Question

Let \(B=\left\\{\mathbf{f}_{1}, \mathbf{f}_{2}, \ldots, \mathbf{f}_{n}\right\\}\) be an or- thonormal basis of an inner product space \(V\). Given \(T: V \rightarrow V,\) define \(T^{\prime}: V \rightarrow V\) by $$ \begin{aligned} T^{\prime}(\mathbf{v}) &=\left\langle\mathbf{v}, T\left(\mathbf{f}_{1}\right)\right\rangle \mathbf{f}_{1}+\left\langle\mathbf{v}, T\left(\mathbf{f}_{2}\right)\right\rangle \mathbf{f}_{2}+\cdots+\left\langle\mathbf{v}, T\left(\mathbf{f}_{n}\right)\right\rangle \mathbf{f}_{n} \\ &=\sum_{i=1}^{n}\left\langle\mathbf{v}, T\left(\mathbf{f}_{i}\right)\right\rangle \mathbf{f}_{i} \end{aligned} $$ a. Show that \((a T)^{\prime}=a T^{\prime}\). b. Show that \((S+T)^{\prime}=S^{\prime}+T^{\prime}\). c. Show that \(M_{B}\left(T^{\prime}\right)\) is the transpose of \(M_{B}(T)\). d. Show that \(\left(T^{\prime}\right)^{\prime}=T,\) using part (c). [Hint: \(M_{B}(S)=M_{B}(T)\) implies that \(\left.S=T .\right]\) e. Show that \((S T)^{\prime}=T^{\prime} S^{\prime}\), using part (c). f. Show that \(T\) is symmetric if and only if \(T=T^{\prime} .\) [Hint: Use the expansion theorem and Theorem 10.3.3.] \(\mathrm{g}\). Show that \(T+T^{\prime}\) and \(T T^{\prime}\) are symmetric, using parts (b) through (e). h. Show that \(T^{\prime}(\mathbf{v})\) is independent of the choice of orthonormal basis \(B\). [Hint: If \(D=\left\\{\mathbf{g}_{1}, \ldots, \mathbf{g}_{n}\right\\}\) is also orthonormal, use the fact that \(\mathbf{f}_{i}=\sum_{j=1}^{n}\left\langle\mathbf{f}_{i}, \mathbf{g}_{j}\right\rangle \mathbf{g}_{j}\) for each \(\left.i .\right]\)

Short answer

The problem demonstrates linear operations and basis independence on an operator transpose.

Step-by-step solution

Prove (a T)' = a T'

To show \((a T)'=a T'\), we start by considering the definition of \(T'\). For any vector \(\mathbf{v} \in V\), \( (a T)'(\mathbf{v}) = \left\langle \mathbf{v}, aT(\mathbf{f}_i) \right\rangle \mathbf{f}_i \). Since scalar multiplication can factor out of the inner product, \( (a T)'(\mathbf{v}) = a \left\langle \mathbf{v}, T(\mathbf{f}_i) \right\rangle \mathbf{f}_i \). Thus, \((a T)' = a T'\) as desired.

Prove (S+T)' = S' + T'

Let \(S'\) and \(T'\) be defined similarly to \(T'\). For the operator \((S+T)'\), we have: \((S+T)'(\mathbf{v}) = \sum_{i=1}^{n} \left\langle \mathbf{v}, (S+T)(\mathbf{f}_i) \right\rangle \mathbf{f}_i\). Using properties of inner products, this splits into \(\sum_{i=1}^{n} (\left\langle \mathbf{v}, S(\mathbf{f}_i) \right\rangle + \left\langle \mathbf{v}, T(\mathbf{f}_i) \right\rangle) \mathbf{f}_i\). Distributing the sum yields \(\sum_{i=1}^{n} \left\langle \mathbf{v}, S(\mathbf{f}_i) \right\rangle \mathbf{f}_i + \sum_{i=1}^{n} \left\langle \mathbf{v}, T(\mathbf{f}_i) \right\rangle \mathbf{f}_i\), which is \(S'(\mathbf{v}) + T'(\mathbf{v})\). Thus, \((S+T)' = S' + T'\).

Demonstrate M_B(T') is Transpose of M_B(T)

To show that \(M_{B}(T')\) is the transpose of \(M_{B}(T)\), we examine the components of these matrices. In basis \(B\), the \((i,j)\) entry of \(M_{B}(T)\) is \(\left\langle \mathbf{f}_i, T(\mathbf{f}_j) \right\rangle\). For \(M_{B}(T')\), the \((j,i)\) entry is \(\left\langle T'(\mathbf{f}_i), \mathbf{f}_j \right\rangle\). By the definition of \(T'\), it follows that \(\left\langle T'(\mathbf{f}_i), \mathbf{f}_j \right\rangle = \left\langle \mathbf{f}_i, T(\mathbf{f}_j) \right\rangle\). This confirms that \(M_{B}(T')\) is the transpose of \(M_{B}(T)\).

Prove (T')' = T

Using the transpose property from Step 3, we know \(M_B((T')') = (M_B(T'))^T = (M_B(T)^T)^T\). This equals \(M_B(T)\) because the double transpose returns to the original matrix. By the hint, if matrices are equal in all bases, operators are equal. Therefore, \((T')' = T\).

Prove (ST)' = T'S'

For \((ST)'\), the \((i,j)\) entry in \(M_B((ST)')\) is \(\left\langle \mathbf{f}_i, ST(\mathbf{f}_j) \right\rangle \), which expands to the sum \(\sum_k \left\langle \mathbf{f}_i, S(\mathbf{f}_k) \right\rangle \left\langle \mathbf{f}_k, T(\mathbf{f}_j) \right\rangle \). The transpose properties developed thus far give us \(\left\langle \mathbf{f}_j, S' T'(\mathbf{f}_i) \right\rangle \), indicating that \((ST)' = T'S'\).

Prove T is symmetric if and only if T = T'

Recall that an operator is symmetric if it equals its adjoint \(T^*\). Since \(T'\) is the transpose in our context, \(T=T'\) implies symmetry. Conversely, symmetry (\(T^*=T\)) implies \(M_B(T') = M_B(T)^T\), which is equivalent by definition to \(M_B(T)\), so \(T=T'\).

Prove T+T' and TT' are symmetric

From \((T + T')' = T' + (T')' = T' + T\), by symmetry, \(T+T'\) is symmetric. For \(TT'\), \((TT')' = (T')'T' = TT''\), which again equals \(TT'\) using Step 4, confirming symmetry.

Show T'(v) is Independent of Orthonormal Basis B

If \(D=\left\{\mathbf{g}_1, \ldots, \mathbf{g}_n\right\}\) is another orthonormal basis, express \(\mathbf{f}_i\) in terms of \(\mathbf{g}_j\) through \(\mathbf{f}_i = \sum_j \left\langle \mathbf{f}_i, \mathbf{g}_j \right\rangle \mathbf{g}_j\). Substituting this into the expression for \(T'(\mathbf{v})\) retains the result due to the properties of orthonormal bases and linearity of the inner product, confirming that \(T'\) is basis-invariant.

Key concepts

Inner Product Space

An inner product space is a vector space equipped with an additional structure called an "inner product." This inner product is a mathematical operation that takes two vectors and returns a scalar, often denoted as \( \langle \mathbf{u}, \mathbf{v} \rangle \) for vectors \( \mathbf{u} \) and \( \mathbf{v} \). The inner product needs to satisfy four main properties:

• Conjugate symmetry: \( \langle \mathbf{u}, \mathbf{v} \rangle = \overline{ \langle \mathbf{v}, \mathbf{u} \rangle } \)
• Linearity in the first argument: \( \langle a\mathbf{u} + b\mathbf{w}, \mathbf{v} \rangle = a \langle \mathbf{u}, \mathbf{v} \rangle + b \langle \mathbf{w}, \mathbf{v} \rangle \)
• Positive-definiteness: \( \langle \mathbf{v}, \mathbf{v} \rangle \geq 0 \) with equality if and only if \( \mathbf{v} = \mathbf{0} \)
• Scalar multiplication compatibility with conjugation: \( \langle \mathbf{u}, a \mathbf{v} \rangle = a \langle \mathbf{u}, \mathbf{v} \rangle \)

These properties allow us to generalize many concepts from Euclidean space, like angles and length, to more abstract vector spaces. Inner product spaces, therefore, provide a foundational structure for understanding geometry and functional analysis in linear algebra.

Orthonormal Basis

An orthonormal basis for a vector space is a basis where all vectors are unit vectors (norm equal to one) and are mutually orthogonal. This means each pair of different basis vectors \( \mathbf{f}_i \) and \( \mathbf{f}_j \) satisfies \( \langle \mathbf{f}_i, \mathbf{f}_j \rangle = 0 \) for \( i eq j \), and \( \langle \mathbf{f}_i, \mathbf{f}_i \rangle = 1 \).

• An orthonormal basis simplifies many mathematical operations, including simplifying the computation of a vector’s representation in the space.
• Any vector \( \mathbf{v} \) in the space can be expressed as a linear combination of the orthonormal basis vectors: \( \mathbf{v} = \sum a_i \mathbf{f}_i \), where \( a_i = \langle \mathbf{v}, \mathbf{f}_i \rangle \).
• Because of the perpendicular nature of basis vectors, projections onto these vectors become straightforward.

Orthonormal bases are particularly important in minimizing computational errors and facilitating simpler numerical calculations in linear algebra problems.

Matrix Transpose

The transpose of a matrix is an operation that flips a matrix over its diagonal. If \( M \) is a matrix, its transpose, denoted as \( M^T \), switches the row and column indices of each element, effectively exchanging rows with columns.

• For a matrix \( A = [a_{ij}] \), the transposed matrix \( A^T \) becomes \( [a_{ji}] \).
• Transpose properties include: \( (A^T)^T = A \), \( (AB)^T = B^T A^T \), and \( (A+B)^T = A^T + B^T \).
• Transposing a matrix changes its orientation but preserves the original data’s relationships.

Transposition is widely used in various linear algebra operations, such as solving systems of equations, and plays a crucial role in defining concepts like symmetric matrices.

Symmetric Operators

A symmetric operator in the context of linear spaces is a linear transformation \( T \) such that for every pair of vectors \( \mathbf{u} \) and \( \mathbf{v} \), \( \langle T(\mathbf{u}), \mathbf{v} \rangle = \langle \mathbf{u}, T(\mathbf{v}) \rangle \). An equivalent condition in terms of matrices is that a matrix is symmetric if it is equal to its transpose, meaning \( A = A^T \).

• Symmetric operators and matrices have real eigenvalues, and their eigenvectors corresponding to distinct eigenvalues are orthogonal.
• They appear in many settings, such as quadratic forms where symmetry is naturally desired.
• In the context of function spaces, symmetric operators often correspond to self-adjoint operators, crucial in quantum mechanics and other applications.

Understanding symmetric operators is important because they simplify many mathematical problems and provide more predictable behavior in applied scenarios.