text{rank}(T) = 2$. By the Rank-Nullity Theorem, $\text{nullity}(T) = 3 - 2 = 1$. To find the basis for the kernel, we solve $T(x, y, z) = (0, 0, 0)$. The resulting system yields $x = -2z$ and $y = 4z$. Thus $(x, y, z) = z(-2, 4, 1)$, and a basis for $\text{ker}(T)$ is $\text{B} = \{(-2, 4, 1)\}$.

Rank of Matrices

The rank of a matrix is a fundamental property related to its invertible sub-structures. The column rank of a matrix $A \in M_{m \times n}(F)$ is the maximum number of linearly independent column vectors in $A$. The row rank is the maximum number of linearly independent row vectors. Example 2.5.2 and 2.5.4 compute these for a specific matrix $A$. For: $A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 0 & 3 \\ 1 & 2 & -1 \end{pmatrix}$ By applying row operations such as $R_2 - 2R_1$ and $R_3 - R_1$, the matrix is reduced to a form showing only two rows are linearly independent. Thus, the row rank is 2. Similarly, the column rank is 2. Theorem 2.5.5 states that row equivalent matrices have the same column rank, and Theorem 2.5.6 confirms that for any matrix $A$, $\text{rank}(A) = \text{column rank}(A) = \text{row rank}(A)$.

Solution of Systems of Linear Equations

Systems of linear equations can be subdivided into homogeneous and non-homogeneous categories. A homogeneous system is of the form $AX = 0$. Theorem 2.6.1 states that the solutions form a subspace of $F^{(n)}$ with dimension $n - \text{rank}(A)$. A non-zero solution exists if and only if $n > \text{rank}(A)$. If $n = \text{rank}(A)$, only the trivial zero solution exists. Example 2.6.2 shows a system where $n = 4$ and $\text{rank}(A) \leq 3$, thus non-zero solutions exist. Example 2.6.3 shows a system where $n = 3$ and $\text{rank}(A) = 3$, meaning only the zero solution exists.

A non-homogeneous system is of the form $AX = B$. Theorem 2.6.4 states that a solution exists if and only if $\text{rank}(A|B) = \text{rank}(A)$, where $(A|B)$ is the augmented matrix. Example 2.6.5 (first instance) provides a system where $\text{rank}(A|B) = 3 = \text{rank}(A)$, and the solution is found to be $x_1=1, x_2=2, x_3=1, x_4=-3, x_5=1$. However, in a second instance of Example 2.6.5, a system is presented where $\text{rank}(A|B) = 3$ but $\text{rank}(A) = 2$. Since the ranks are not equal, the system is inconsistent and has no solution.