Question

In each case either prove the statement or give an example in which it is false. Throughout, let \(T: V \rightarrow W\) be a linear transformation where \(V\) and \(W\) are finite dimensional. a. If \(V=W,\) then ker \(T \subseteq \operatorname{im} T\). b. If \(\operatorname{dim} V=5, \operatorname{dim} W=3,\) and \(\operatorname{dim}(\operatorname{ker} T)=2\), then \(T\) is onto. c. If \(\operatorname{dim} V=5\) and \(\operatorname{dim} W=4,\) then \(\operatorname{ker} T \neq\\{\mathbf{0}\\}\). d. If ker \(T=V,\) then \(W=\\{\mathbf{0}\\}\). e. If \(W=\\{\mathbf{0}\\},\) then \(\operatorname{ker} T=V\). f. If \(W=V,\) and \(\operatorname{im} T \subseteq \operatorname{ker} T,\) then \(T=0\). g. If \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\right\\}\) is a basis of \(V\) and \(T\left(\mathbf{e}_{1}\right)=\mathbf{0}=T\left(\mathbf{e}_{2}\right),\) then \(\operatorname{dim}(\operatorname{im} T) \leq 1\) h. If \(\operatorname{dim}(\operatorname{ker} T) \leq \operatorname{dim} W,\) then \(\operatorname{dim} W \geq \frac{1}{2} \operatorname{dim} V\). i. If \(T\) is one-to-one, then \(\operatorname{dim} V \leq \operatorname{dim} W\). j. If \(\operatorname{dim} V \leq \operatorname{dim} W,\) then \(T\) is one-to- one. k. If \(T\) is onto, then \(\operatorname{dim} V \geq \operatorname{dim} W\). 1\. If \(\operatorname{dim} V \geq \operatorname{dim} W,\) then \(T\) is onto. m. If \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{k}\right)\right\\}\) is independent, then \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right\\}\) is independent. n. If \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right\\}\) spans \(V,\) then \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{k}\right)\right\\}\) spans \(W\)

Short answer

Statements a, d, j, l, and n are false; statements b, c, e, f, g, h, i, k, and m are true.

Step-by-step solution

Part a: Prove or Disprove

We are asked if for a linear transformation \( T: V \to W \) with \( V = W \), \( \ker T \subseteq \operatorname{im} T \) holds. Consider the transformation \( T: V \to V \) that maps every vector to the zero vector. In this case, \( \ker T = V \) because all vectors map to zero, but \( \operatorname{im} T = \{\mathbf{0}\} \). Clearly, \( V subseteq \{\mathbf{0}\} \). Hence, the statement is false.

Part b: Linear Transformation Dimension Properties

Given \( \operatorname{dim} V = 5 \), \( \operatorname{dim} W = 3 \), and \( \operatorname{dim}(\operatorname{ker} T) = 2 \), use the rank-nullity theorem: \( \operatorname{dim}(\ker T) + \operatorname{rank}(T) = \operatorname{dim}(V) \). Substituting the values, \( 2 + \operatorname{rank}(T) = 5 \) gives \( \operatorname{rank}(T) = 3 \). Since the rank of \( T \) equals the dimension of \( W \), \( T \) is onto. Thus, the statement is true.

Part c: Kernel of T is Nonzero

With \( \operatorname{dim} V = 5 \) and \( \operatorname{dim} W = 4 \), by the rank-nullity theorem, \( \operatorname{dim}(\ker T) + \operatorname{rank}(T) = 5 \). The rank, being \( \leq 4 \), implies \( \operatorname{dim}(\ker T) \geq 1 \). Therefore, \( \ker T eq \{\mathbf{0}\} \), so the statement is true.

Part d: Validation of V is Ker T

If \( \ker T = V \), every vector in \( V \) maps to zero, hence the image of \( T \) is \( \{\mathbf{0}\} \). No condition in the statement requires \( W \) to be trivial, leading to \( W \) not necessarily being \( \{\mathbf{0}\} \). Therefore, the statement is false.

Part e: All of V is Kernel

For any transformation \( T: V \to \{\mathbf{0}\} \), each vector in \( V \) maps to the zero element in \( W \). Consequently, \( \operatorname{ker} T = V \). Thus, the statement is true.

Part f: Imaging is completely within Kernel

If \( \operatorname{im} T \subseteq \operatorname{ker} T \), then for any \( \mathbf{v} \in V \), \( T(T(\mathbf{v})) = \mathbf{0} \). Further, when \( W = V \) meaning \( \operatorname{dim}(\operatorname{ker} T) \geq \operatorname{dim}(\operatorname{im} T) \), it follows \( T(V) = \{\mathbf{0}\} \), indicating \( T \) as the zero transformation since its effect nullifies the space. Thus, the statement holds true.

Part g: Basis with Image Constraint

Here, since \( T(\mathbf{e}_1) = \mathbf{0} \) and \( T(\mathbf{e}_2) = \mathbf{0} \), these two basis elements of \( V \) contribute nothing to the image of \( T \). Thus, the image can only be spanned possibly by \( T(\mathbf{e}_3) \). Hence, the image has dimension at most 1, making the statement true.

Part h: Dimensions Comparison

With \( \operatorname{dim}(\ker T) \leq \operatorname{dim}(W) \), and using \( \operatorname{dim}V = \operatorname{rank}(T) + \operatorname{dim}(\ker T) \), imply \( \operatorname{rank}(T) = \operatorname{dim}V - \operatorname{dim}(\ker T) \leq \operatorname{dim}V - \operatorname{dim}W \). Therefore, \( \operatorname{dim}W \geq \frac{1}{2} \operatorname{dim}V \) holds. Hence, the statement is true.

Part i: One-to-One ensures Dimensional Ordering

Since \( T \) is one-to-one, by dimensional equality from the rank-nullity theorem where \( \operatorname{dim}(\ker T) \) is zero, we derive \( \operatorname{rank}(T) = \operatorname{dim}V \). The rank, representing active images, rests \( \leq \operatorname{dim}W \). Hence, \( \operatorname{dim}V \leq \operatorname{dim}W \), rendering this statement true.

Part j: Dimensional comparison not ensuring Injectivity

Though \( \operatorname{dim}V \leq \operatorname{dim}W \), no inherent property demands injectivity (one-to-one). For example: mapping all in \( V \) to a subset in \( W \) performs without contradicting dimensions. Therefore, without further specifics, incurrence of one-to-one isn't confirmed, designating the statement false.

Part k: Surjectivity indicates Dimensional Order

If \( T \) is onto, then all elements of \( W \) are images of elements from \( V \). Hence, \( \operatorname{rank}(T) = \operatorname{dim}(W) \). From rank-nullity theorem, \( \operatorname{dim}(V) = \operatorname{rank}(T) + \operatorname{dim}(\ker T) \geq \operatorname{dim}(W) \). Therefore, the statement is correct.

Part l: Dimension Implication of Surjectivity

That \( \operatorname{dim}V \geq \operatorname{dim}W \), doesn't directly confer \( T \) as onto unless rank aligns equally with \( \operatorname{dim}W \) through exact mappings. Surjective condition necessitates function surmount beyond and within dimensional allowances. Without further data given, falsehood ascribed.

Part m: Independence Implies Independence

Transformational linear independence supposes original independence through injectivity of representation, equivocally rendering V’s subbasis. Thus on affirmed ground, correct identity of independence transfers preserved. Therefore, statement aligns true.

Part n: Spanning V implies Spanning W

Within query upon span, transformative encapsulation alone indirectly exhausts dynamics, disallowing immediate contagion as spanning deficit potentiality. Unstated conditions for \( T \) to transmit span, thus ambiguous, designating false assertion.

Key concepts

Rank-Nullity Theorem

When trying to understand linear transformations, one powerful tool you can use is the Rank-Nullity Theorem. This theorem connects three important concepts: the dimension of the kernel, the rank of the transformation, and the dimension of the vector space. It's expressed by the equation:

• \[\operatorname{dim}( ext{ker } T) + \operatorname{rank}(T) = \operatorname{dim}(V) \]

This equation helps us understand how the transformation affects the vector space. The dimension of the kernel, \(\operatorname{dim}( ext{ker } T)\), tells us how many vectors are mapped to zero.
The rank, \(\operatorname{rank}(T)\), is the dimension of the image, \(\operatorname{im} T\), which consists of vectors actually "reached" by \(T\).
In practical terms, Rank-Nullity Theorem provides insights into many properties of transformations. For example, if you know the kernel's dimension and the space's dimension, you can deduce the image's dimension. This is useful in determining whether a transformation is one-to-one or onto. In parts of the exercise where transformation dimensions are considered, using this theorem is crucial. A common mistake is overlooking that the sum of dimensions of the kernel and image must always equal the dimension of the original space.

Kernel

The kernel of a linear transformation \(T: V \rightarrow W\) is a foundational concept to grasp. It is the set of vectors in \(V\) that are mapped to the zero vector in \(W\). In simpler terms, it's those vectors that lose all identity and become zero after transformation.
Let's explore what this means, typically denoted as:

• \[\ker T = \{ \mathbf{v} \in V \mid T(\mathbf{v}) = \mathbf{0} \} \]

The kernel provides essential information about a transformation. A trivial kernel, \( \ker T = \{\mathbf{0}\}\), implies the transformation is injective or one-to-one, meaning different input vectors produce different output vectors.
By analyzing the kernel, you can deduce much about how your linear transformation functions. For instance, in the exercise, knowing the dimension of the kernel allows students to draw conclusions about the transformation, whether it's onto or one-to-one. It also helps in verifying how much of the vector space is dampened or nullified by the transformation.

Image

In the study of linear transformations, the image is a crucial concept that helps us understand what a transformation truly 'projects' or 'outputs.' The image of a transformation \(T: V \rightarrow W\) is the set of all possible outputs \(T(\mathbf{v})\) as \(\mathbf{v}\) varies over the entire vector space \(V\).
Symbolically, it's written as:

• \[\operatorname{im} T = \{ T(\mathbf{v}) \mid \mathbf{v} \in V \}\]

The dimension of the image, known as the rank of the transformation, shows how many dimensions of \(W\) are effectively utilized by \(T\). It gives insight into whether or not \(T\) is onto (surjective), meaning every element of \(W\) can be expressed as \(T(\mathbf{v})\) for some \(\mathbf{v}\) from \(V\). In parts of the exercise, understanding the image aids in determining whether transformations span the entire target space.
This tool aids in visualizing which outputs are "hit" by the transformation. If the image covers all of \(W\), the transformation is known as surjective. If an image is limited, it highlights constraints within the transformation and aids in exploring solution possibilities of different vector space interactions.