Question

If \(T: V \rightarrow V\) is any linear operator, show that ker \(T\) and im \(T\) are \(T\) -invariant subspaces.

Short answer

Ker(T) and im(T) are invariant under T because applying T keeps vectors within those subspaces.

Step-by-step solution

Understand Kernel and Image

The kernel (ker) of a linear operator \(T: V \rightarrow V\) is the set of vectors in \(V\) that map to the zero vector. The image (im) is the set of vectors \(T\) maps to from \(V\). We need to show these are \(T\)-invariant, meaning applying \(T\) to these subspaces results in vectors still within the subspace.

Prove Kernel is T-Invariant

Choose an arbitrary vector \(v \in \text{ker}(T)\), meaning \(T(v) = 0\). We need to show \(T(v)\) still lies within \(\text{ker}(T)\) when \(T\) is applied again. Since \(T(v) = 0\), applying \(T\) again gives \(T(T(v)) = T(0) = 0\). Thus, \(T(v)\) remains in the kernel, proving \(\text{ker}(T)\) is invariant under \(T\).

Prove Image is T-Invariant

Take any vector \(w \in \text{im}(T)\). This implies there exists some vector \(v \in V\) such that \(T(v) = w\). We need to show \(T(w)\) is still in \(\text{im}(T)\). Since \(w = T(v)\), applying \(T\) gives \(T(w) = T(T(v))\), which is the image of a vector \(u = T(v)\). Since \(u = T(x)\) for some \(x \in V\), \(T(u)\) stays within \(\text{im}(T)\), confirming \(\text{im}(T)\) is \(T\)-invariant.

Conclusion

From these arguments, both the kernel and image of \(T\) are \(T\)-invariant subspaces as applying the linear operator \(T\) to any vector in these subspaces results in vectors remaining in the same subspace.

Key concepts

Linear Operator

A linear operator is a function that maps one vector space to another while respecting the operations of vector addition and scalar multiplication. Specifically, if we have a linear operator \( T: V \rightarrow V \), this means it acts on vectors from the vector space \( V \) and returns vectors in that same space.

• Additivity: For any vectors \( u \) and \( v \) in \( V \), \( T(u + v) = T(u) + T(v) \).
• Homogeneity: For any vector \( v \) in \( V \) and scalar \( c \), \( T(cv) = cT(v) \).

Understanding linear operators is fundamental in linear algebra because they encapsulate how transformations work within vector spaces. They preserve the structure of the space, making them key tools in understanding both theoretical and applied aspects of linear systems. For example, they are used to model rotations, reflections, and more complex transformations in various dimensions.
Studying linear operators helps define and solve equations, understanding dimensions, and analyzing the behavior of transformations, crucial for both pure and applied mathematics.

Kernel and Image

When analyzing a linear operator \( T: V \rightarrow V \), two critical subsets are its kernel and image. These are foundational in understanding the behavior of linear transformations.
Kernel: The kernel of \( T \), written as \( \text{ker}(T) \), consists of all vectors \( v \in V \) such that \( T(v) = 0 \). In simpler terms, it's the set of vectors that the operator maps to the zero vector. The kernel tells us about the solutions to the homogeneous equation \( T(v) = 0 \), and the size of the kernel can indicate the amount of freedom or redundancy within the system.
Image: The image of \( T \), written as \( \text{im}(T) \), is the set of all output vectors that \( T \) can produce from vectors in \( V \). It reflects the actual range of the transformation, showing what vectors can be reached or mapped to by the operator from its domain.

• Both kernel and image are subspaces of \( V \).
• Their relationship helps us understand the rank-nullity theorem which links dimensions of a vector space through these subspaces.

In summary, the kernel and image give insight into the characteristics and power of a linear transformation, such as whether it's injective (kernel is trivial) or surjective (image is entire codomain). They are foundational concepts for solving systems of linear equations and more.

Invariant Subspace

An invariant subspace of a vector space under a linear operator is a subset of the vector space that remains unchanged when the operator is applied. In the context of linear algebra, for a subspace \( W \subset V \) to be invariant under a linear operator \( T: V \rightarrow V \), it must satisfy the condition that for every vector \( w \in W \), \( T(w) \) is also in \( W \).
Invariant subspaces are significant since they allow us to "break down" a large transformation into smaller, more manageable parts. This decomposition often leads to simpler computation and deeper understanding of the transformation.

• Example: Eigenvectors, which are vectors that only scale under a linear transformation, often form the basis of invariant subspaces.
• By locating these invariant subspaces, we can diagonalize matrices where possible, simplifying matrix operations and solving differential equations.

In practice, discovering invariant subspaces can simplify theoretical problems and enhance computational efficiency. They provide insights into the operator's structure, allowing mathematicians and engineers to implement more efficient solutions by targeting these subspaces.