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Chapter 3: Problem 25

Prove that if \(V\) is finite dimensional with \(\operatorname{dim} V>1\), then the set of noninvertible operators on \(V\) is not a subspace of \(\mathcal{L}(V)\)

Short Answer

Expert verified
In conclusion, the set of noninvertible operators on a finite-dimensional vector space, V, is not a subspace of the space of linear operators, L(V), because the sum of noninvertible operators is noninvertible and the scalar product of a noninvertible operator is noninvertible as well. Thus, the set of noninvertible operators does not meet the requirements for being a subspace.

Step by step solution

01

Sum of noninvertible operators

Let's consider two operators, A and B, such that A is a zero operator and B is a noninvertible operator. We know that A is noninvertible, since det(A)=0, and an operator is invertible if and only if its determinant is nonzero. Additionally, we know that B is noninvertible by definition. Now, let's consider their sum: \(C = A+B\) If the sum of A and B is also noninvertible, then the set of noninvertible operators could still potentially be a subspace. If the sum is invertible, however, then the set is not a subspace. Let's calculate the determinant of C: \(det(C) = det(A+B)\) Since A is a zero operator, this simplifies to: \(det(C) = det(0+B) = det(B)\) We know that B is noninvertible, so the determinant of B is zero. Thus, the determinant of C is also zero, which implies that the sum of two noninvertible operators is also noninvertible.
02

Scalar product of a noninvertible operator

Now, let's consider an arbitrary scalar, k, and a noninvertible operator, B. We want to show whether or not their product, kB, is also noninvertible. If this product is noninvertible, then the set of noninvertible operators is still potentially a subspace. If the product is invertible, however, then the set is not a subspace. Let's calculate the determinant of kB: \(det(kB) = k^n det(B)\), where n is the dimension of V, since it's a finite-dimensional vector space. Since the operator B is noninvertible, we know that det(B) = 0. However, multiplying this by k will still yield a determinant of zero (if k is nonzero), which implies that the scalar product of a noninvertible operator is also noninvertible: \(det(kB) = k^n (0) = 0\) However, if k is equal to zero, the scalar product becomes a zero operator, and the determinant is still zero, making the scalar product noninvertible. #Conclusion# Based on the above two steps, we have seen that the sum of noninvertible operators is noninvertible and the scalar product of a noninvertible operator is noninvertible as well. Therefore, the set of noninvertible operators is not a subspace of L(V).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Finite-Dimensional Vector Spaces
In linear algebra, a vector space is termed finite-dimensional if its vectors can be completely described using a finite number of coordinates. This essentially means that there exists a finite basis for the space; for any vector within this space, it can be expressed as a finite linear combination of these basis vectors.

To give an example, the three-dimensional space we are familiar with is a finite-dimensional vector space, where every point can be defined using just three coordinates: length, width, and height (or x, y, and z in Cartesian coordinates). In the context of the exercise, the dimension of the vector space, denoted as \(\operatorname{dim} V\), refers to the number of vectors in the basis of vector space V.

Understanding the concept of finite-dimensional vector spaces is crucial when considering linear operators, as these operators act on vectors in the space, and their properties are deeply tied to the dimensions of the space.
Subspace Criteria
A subspace is a subset of a vector space that is itself a vector space under the same addition and scalar multiplication of its parent space. To qualify as a subspace, the subset must satisfy three criteria:
  • It must contain the zero vector.
  • It must be closed under addition, meaning if you take any two vectors in the subset, their sum must also be in the subset.
  • It must be closed under scalar multiplication; multiplying any vector in the subset by a scalar must yield another vector within the subset.

The exercise provided presents a situation where noninvertible operators are assessed to determine if they form a subspace within all linear operators acting on a finite-dimensional space. We conclude that noninvertible operators don't form a subspace because, while they include the zero vector (a noninvertible operator), closure under addition is not guaranteed as seen in the step-by-step solution when noninvertible operators are combined.
Determinant of a Linear Operator
The determinant is a unique scalar value associated with a linear operator (or matrix) that provides critical information about the operator. For instance, the determinant can disclose whether an operator is invertible or not; a nonzero determinant implies an invertible operator while a zero determinant indicates a noninvertible operator.

In the context of our exercise, the zero determinant of the sum and scalar multiple of noninvertible operators (\(A+B\) and \(kB\)) confirms that these operations do not result in an invertible operator. This property of determinants being zero for noninvertible operators played a central role in proving that noninvertible operators do not form a subspace, as one of the key properties of a subspace is to be closed under scalar multiplication and addition.
Linear Algebra
Linear algebra is the area of mathematics that deals with vectors, vector spaces, linear mappings, and systems of linear equations. It provides the foundation for many areas of mathematics and applications in science and engineering. From computer graphics to solving complex optimization problems or understanding dynamic systems, linear algebra is fundamental.

The exercise concerning noninvertible operators challenges the student to apply concepts from linear algebra, such as vector spaces, determinants, and subspace criteria. Understanding the characteristics of linear operators and how they interact within the structure of a vector space is essential in higher mathematics and many applied disciplines.

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Most popular questions from this chapter

Prove that if \(S_{1}, \ldots, S_{n}\) are injective linear maps such that \(S_{1} \ldots S_{n}\) makes sense, then \(S_{1} \ldots S_{n}\) is injective.

Suppose that \(V\) and \(W\) are both finite dimensional. Prove that there exists a surjective linear map from \(V\) onto \(W\) if and only if \(\operatorname{dim} W \leq \operatorname{dim} V\)

Suppose \(\left(\nu_{1}, \ldots, \nu_{n}\right)\) is a basis of \(V .\) Prove that the function \(T: V \rightarrow \operatorname{Mat}(n, 1, \mathbf{F})\) defined by $$T v=\mathcal{M}(v)$$ is an invertible linear map of \(V\) onto Mat \((n, 1, \mathbf{F}) ;\) here \(\mathcal{M}(v)\) is the matrix of \(v \in V\) with respect to the basis \(\left(v_{1}, \ldots, v_{n}\right)\)

Suppose \(n\) is a positive integer and \(a_{i, j} \in \mathbf{F}\) for \(i, j=1, \ldots, n\) Prove that the following are equivalent: (a) The trivial solution \(x_{1}=\cdots=x_{n}=0\) is the only solution to the homogeneous system of equations $$\begin{array}{c}\sum_{k=1}^{n} a_{1, k} x_{k}=0 \\\\\vdots \\\\\sum_{k=1}^{n} a_{n, k} x_{k}=0\end{array}$$ (b) For every \(c_{1}, \ldots, c_{n} \in \mathbf{F},\) there exists a solution to the sys tem of equations $$\begin{array}{c}\sum_{k=1}^{n} a_{1, k} x_{k}=c_{1} \\ \vdots \\\\\sum_{k=1}^{n} a_{n, k} x_{k}=c_{n}\end{array}$$ Note that here we have the same number of equations as variables.

Suppose that \(V\) and \(W\) are finite dimensional and that \(U\) is a subspace of \(V .\) Prove that there exists \(T \in \mathcal{L}(V, W)\) such that null \(T=U\) if and only if \(\operatorname{dim} U \geq \operatorname{dim} V-\operatorname{dim} W\)
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