Question

Let \(S\) and \(T\) be linear operators on \(V\) and assume that \(S T=T S\). a. Show that \(\mathrm{im} S\) and ker \(S\) are \(T\) -invariant. b. If \(U\) is \(T\) -invariant, show that \(S(U)\) is \(T\) -invariant.

Short answer

Both \( \text{im} S \) and \( \ker S \) are \( T \)-invariant, and \( S(U) \) is \( T \)-invariant if \( U \) is \( T \)-invariant.

Step-by-step solution

Understanding the Problem

We need to show that the image and kernel of a linear operator \( S \) are invariant under another operator \( T \), given that \( S \) and \( T \) commute. Additionally, if a subspace \( U \) is invariant under \( T \), we need to show that \( S(U) \) is also invariant under \( T \).

Proving Image Invariance (a)

To prove that \( \text{im} S \) is \( T \)-invariant, consider any element \( w = S(v) \in \text{im} S \). Since \( S \) and \( T \) commute, we have \( T(S(v)) = S(T(v)) \), which is still in \( \text{im} S \). Thus, \( T \text{im} S \subseteq \text{im} S \), demonstrating \( \text{im} S \) is \( T \)-invariant.

Proving Kernel Invariance (a)

To prove that \( \ker S \) is \( T \)-invariant, consider any \( v \in \ker S \). By definition, \( S(v) = 0 \). We need \( T(v) \in \ker S \), i.e., \( S(T(v)) = 0 \). Since \( S(T(v)) = T(S(v)) = T(0) = 0 \), \( T(v) \) indeed lies in \( \ker S \). Thus, \( \ker S \) is \( T \)-invariant.

Showing Invariance of S(U) (b)

If \( U \) is \( T \)-invariant, for any \( u \in U \), \( T(u) \in U \). We need to show \( S(U) \) is \( T \)-invariant. Take \( S(u) \in S(U) \); then \( T(S(u)) = S(T(u)) \) since \( S \) and \( T \) commute, and \( T(u) \in U \). Hence, \( S(T(u)) \in S(U) \), proving \( S(U) \) is \( T \)-invariant.

Key concepts

Image Invariance

In the world of linear algebra, understanding how linear operators interact is fundamental. The concept of image invariance refers to how the image of a linear operator remains unchanged under another operator. Given two operators, say \( S \) and \( T \), if these operators commute, meaning \( S T = T S \), this property comes into play.

• To demonstrate image invariance, we start with an element \( w = S(v) \), from the image of \( S \).
• Since the operators commute, \( T(S(v)) = S(T(v)) = w' \), where \( w' \) still resides in the image of \( S \).
• This process proves that applying \( T \) does not take elements out of the image of \( S \). Hence, the image of \( S \) is invariant under \( T \).

Image invariance is essential as it reassures that certain aspects of the vector transformations remain consistent despite applying another operator.

Kernel Invariance

Kernel invariance involves a similar idea but focuses on the kernel of an operator. The kernel of a linear operator \( S \) consists of all elements that \( S \) maps to the zero vector. When we say the kernel is invariant under another operator \( T \), it means that applying \( T \) to elements of the kernel does not move them out of the kernel.

• Consider an element \( v \) in the kernel of \( S \).
• Since \( S(v) = 0 \), we check if \( T(v) \) remains in the kernel by seeing if \( S(T(v)) = 0 \).
• Using the commutative property, \( T(S(v)) = S(T(v)) \), we find \( S(T(v)) = T(0) = 0 \).

This shows that \( T(v) \) still maps to zero, ensuring the kernel is invariant under \( T \). This invariance is useful in operations involving null spaces and evaluating linear maps.

Commutative Operators

Commutative operators are the backbone of discussing invariance in linear operators. Commutativity simply means that the order in which operators are applied does not matter. In formulas, \( S T = T S \).

• To grasp this, consider how we apply \( S \) followed by \( T \) to a vector, and vice-versa, the result must be identical.
• This property significantly influences both image and kernel invariance, making certain subspaces resistant to changes or displacements.

Commutativity simplifies dealing with complex linear operations since it reduces potential discrepancies from the sequence of operations. This property is crucial when achieving sophisticated manipulations of vector spaces.

Subspace Invariance

Subspace invariance is about the stability of subspaces under transformations. When a subspace \( U \) is \( T \)-invariant, it means \( T(u) \) remains in \( U \) for any vector \( u \) in \( U \). If \( U \) is \( T \)-invariant, then transformations applied by \( S \), such as \( S(U) \), will also be stable under \( T \).

• When you take an element \( S(u) \) within \( S(U) \), due to commutativity, \( T(S(u)) \) can be computed as \( S(T(u)) \).
• Given \( T(u) \) stays in \( U \), \( S(T(u)) \) must lie in \( S(U) \) because \( S \) maps elements of \( U \) to \( S(U) \).

Understanding subspace invariance helps in assessing the robustness of vector subspaces under multiple transformations. Such insights are critical in fields like quantum mechanics and computer graphics, where vector spaces undergo complex interactions.