Explain the image I sent you---This process defines a unique linear transformation, that is, if $S, T \in \mathcal{L}(V, W)$ satisfying $S(v_i) = T(v_i)$ for all $v_i \in \mathcal{B}$, then $S=T$.
**Example 2.8.** Let $T: \mathbb{R}^2 \mapsto \mathbb{R}^2$ be a linear mapping such that $T(1,0) = (1,2)$ and $T(0,1) = (-3,1)$.
Then, for any $(x,y) \in \mathbb{R}^2$,
$T(x,y) = T(x(1,0) + y(0,1))$
$= xT(1,0) + yT(0,1)$
$= x(1,2) + y(-3,1)$
$= (x, 2x) + (-3y, y)$
$= (x - 3y, 2x + y)$
This implies $T(x,y) = (x - 3y, 2x + y)$.
**Definition 2.9.** Let $T : V \mapsto W$ be linear mapping. Then
1. the dimension of the kernel of $T$ is called the nullity of $T$ and it is denoted by Nullity($T$); and
2. the dimension of the image of $T$ is called the rank of $T$ and it is denoted by Rank($T$).
**Example 2.10.** Let $T: \mathbb{R}^4 \mapsto \mathbb{R}^3$ be the linear transformation given by $T(a, b, c, d) = (a + d, b, d)$. Then determine the nullity and rank of $T$.
Let $V$ and $W$ be vector spaces over a field $F$. If $T \in \mathcal{L}(V, W)$ and $S \subseteq V$ we write $T(S) = \{T(v) | v \in S\}$.
**Theorem 2.11.** Let $V$ and $W$ be vector spaces over a field $F$. Let $T \in \mathcal{L}(V, W)$ be an isomorphism and $S \subseteq V$. Then
1. $S$ spans $V$ if and only if $T(S)$ spans $W$.
2. $S$ is linearly independent in $V$ if and only if $T(S)$ is linearly independent in $W$.
3. $S$ is a basis for $V$ if and only if $T(S)$ is a basis for $W$.
An isomorphism can be characterized as a linear transformation $\mathcal{L}(V, W)$ that maps a basis for $V$ to a basis for $W$.
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Linear transformations have a fundamental uniqueness property. If we know how a linear transformation acts on every vector in a basis of the domain space, then the transformation is completely determined. This means that if two linear transformations S and T agree on all basis vectors, then they must be identical transformations.
Let's work through Example 2.8 to see how the uniqueness property works in practice. We have a linear transformation T from R-squared to R-squared, and we know its values on the standard basis vectors. T maps (1,0) to (1,2) and (0,1) to (-3,1). Using linearity, we can find T of any vector (x,y) by expressing it as a linear combination of basis vectors and applying the transformation.
Now let's explore two fundamental concepts for linear transformations: nullity and rank. The nullity of a transformation T is the dimension of its kernel, which consists of all vectors that map to the zero vector. The rank of T is the dimension of its image, which is the set of all possible output vectors. These concepts help us understand the structure and properties of linear transformations.
Let's work through Example 2.10 to compute the nullity and rank of a specific transformation. We have T from R-4 to R-3 defined by T(a,b,c,d) equals (a+d, b, d). To find the nullity, we solve for the kernel by setting the output to zero. This gives us the system: a+d=0, b=0, d=0. Solving this, we get a=0, b=0, d=0, and c can be any value. So the kernel is spanned by (0,0,1,0), giving nullity 1. The rank is 3 since the transformation maps to a 3-dimensional space.
Finally, let's examine Theorem 2.11, which describes the remarkable properties of isomorphisms. An isomorphism is a bijective linear transformation that preserves all linear algebraic structure. The theorem states three equivalent conditions: a subset spans the domain if and only if its image spans the codomain; linear independence is preserved in both directions; and basis sets map to basis sets. This leads to a beautiful characterization: a linear transformation is an isomorphism if and only if it maps any basis of the domain to a basis of the codomain.