Question

Consider $$ V=\left\\{\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \mid a+c=b+d\right\\} $$ a. Consider \(S: \mathbf{M}_{22} \rightarrow \mathbb{R}\) with \(S\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]=a+c-\) \(b-d\). Show that \(S\) is linear and onto and that \(V\) is a subspace of \(\mathbf{M}_{22}\). Compute \(\operatorname{dim} V\). b. Consider \(T: V \rightarrow \mathbb{R}\) with \(T\left[\begin{array}{ll}a & b \\\ c & d\end{array}\right]=a+c\). Show that \(T\) is linear and onto, and use this information to compute \(\operatorname{dim}(\) ker \(T)\).

Short answer

dim(V) = 3, dim(ker(T)) = 2.

Step-by-step solution

Understanding the Set V

The set \( V \) consists of all \( 2 \times 2 \) matrices \( \left[\begin{array}{cc} a & b \ c & d \end{array}\right] \) such that \( a + c = b + d \). This condition means that the sum of the entries in the first column equals the sum of the entries in the second column.

Showing S is Linear

A map \( S : \mathbf{M}_{22} \rightarrow \mathbb{R} \) defined by \( S \left[\begin{array}{cc} a & b \ c & d \end{array}\right] = a+c-b+d \) is linear if for any matrices \( A, B \) and scalar \( k \), \( S(A+B) = S(A) + S(B) \) and \( S(kA) = kS(A) \). Substituting, we get:- \( S \left( A + B \right) = (a_1+b_1) + (c_1+d_1) - (b_1+d_1) \) which simplifies to \( S(A) + S(B) \).- \( S(kA) = k(a+c-b+d) = kS(A) \).Hence, \( S \) is linear.

Showing S is Onto

The map \( S \) is onto if for every real number \( y \), there is a matrix \( A \in \mathbf{M}_{22} \) such that \( S(A) = y \). Choose \( A = \left[ \begin{array}{cc} y & 0 \ 0 & 0 \end{array} \right] \). Then \( S(A) = y - 0 + 0 - 0 = y \). Hence, \( S \) is onto.

Showing V is a Subspace

A set \( V \) is a subspace of \( \mathbf{M}_{22} \) if it is closed under addition and scalar multiplication, and contains the zero vector. For any matrices \( A = \left[ \begin{array}{cc} a & b \ c & d \end{array} \right] \) and \( B = \left[ \begin{array}{cc} e & f \ g & h \end{array} \right] \) in \( V \), if \( a+c = b+d \) and \( e+g = f+h \), then \( (a+e) + (c+g) = (b+f) + (d+h) \). Thus, \( A + B \in V \). For any scalar \( k \), \( kA = \left[ \begin{array}{cc} ka & kb \ kc & kd \end{array} \right] \) satisfies \( ka + kc = kb + kd \), so \( kA \in V \). Thus, \( V \) is a subspace.

Calculating \( \operatorname{dim} V \)

Since \( S \) is onto, by the rank-nullity theorem, \( \operatorname{dim} V = \text{rank}(S) \). The kernel of \( S \) consists of all matrices \( A \) such that \( a+c=b+d \), hence dim(ker(S)) = 1. With \( \operatorname{dim} \mathbf{M}_{22} = 4 \), we have \( \operatorname{dim} V = 4 - 1 = 3 \).

Analyzing T

The map \( T: V \rightarrow \mathbb{R} \) defined by \( T \left[ \begin{array}{cc} a & b \ c & d \end{array} \right] = a+c \) is linear if \( T(A+B) = T(A) + T(B) \) and \( T(kA) = kT(A) \). Both conditions follow similarly to S as the map is defined by summing direct matrix elements.

Showing T is Onto

The map \( T \) is onto because for any real \( y \), choose the matrix \( \left[ \begin{array}{cc} y/2 & 0 \ y/2 & 0 \end{array} \right] \). Then \( T(\text{Matrix}) = y/2 + y/2 = y \). Hence, \( T \) is onto.

Calculating \( \operatorname{dim} \ker(T) \)

The dimension of the kernel is the difference between the dimension of \( V \) and the rank of \( T \), which is 1 (since T is onto to \( \mathbb{R} \), it spans a 1-dimensional range). Thus, \( \operatorname{dim} \ker(T) = 3 - 1 = 2 \).

Key concepts

Matrix Subspaces

Matrix subspaces are an essential topic in linear algebra. They extend the concept of vector spaces to matrices. A subspace is a subset that is closed under addition and scalar multiplication and includes the zero vector. In other words, if you add any two matrices from the subspace or multiply them by a scalar, you'll still have a matrix in the subspace. This ensures the subspace maintains its structure without "stepping outside" its boundaries.
V in our exercise, is such a subspace in the matrix space \( \mathbf{M}_{22} \). The defining condition \( a+c = b+d \) means any combination of these elements keeping this property will remain in the subspace. This property guarantees V is legitimate as a subspace under standard operations.
Understanding matrix subspaces provides a foundation for complex algebraic structures and transformations. It allows us to analyze various properties of matrices by considering smaller, controlled subsets.

Linear Transformations

Linear transformations are functions that map one vector space to another while keeping operations of addition and scalar multiplication intact. If a function \( T \) is linear, then for any matrices \( A \) and \( B \), and scalar \( k \), the transformation satisfies:

• \( T(A+B) = T(A) + T(B) \)
• \( T(kA) = kT(A) \)

This exercise uses two linear transformations, \( S \) and \( T \), to map matrices to the real numbers. Both demonstrate the core idea of linearity by showing how operations within matrices translate directly to operations in the reals.
Specifically, transformation \( S \) checks by eliminating specific elements to produce a scalar outcome. Similarly, \( T \) collects values from a matrix and adds them based on their specific location criteria. Linear transformations like these are pivotal as they simplify complex matrix structures while preserving core properties.

Rank-Nullity Theorem

The rank-nullity theorem is a fundamental concept in linear algebra that relates the dimensions of different subspaces involved with a linear map. It connects the rank and nullity of a linear transformation \( T: V \rightarrow W \) to the dimension of the original space \( V \).
The theorem states:
\[ \text{dim}(V) = \text{rank}(T) + \text{nullity}(T) \]
Here, rank is the dimension of the image of \( T \) (all output results possible through \( T \)), while nullity is the dimension of the kernel (the set of all inputs that map to zero).
In this problem, we see these concepts applied: \( \operatorname{dim} V \) combines the insight from \( S \) being onto, and the calculation of the kernel, demonstrating how Solving for these dimensions confirms the space's structure and transformation abilities.
Understanding rank, nullity, and how they relate brings clarity about the linear map's efficiency and the capacity to span subspaces within a broader vector space.