MA25C02 | Linear Algebra Example 7: Proving 2x2 Matrices \[\begin{bmatrix} a & a+b \\ a+b & b \end{bmatrix}, a, b \in \mathbb{R}\] as Vector Spaces! 📚✨📐 (Reg 2025)

Example7: Vector Space Verification

▶ Tutorial Video: Vector Spaces-Example 7

Problem: Determine whether the set of all \(2 \times 2\) matrices of the form \[\begin{bmatrix} a & a+b \\ a+b & b \end{bmatrix}, a, b \in \mathbb{R}\] with respect to standard matrix addition and scalar multiplication is a vector space or not. If not, list all the axioms that fail to hold.

Let \(W = \left\{ \begin{bmatrix} a & a+b \\ a+b & b \end{bmatrix} \;\middle|\; a, b \in \mathbb{R} \right\}\) denote the set of matrices. Let:

\[A = \begin{bmatrix} a_1 & a_1+b_1 \\ a_1+b_1 & b_1 \end{bmatrix}, \quad B = \begin{bmatrix} a_2 & a_2+b_2 \\ a_2+b_2 & b_2 \end{bmatrix}, \quad C = \begin{bmatrix} a_3 & a_3+b_3 \\ a_3+b_3 & b_3 \end{bmatrix}\]

Let \(k, m \in \mathbb{R}\) be arbitrary scalars.

Axiom Checks

1. 1. Closure under Addition:
   \(A + B = \begin{bmatrix} a_1 + a_2 & (a_1+a_2) + (b_1+b_2) \\ (a_1+a_2) + (b_1+b_2) & b_1 + b_2 \end{bmatrix}\).
   Let \(a' = a_1 + a_2\) and \(b' = b_1 + b_2\). Then \(A + B \in W\). ✅


2. 2. Commutativity:
   \(A + B = B + A\) follows from real number addition properties. ✅


3. 3. Associativity of Addition:
   \((A + B) + C = A + (B + C)\) holds. ✅


4. 4. Additive Identity:
   The zero matrix \(\mathbf{0} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\) belongs to \(W\) (where \(a=0, b=0\)). ✅


5. 5. Additive Inverse:
   For every \(A \in W\), \(-A = \begin{bmatrix} -a & -(a+b) \\ -(a+b) & -b \end{bmatrix}\) is also in \(W\). ✅


6. 6. Closure under Scalar Multiplication:
   \(kA = \begin{bmatrix} ka_1 & k(a_1+b_1) \\ k(a_1+b_1) & kb_1 \end{bmatrix}\).
   Since \(ka_1 + kb_1 = k(a_1+b_1)\), \(kA \in W\). ✅


7. 7. Distributivity (Scalar over Vector):
   We need to show \(k(A + B) = kA + kB\).
   Left Hand Side: \[k(A+B) = k \begin{bmatrix} a_1+a_2 & (a_1+a_2)+(b_1+b_2) \\ (a_1+a_2)+(b_1+b_2) & b_1+b_2 \end{bmatrix}\] \[= \begin{bmatrix} k(a_1+a_2) & k(a_1+a_2+b_1+b_2) \\ k(a_1+a_2+b_1+b_2) & k(b_1+b_2) \end{bmatrix}\]
   Right Hand Side: \[kA + kB = \begin{bmatrix} ka_1 & k(a_1+b_1) \\ k(a_1+b_1) & kb_1 \end{bmatrix} + \begin{bmatrix} ka_2 & k(a_2+b_2) \\ k(a_2+b_2) & kb_2 \end{bmatrix}\] \[= \begin{bmatrix} ka_1+ka_2 & k(a_1+b_1)+k(a_2+b_2) \\ k(a_1+b_1)+k(a_2+b_2) & kb_1+kb_2 \end{bmatrix}\] Since \(k(a_1+a_2) = ka_1+ka_2\), both sides are equal. ✅


8. 8. Distributivity (Vector over Scalar):
   We need to show \((k + m)A = kA + mA\). \[(k+m)A = \begin{bmatrix} (k+m)a_1 & (k+m)(a_1+b_1) \\ (k+m)(a_1+b_1) & (k+m)b_1 \end{bmatrix}\] By distributive law of real numbers: \[= \begin{bmatrix} ka_1+ma_1 & k(a_1+b_1)+m(a_1+b_1) \\ k(a_1+b_1)+m(a_1+b_1) & kb_1+mb_1 \end{bmatrix}\] \[= \begin{bmatrix} ka_1 & k(a_1+b_1) \\ k(a_1+b_1) & kb_1 \end{bmatrix} + \begin{bmatrix} ma_1 & m(a_1+b_1) \\ m(a_1+b_1) & mb_1 \end{bmatrix} = kA + mA\] ✅


9. 9. Compatibility of Scalars:
   We show \(k(mA) = (km)A\). \[k(mA) = k \begin{bmatrix} ma_1 & m(a_1+b_1) \\ m(a_1+b_1) & mb_1 \end{bmatrix}\] \[= \begin{bmatrix} k(ma_1) & k(m(a_1+b_1)) \\ k(m(a_1+b_1)) & k(mb_1) \end{bmatrix}\] By associativity of multiplication in \(\mathbb{R}\): \[= \begin{bmatrix} (km)a_1 & (km)(a_1+b_1) \\ (km)(a_1+b_1) & (km)b_1 \end{bmatrix} = (km)A\] ✅


10. 10. Identity of Scalar Multiplication:
    Let \(k = 1\). \[1 \cdot A = \begin{bmatrix} 1(a_1) & 1(a_1+b_1) \\ 1(a_1+b_1) & 1(b_1) \end{bmatrix} = \begin{bmatrix} a_1 & a_1+b_1 \\ a_1+b_1 & b_1 \end{bmatrix} = A\] ✅

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Conclusion

All ten axioms are satisfied.
Therefore, the set \(W\) is a vector space.

Note: \(W\) is a 2-dimensional subspace of \(M_{2\times 2}(\mathbb{R})\) spanned by: \[W = \operatorname{span}\left\{ \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \right\}\]