Problem 4 - Differential Equations and Linear Algebra

Question

Find \(\operatorname{Ker}(T)\) and \(\operatorname{Rng}(T),\) and give a geometrical description of each. Also, find \(\operatorname{dim}[\operatorname{Ker}(T)]\) and \(\operatorname{dim} [\operatorname{Rng}(T)],\) and verify Theorem 6.3.8. \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) defined by \(T(\mathbf{x})=A \mathbf{x},\) where $$A=\left[\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{array}\right].$$

Step-by-Step Solution

Answer

The kernel of the linear transformation \(T\) is \(\operatorname{Ker}(T)=\{ \mathbf{0}\}\), which is just a single point (the origin) in \(\mathbb{R}^3\), and its dimension is 0. The range of \(T\) is the entire \(\mathbb{R}^3\), and its dimension is 3. Theorem 6.3.8, which states that \(\operatorname{dim}[\operatorname{Ker}(T)] + \operatorname{dim}[\operatorname{Rng}(T)] = \operatorname{dim}(\mathbb{R}^{3})\), holds true in this case since \(0 + 3 = 3\).

Step 1: Find the Kernel of T

First, we have to find the null space of the matrix \(A\), which corresponds to \(\operatorname{Ker}(T)\). This means that we want to find all vectors \(\mathbf{x}\) that satisfy the equation \(A\mathbf{x}=\mathbf{0}\). To do this, we will form the augmented matrix and row reduce it to its reduced row echelon form: $$\left[\begin{array}{rrr|r} 1 & -1 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 2 & -1 & 1 & 0 \end{array}\right] \to \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right] \to \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & -1 & 0 \end{array}\right].$$ The general solution is: $$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.$$ Thus, the kernel of \(T\) is \(\operatorname{Ker}(T)=\{ \mathbf{0}\},\) which is a single point.

Step 2: Find the Range of T

Next, we need to find the column space of the matrix \(A\), which corresponds to \(\operatorname{Rng}(T)\). The column space is spanned by the columns of \(A\). However, since the matrix is in reduced row echelon form, the column space of the original matrix is equal to the column space of the row reduced matrix. The columns of the row reduced matrix are linearly independent, so the column space of \(A\) is just the span of these three columns: $$\operatorname{Rng}(T) = \operatorname{span}\left( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix}\right).$$

Step 3: Geometrical Description of Kernel and Range

Now we can describe the kernel and range of \(T\) geometrically. The kernel of \(T\) is \(\operatorname{Ker}(T)=\{ \mathbf{0}\},\) which is just a single point in \(\mathbb{R}^3\). Geometrically, this is just the origin. The range of \(T\) is the span of three linearly independent vectors in \(\mathbb{R}^3\), which means that it fills the entire space. Geometrically, \(\operatorname{Rng}(T)\) is the entire \(\mathbb{R}^3\).

Step 4: Dimensions of Kernel and Range

The dimension of the kernel of \(T\) is equal to the number of free variables in the general solution. Since there are no free variables in this case, \(\operatorname{dim}[\operatorname{Ker}(T)] = 0\). The dimension of the range of \(T\) is the number of linearly independent columns in \(A\). Since all columns are linearly independent, \(\operatorname{dim}[\operatorname{Rng}(T)] = 3\).

Step 5: Verify Theorem 6.3.8

Theorem 6.3.8 states that \(\operatorname{dim}[\operatorname{Ker}(T)] + \operatorname{dim}[\operatorname{Rng}(T)] = \operatorname{dim}(\mathbb{R}^{3})\). In this case, we found that \(\operatorname{dim}[\operatorname{Ker}(T)] = 0\) and \(\operatorname{dim}[\operatorname{Rng}(T)] = 3\). Therefore, we can verify the theorem as follows: $$0 + 3 = 3.$$ The theorem holds true in this case, as the dimensions of the kernel and range sum to the dimension of the domain (\(\mathbb{R}^3\)).

Key Concepts

Kernel of a Transformation

The kernel of a transformation, often denoted as \(\operatorname{Ker}(T)\), is the set of all vectors that map to the zero vector under the transformation. In simpler terms, it tells us which inputs result in no effect from the transformation. For the matrix \(A\) in the original exercise, this means finding all vectors \(\mathbf{x}\) such that \(A\mathbf{x} = \mathbf{0}\).
To identify these vectors, we solve the equation by forming an augmented matrix and reducing it to row echelon form. When we see that only the zero vector satisfies this condition, the kernel is trivial, meaning it only includes the zero vector itself.
This result geometrically signifies that no directions in space are flattened to a single point, other than the origin. In this case, the kernel of \(T\) is just \(\{\mathbf{0}\}\), the origin in \(\mathbb{R}^3\).

Range of a Transformation

The range of a transformation, or \(\operatorname{Rng}(T)\), represents all possible outputs from the transformation. When mapped using matrix \(A\), it encompasses all the vectors formed by linear combinations of the columns of \(A\). This is known as the column space.
The exercise shows how \(\operatorname{Rng}(T)\) is determined by the column space of the row-reduced form of the matrix. Here, the column vectors are linearly independent, highlighting that they span the entire three-dimensional space \(\mathbb{R}^3\).
Geometrically, if the range covers the whole space, any vector in \(\mathbb{R}^3\) can be reached by transforming some vector via matrix \(A\). Thus, \(\operatorname{Rng}(T) = \mathbb{R}^3\).

Dimension Theorem

The Dimension Theorem, an essential concept in linear algebra, connects the kernel and the range. It states that the sum of the dimensions of the kernel and range equals the dimension of the domain space.
In simpler terms:

• The dimension of \(\operatorname{Ker}(T)\) indicates directions squashed to zero by the matrix \(A\).
• The dimension of \(\operatorname{Rng}(T)\) reveals the number of independent directions or dimensions preserved.

For the exercise, since \(\operatorname{Ker}(T) = \{\mathbf{0}\}\), its dimension is 0. Meanwhile, \(\operatorname{Rng}(T)\) has a dimension of 3, matching \(\mathbb{R}^3\). The theorem succinctly verifies as \(0 + 3 = 3\), which proves the theorem correct for this transformation.

Null Space

The null space of a matrix \(A\), corresponding to the kernel when discussing transformations, is the set of vectors that result in the zero vector when multiplied by \(A\). In the context of \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\), it's a collection of all solutions to the homogeneous equation \(A\mathbf{x} = \mathbf{0}\).
With our specific exercise matrix, performing row reduction reveals that the only solution is the zero vector \(\mathbf{0}\).
This implies that the null space is just \(\{\mathbf{0}\}\), a trivial null space, emphasizing that \(A\) doesn't collapse any spatial dimensions beyond just reducing everything to the origin itself.

Column Space

The column space of a matrix \(A\) is all possible linear combinations of its column vectors. It effectively reveals how the transformation \(T\) projects or transforms vectors in space.
In our example, determining the column space involves inspecting the span of the linearly independent columns of \(A\). Upon transforming to reduced row echelon form, we see that they span \(\mathbb{R}^3\), confirming the range spans the entire space.
This means every vector in \(\mathbb{R}^3\) can be constructed from these columns, highlighting that there are no restrictions on the directions that \(T\) can produce, further emphasizing the dimension of the range is complete at 3.