Problem 10
Question
Decide whether or not the given mapping \(T\) is a linear transformation. Justify your answers. For each mapping that is a linear transformation, decide whether or not \(T\) is one-to-one, onto, both, or neither, and find a basis and dimension for \(\operatorname{Ker}(T)\) and \(\operatorname{Rng}(T)\) $$\begin{aligned} &T: \mathbb{R}^{3} \rightarrow M_{2}(\mathbb{R}) \text { defined by }\\\ &T\left(\left(x_{1}, x_{2}, x_{3}\right)\right)=\left[\begin{array}{cc} 0 & x_{1}-x_{2}+x_{3} \\ -x_{1}+x_{2}-x_{3} & 0 \end{array}\right] \end{aligned}$$
Step-by-Step Solution
Verified Answer
The given transformation \(T\) is a linear transformation, which is one-to-one but not onto. The basis and dimension of the kernel are the empty set and 0, respectively, while the basis of the range is given by \(\left\{ \left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right], \left[\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right] \right\}\) and its dimension is 2.
1Step 1: Check if the mapping is linear
We need to check if the given mapping T satisfies the conditions of being a linear transformation. These conditions are:
1. Additivity: \(T(u+v) = T(u) + T(v)\) for any vectors \(u\) and \(v\)
2. Homogeneity: \(T(kv) = kT(v)\) for any scalar \(k\) and any vector \(v\)
To check the conditions, let \(u = (u_1, u_2, u_3)\) and \(v = (v_1, v_2, v_3)\) in \(\mathbb{R}^3\), and let \(k\) be a scalar. Then,
Additivity:
\(T(u+v) = T((u_1+v_1, u_2+v_2, u_3+v_3)) = \left[\begin{array}{cc}
0 & (u_1+v_1)-(u_2+v_2)+(u_3+v_3) \\
-(u_1+v_1)+(u_2+v_2)-(u_3+v_3) & 0
\end{array}\right]\)
We also have,
\(T(u) + T(v) = \left[\begin{array}{cc}
0 & u_1-u_2+u_3 \\
-u_1+u_2-u_3 & 0
\end{array}\right] +
\left[\begin{array}{cc}
0 & v_1-v_2+v_3 \\
-v_1+v_2-v_3 & 0
\end{array}\right] = \left[\begin{array}{cc}
0 & (u_1+v_1)-(u_2+v_2)+(u_3+v_3) \\
-(u_1+v_1)+(u_2+v_2)-(u_3+v_3) & 0
\end{array}\right]\)
Since \(T(u+v) = T(u) + T(v)\), additivity condition is satisfied.
Homogeneity:
\(T(kv) = T((k v_1, k v_2, k v_3)) = \left[\begin{array}{cc}
0 & k v_1 - k v_2 + k v_3 \\
- k v_1 + k v_2 - k v_3 & 0
\end{array}\right]\)
We also have,
\(kT(v) = k\left[\begin{array}{cc}
0 & v_1 - v_2 + v_3 \\
- v_1 + v_2 - v_3 & 0
\end{array}\right] = \left[\begin{array}{cc}
0 & k v_1 - k v_2 + k v_3 \\
- k v_1 + k v_2 - k v_3 & 0
\end{array}\right]\)
Since \(T(kv) = kT(v)\), homogeneity condition is also satisfied. Hence, the given mapping T is a linear transformation.
2Step 2: Determine if T is one-to-one, onto, both or neither
Now that we know T is a linear transformation, let's analyze whether it's one-to-one, onto, both or neither.
One-to-one (injective) if and only if \(T(u) = T(v)\) implies \(u=v\).
Onto (surjective) if and only if for every matrix in \(M_{2}(\mathbb{R})\), there exists a vector in \(\mathbb{R}^{3}\) such that T maps that vector to that matrix.
Let's find the kernel of T, which will help in determining if T is one-to-one or not.
Kernel:
\(\operatorname{ker}(T) = \{v \in \mathbb{R}^3 | T(v) = 0\}\)
The kernel of T is all vectors \(v\) in \(\mathbb{R}^3\) such that
\(T(v) = \left[\begin{array}{cc}
0 & v_1 - v_2 + v_3 \\
- v_1 + v_2 - v_3 & 0
\end{array}\right] = \left[\begin{array}{cc}
0 & 0 \\
0 & 0
\end{array}\right]\)
This is equivalent to the system of equations:
\(v_1 - v_2 + v_3 = 0\)
\(- v_1 + v_2 - v_3 = 0\)
It's clear that the only solution to this system is the trivial solution, \(v = (0,0,0)\). Therefore, \(\operatorname{ker}(T) = \{(0,0,0)\}\), which means T is one-to-one.
Now let's determine if T is onto.
To do this, observe that any element of \(M_{2}(\mathbb{R})\) looks like this:
\(\left[\begin{array}{cc}
a & b \\
c & d
\end{array}\right]\)
Observe that, based on the definition of T, the diagonal elements will always be 0. Therefore, we can't find any matrix with non-zero diagonal elements in the range of T.
So T is not onto.
Thus, the given linear transformation T is one-to-one but not onto.
3Step 3: Finding basis and dimension for the kernel and range of T
Now that we know T is one-to-one and not onto, let's find the basis and dimension for the kernel and range of T.
Kernel:
We already found the kernel of T to be \(\operatorname{ker}(T) = \{(0,0,0)\}\). The basis for the kernel of T is the empty set, and the dimension is 0, since the kernel contains only the zero vector.
Range:
The generic image of T can be written as:
\(T\left(\left(x_{1}, x_{2}, x_{3}\right)\right)=\left[\begin{array}{cc}
0 & x_{1}-x_{2}+x_{3} \\\
-x_{1}+x_{2}-x_{3}& 0
\end{array}\right] = \left[\begin{array}{cc}
0 & a_1 \\\
a_2 & 0
\end{array}\right]\), with \( a_1 = x_1 -x_2 + x_3\) and \(a_2 = -x_1 + x_2 - x_3\)
Any matrix in the range of T looks like \(\left[\begin{array}{cc}
0 & a_1 \\
a_2 & 0
\end{array}\right]\), with \(a_1, a_2 \in \mathbb{R}\).
A basis for the range of T is given by:
\(\operatorname{Rng}(T) = \left\{ \left[\begin{array}{cc}
0 & 1 \\
0 & 0
\end{array}\right], \left[\begin{array}{cc}
0 & 0 \\
1 & 0
\end{array}\right] \right\}\)
The dimension of the range of T is 2, which is the number of elements in the basis.
In conclusion, for the given linear transformation T, it is one-to-one, not onto, the kernel has a basis of the empty set and dimension 0, and the range has a basis of two matrices and dimension 2.
Key Concepts
Basis and DimensionOne-to-One MappingOnto MappingKernel of a TransformationRange of a Transformation
Basis and Dimension
Understanding the ideas of basis and dimension is fundamental in linear algebra, as they provide a framework for talking about the structure of vector spaces. A basis of a vector space is a set of vectors that is linearly independent and spans the entire space. This means that every vector in the space can be expressed as a linear combination of the basis vectors, and none of the basis vectors can be written as a combination of the others.
The dimension of a vector space is simply the number of vectors in any of its bases, which by definition must all have the same number of elements. If a vector space has a finite basis of n vectors, we say that the space is n-dimensional. For example, \( \text{Ker}(T) \), the kernel of the transformation in our exercise, is 0-dimensional because its only member is the zero vector, often shown as \( \text{{dim}}(\text{Ker}(T)) = 0 \). This also reveals that T is one-to-one, since only the zero vector is mapped to the zero matrix. On the other hand, the range of T, \( \text{Rng}(T) \), is spanned by a basis of two elements, meaning its dimension is 2, \( \text{{dim}}(\text{Rng}(T)) = 2 \).
The dimension of a vector space is simply the number of vectors in any of its bases, which by definition must all have the same number of elements. If a vector space has a finite basis of n vectors, we say that the space is n-dimensional. For example, \( \text{Ker}(T) \), the kernel of the transformation in our exercise, is 0-dimensional because its only member is the zero vector, often shown as \( \text{{dim}}(\text{Ker}(T)) = 0 \). This also reveals that T is one-to-one, since only the zero vector is mapped to the zero matrix. On the other hand, the range of T, \( \text{Rng}(T) \), is spanned by a basis of two elements, meaning its dimension is 2, \( \text{{dim}}(\text{Rng}(T)) = 2 \).
One-to-One Mapping
A one-to-one mapping, also called an injective function, guarantees that distinct inputs produce distinct outputs. For a linear transformation T from one vector space to another, being one-to-one means that if \( T(u) = T(v) \), then necessarily \( u = v \).
In the context of our example, we check this by looking at the kernel of T. Since the kernel only contains the zero vector, there're no other vectors that map to the zero matrix, confirming that T is indeed one-to-one. In technical terms, the fact that \( \text{Ker}(T) \) is trivial (contains only the zero vector) is what assures us of the injective property of the transformation.
In the context of our example, we check this by looking at the kernel of T. Since the kernel only contains the zero vector, there're no other vectors that map to the zero matrix, confirming that T is indeed one-to-one. In technical terms, the fact that \( \text{Ker}(T) \) is trivial (contains only the zero vector) is what assures us of the injective property of the transformation.
Onto Mapping
Conversely, an onto mapping, or a surjective function, describes a transformation where every possible output is achieved by some input. For a linear transformation T: \( \text{\(U\)} \) to \( \text{\(V\)} \), T is said to be onto if for every element \( v \) in V, there exists at least one element \( u \) in U such that \( T(u) = v \).
In our exercise, we determine that T is not onto because there exist matrices in \( M_{2}(\mathbb{R}) \) whose diagonal elements are not zero - these cannot be the output of the given transformation T. Thus, the range of T does not include all possible 2 by 2 matrices, making the transformation not onto.
In our exercise, we determine that T is not onto because there exist matrices in \( M_{2}(\mathbb{R}) \) whose diagonal elements are not zero - these cannot be the output of the given transformation T. Thus, the range of T does not include all possible 2 by 2 matrices, making the transformation not onto.
Kernel of a Transformation
The kernel of a transformation T, denoted by \( \text{Ker}(T) \), represents all the inputs that map to the zero vector (or zero matrix in the case of transformations to \( M_{2}(\mathbb{R}) \)). Mathematically, \( \text{Ker}(T) = \{v \text{ in } U : T(v) = 0\} \).
In the solved example, we find that the kernel of T consists only of the zero vector in \( \text{\( \mathbb{R}^3\)} \). This reveals an important property – since the kernel is trivial, it tells us the transformation is one-to-one. A non-trivial kernel, one that contains vectors other than the zero vector, would mean the transformation is not one-to-one, as it would imply that there are multiple inputs for a single output (the zero vector).
In the solved example, we find that the kernel of T consists only of the zero vector in \( \text{\( \mathbb{R}^3\)} \). This reveals an important property – since the kernel is trivial, it tells us the transformation is one-to-one. A non-trivial kernel, one that contains vectors other than the zero vector, would mean the transformation is not one-to-one, as it would imply that there are multiple inputs for a single output (the zero vector).
Range of a Transformation
The range of a transformation T, referred to as \( \text{Rng}(T) \), includes all possible outputs of T. It's the set of vectors in the target space that T can produce from the input space. The range is pivotal when determining if T is onto.
For our mapping, we conclude that the range is the set of all matrices in \( M_{2}(\mathbb{R}) \) of the form \( \text{{left[}}\begin{array}{cc} 0 & a_1 \ a_2 & 0 \end{array}\text{{right]}} \). The basis for this range consists of two matrices, which shows us that any matrix in the range can be expressed as a combination of these two. Though the range is rich enough for a 2-dimensional vector space, it doesn't cover all 2 by 2 matrices, indicating that T is not onto.
For our mapping, we conclude that the range is the set of all matrices in \( M_{2}(\mathbb{R}) \) of the form \( \text{{left[}}\begin{array}{cc} 0 & a_1 \ a_2 & 0 \end{array}\text{{right]}} \). The basis for this range consists of two matrices, which shows us that any matrix in the range can be expressed as a combination of these two. Though the range is rich enough for a 2-dimensional vector space, it doesn't cover all 2 by 2 matrices, indicating that T is not onto.
Other exercises in this chapter
Problem 9
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