Problem 4 - A Book of Abstract Algebra

Question

Prove that \(\|_{2}(R)\), the set of all \(2 \times 2\) matrices of real numbers, with matrix addition and the scalar multiplication $$ k\left(\begin{array}{ll} a & b \\ c & d \end{array}\right)=\left(\begin{array}{ll} k a & k b \\ k c & k d \end{array}\right) $$ is a vector space over \(\mathbb{R}\).

Step-by-Step Solution

Answer

The set of \(2 \times 2\) matrices forms a vector space over \(\mathbb{R}\).

Step 1: Check Vector Addition Closure

Ensure that the sum of any two \(2 \times 2\) matrices results in another \(2 \times 2\) matrix.\(\begin{pmatrix}a & b \ c & d\end{pmatrix} + \begin{pmatrix}e & f \ g & h\end{pmatrix} = \begin{pmatrix}a+e & b+f \ c+g & d+h\end{pmatrix}\). Since the result is also a \(2 \times 2\) matrix, the vector addition is closed.

Step 2: Verify Associative Property of Addition

Check if matrix addition is associative.\((U + V) + W = U + (V + W)\), for matrices \(U = \begin{pmatrix}a & b \ c & d\end{pmatrix}\), \(V = \begin{pmatrix}e & f \ g & h\end{pmatrix}\), and \(W = \begin{pmatrix}p & q \ r & s\end{pmatrix}\). Compute both sides to ensure equivalence.

Step 3: Check Commutative Property of Addition

Confirm that matrix addition is commutative.\(U + V = V + U\). Check for matrices \(U = \begin{pmatrix}a & b \ c & d\end{pmatrix}\) and \(V = \begin{pmatrix}e & f \ g & h\end{pmatrix}\) that \(\begin{pmatrix}a+e & b+f \ c+g & d+h\end{pmatrix} = \begin{pmatrix}e+a & f+b \ g+c & h+d\end{pmatrix}\). This condition holds true.

Step 4: Identity Element for Addition

Identify the zero matrix as the additive identity. Check that \(U + 0 = U\) for any \(U = \begin{pmatrix}a & b \ c & d\end{pmatrix}\), where \(0\) is \(\begin{pmatrix}0 & 0 \ 0 & 0\end{pmatrix}\). The addition results in \(U\), proving there is an identity matrix.

Step 5: Inverse Elements for Addition

For every matrix \(U\), \(U + (-U) = 0\). The inverse matrix \(-U\) is \(\begin{pmatrix}-a & -b \ -c & -d\end{pmatrix}\). Verify that \(\begin{pmatrix}a & b \ c & d\end{pmatrix} + \begin{pmatrix}-a & -b \ -c & -d\end{pmatrix} = \begin{pmatrix}0 & 0 \ 0 & 0\end{pmatrix}\).

Step 6: Scalar Multiplication Closure

Ensure scalar multiplication results in a matrix in the space. If \(k \in \mathbb{R}\) and \(U = \begin{pmatrix}a & b \ c & d\end{pmatrix}\), then \(kU = \begin{pmatrix}ka & kb \ kc & kd\end{pmatrix}\) is also a \(2 \times 2\) matrix.

Step 7: Associative Property of Scalar Multiplication

Check that \((kl)U = k(lU)\) for scalars \(k, l\) and matrix \(U\). Simplify both \((kl)U = \begin{pmatrix}(kl)a & (kl)b \ (kl)c & (kl)d\end{pmatrix}\) and \(k(lU) = \begin{pmatrix}k(la) & k(lb) \ k(lc) & k(ld)\end{pmatrix}\). These are equivalent.

Step 8: Identity Scalar Multiplication

Verify scalar identity \(1 \cdot U = U\). For \(U\), \(1 \cdot \begin{pmatrix}a & b \ c & d\end{pmatrix} = \begin{pmatrix}a & b \ c & d\end{pmatrix}\), confirming the identity.

Step 9: Distributive Property of Scalar over Matrix Addition

Ensure \(k(U + V) = kU + kV\). Compute both sides for \(U + V = \begin{pmatrix}a+e & b+f \ c+g & d+h\end{pmatrix}\), and verify the scalar distribution.

Step 10: Distributive Property of Scalars over Addition

Prove \((k+l)U = kU + lU\) for \(k, l\) and \(U\). Simplify \((k+l)U = \begin{pmatrix}(k+l)a & (k+l)b \ (k+l)c & (k+l)d\end{pmatrix}\) and verify it equals \(kU + lU\).

Key Concepts

Matrix Addition

Matrix addition is a fundamental concept when working with matrices. It involves adding two matrices together, which results in another matrix of the same dimensions. For two matrices to be added, they must be of the same size. For instance, adding two \(2 \times 2\) matrices involves adding corresponding elements together:

\[\begin{pmatrix} a & b \ c & d \end{pmatrix} + \begin{pmatrix} e & f \ g & h \end{pmatrix} = \begin{pmatrix} a+e & b+f \ c+g & d+h \end{pmatrix}\]

This concept is similar to adding numbers, but we do so element-wise. It's important in verifying if a set forms a vector space, as it needs to show closure under addition. Here, the addition of two \(2 \times 2\) matrices results in another \(2 \times 2\) matrix, thus demonstrating closure.

Scalar Multiplication

Scalar multiplication refers to multiplying a matrix by a scalar (a real number). This operation involves multiplying each element of the matrix by the scalar. For example, given a scalar \(k\) and a \(2 \times 2\) matrix, scalar multiplication is represented as:

\[k \times \begin{pmatrix} a & b \ c & d \end{pmatrix} = \begin{pmatrix} ka & kb \ kc & kd \end{pmatrix}\]

This multiplication is straightforward but essential when determining if a set forms a vector space. The result of scalar multiplication is still a \(2 \times 2\) matrix, thereby showing closure under this operation. It's key in vector spaces as it reflects how matrices behave under transformations influenced by real numbers.

Real Numbers

Real numbers play a crucial role in the concept of vector spaces when considering matrices. They serve as scalars in scalar multiplication and also ensure that all operations remain within the real number system. A real number is any positive or negative number or zero that can be found on the number line.

Key characteristics include:

• Real numbers include rational numbers (like fractions) and irrational numbers (numbers that cannot be expressed as fractions).
• They are used in scalar multiplication to stretch or shrink the matrix along its dimensions.

Ensuring that all elements of a matrix, as well as results of operations such as matrix addition and scalar multiplication, remain within the real number domain is essential for maintaining the integrity of vector spaces over \(\mathbb{R}\).

Associative Property

The associative property is an important characteristic in mathematics and also relevant to operations within vector spaces. It states that how you group numbers or expressions in addition or multiplication does not change their sum or product. In the context of matrix addition, it involves:

\[(U + V) + W = U + (V + W)\]

Where \(U, V,\) and \(W\) are matrices of the same size. This property ensures flexibility in computation, enabling complex calculations to be broken down more easily without affecting the outcome.

Similarly, in scalar multiplication, the property appears as:

\[(kl)U = k(lU)\]

This means multiplying a matrix by a product of scalars \(kl\) gives the same result as multiplying each scalar in sequence. The associative property assures us that the overall structure of operations remains consistent, which is crucial for vector spaces to function under these mathematical laws.