Question

Let \(T: V \rightarrow V\) be an operator satisfying \(T^{2}=c T, c \neq 0\). a. Show that \(V=U \oplus\) ker \(T\), where \(U=\\{\mathbf{u} \mid T(\mathbf{u})=c \mathbf{u}\\}\) [Hint: Compute \(\left.T\left(\mathbf{v}-\frac{1}{c} T(\mathbf{v})\right) .\right]\) b. If \(\operatorname{dim} V=n\), show that \(V\) has a basis \(B\) such that \(M_{B}(T)=\left[\begin{array}{rr}c I_{r} & 0 \\ 0 & 0\end{array}\right],\) where \(r=\operatorname{rank} T\) c. If \(A\) is any \(n \times n\) matrix of rank \(r\) such that \(A^{2}=c A, c \neq 0,\) show that \(A\) is similar to \(\left[\begin{array}{rr}c I_{r} & 0 \\ 0 & 0\end{array}\right]\)

Short answer

The operator splits as \( V = U \oplus \ker T \), where \( M_B(T) = \begin{bmatrix} cI_r & 0 \\ 0 & 0 \end{bmatrix} \), and any matrix satisfying similar properties is similar to this form.

Step-by-step solution

Understanding the given operator

We have an operator \( T: V \to V \) such that \( T^2 = cT \) and \( c eq 0 \). This implies for every vector \( \mathbf{v} \in V \), \( T(T(\mathbf{v})) = cT(\mathbf{v}) \). This will help in finding the structure of the space \( V \).

Show V = U ⊕ ker T

Consider \( \mathbf{v} - \frac{1}{c}T(\mathbf{v}) \). Applying T, we have \( T(\mathbf{v}) - \frac{1}{c}T^2(\mathbf{v}) = T(\mathbf{v}) - \frac{1}{c}cT(\mathbf{v}) = 0 \), implying \( \mathbf{v} - \frac{1}{c}T(\mathbf{v}) \in \ker T \). Thus any \( \mathbf{v} \) can be expressed as \( (\mathbf{v} - \frac{1}{c}T(\mathbf{v})) + \frac{1}{c}T(\mathbf{v}) \). \( \frac{1}{c}T(\mathbf{v}) \in U \), showing \( V = U \oplus \ker T \).

Setup for finding a basis for V

Since \( V = U \oplus \ker T \), choose a basis for \( U \) first, say \( \{\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_r\} \), where \( T(\mathbf{u}_i) = c\mathbf{u}_i \). Similarly, choose a basis for \( \ker T \), say \( \{\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_s\} \). Note that \( r + s = n \).

Construct the basis B for V and find MB(T)

The basis \( B = \{\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_r, \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_s\} \) is a basis for \( V \). With respect to this basis, the matrix \( M_B(T) \) is \( \begin{bmatrix} cI_r & 0 \ 0 & 0 \end{bmatrix} \) as for basis vectors in \( U \), they are mapped as \( T(\mathbf{u}_i) = c\mathbf{u}_i \), while for those in \( \ker T \), \( T(\mathbf{v}_i) = 0 \).

Show similarity of A with \([cI_r, 0; 0, 0] \)

Given \( A^2 = cA \) for an \( n \times n \) matrix \( A \) of rank \( r \), similar reasoning as the earlier parts shows \( \mathbb{R}^n = \ker A \oplus U_A \), where \( U_A = \{ \mathbf{x} \mid A \mathbf{x} = c\mathbf{x} \} \). Choosing the basis w.r.t these subspaces, the matrix \( A \) will have the form \( \begin{bmatrix} cI_r & 0 \ 0 & 0 \end{bmatrix} \) in a similar way as \( T \), establishing similarity.

Key concepts

Vector Spaces

A vector space is a collection of objects called vectors, which can be added together and multiplied by scalars (numbers), and still satisfy certain basic properties. Imagine vectors as arrows that have both direction and magnitude. In linear algebra, vector spaces are foundational because they provide a framework for solving systems of linear equations.

Here are a few essential properties of vector spaces:

• **Additivity:** You can add two vectors to get another vector.
• **Scalar multiplication:** A vector can be multiplied by a scalar, resulting in a different vector.
• **Zero vector:** There is a special vector called the zero vector, which acts as an additive identity.
• **Inverse:** Every vector has an opposite vector (additive inverse) which cancels out its effect.

Considering vectors as elements of the space 'V', if we have a linear operator 'T' on 'V', it transforms vectors within this space while maintaining the vector space properties. When applying 'T', we can explore unique subspaces like the **kernel** of 'T', which is the set of vectors that the transformation 'T' sends to the zero vector.

Linear Transformations

Linear transformations are mappings between vector spaces that preserve vector addition and scalar multiplication. They are crucial for understanding how spaces can be manipulated and transformed while preserving their structure.

Consider a linear transformation defined as a function \(T: V \rightarrow V\). For linear transformations, the following properties hold:

• **Additivity:** \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \)
• **Homogeneity:** \( T(c\mathbf{v}) = cT(\mathbf{v}) \) for any scalar \(c\).

In the exercise, the function \(T\) squares back to itself when multiplied by \(c\), i.e., \(T^{2} = cT\). This unique property allows us to split the vector space \(V\) into two distinct parts - \(U\) and \(\ker T\). The first part, \(U\), contains vectors that scale by \(c\) under \(T\), and the second, the kernel, contains vectors sent to zero by \(T\). These transformations help simplify complex vector operations and are visualized with matrices.

Matrix Similarity

Matrix similarity is a concept that shows how two matrices can represent the same linear transformation, albeit in different bases. When matrices are similar, there exists an invertible matrix \(P\) such that \(P^{-1}AP = B\), where \(A\) and \(B\) are the similar matrices.

This concept is particularly useful because similar matrices share a lot of fundamental properties:

• **The same eigenvalues:** An eigenvalue is a scalar \(\lambda\) such that there exists a non-zero vector \(\mathbf{v}\) where \(A\mathbf{v} = \lambda\mathbf{v}\).
• **Rank:** The number of linearly independent rows (or columns) is the same.
• **Determinant and trace:** Both matrices have the same determinant and trace.

In our exercise, given an operator \(T\) or a matrix \(A\) that satisfies \(T^{2} = cT\) or \(A^{2} = cA\), the transformation is greatly simplified. We find that \(A\) is similar to a matrix made of blocks \([cI_{r}, 0; 0, 0]\), telling us that \(A\) preserves the dimensions and crucial features while being expressed in a basis that highlights its transformation properties more clearly.