Linear Algebra: Vector Space Example 1 📚✨💡 | (x,y)+(p,q) = (x+p+1, y+q+1), k(x,y)=(kx,ky)

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Problem: Vector Space Verification

Determine whether the set of all pairs of real numbers \((x, y)\) with the operations below is a vector space:

• Addition: \((x, y) + (p, q) = (x + p + 1, y + q + 1)\)
• Scalar Multiplication: \(k(x, y) = (kx, ky)\)

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Step-by-Step Solution

I. Closure under Addition

For any \((x,y), (p,q) \in V\), the sum is \((x+p+1, y+q+1)\). Since these are real numbers, the result is in \(V\). ✔ Holds

II. Commutativity

\[ (x, y) + (p, q) = (x + p + 1, y + q + 1) = (p, q) + (x, y) \] ✔ Holds

III. Associativity

Checking \(((u+v)+w)\) vs \((u+(v+w))\): Both result in \((x+p+r+2, y+q+s+2)\). ✔ Holds

IV. Additive Identity

Setting \((x+a+1, y+b+1) = (x, y)\) gives the identity \((-1, -1)\). ✔ Holds

V. Additive Inverse

For any \((x,y)\), the inverse is \((-x-2, -y-2)\). ✔ Holds

VI. Closure under Scalar Multiplication

\(k(x, y) = (kx, ky)\) is always a pair of real numbers. ✔ Holds

VII. Distributivity (Vector Addition)

\(k((x, y) + (p, q)) = (kx + kp + k, ky + kq + k)\)
\(k(x, y) + k(p, q) = (kx + kp + 1, ky + kq + 1)\)
✘ Fails: The terms \(k\) and \(1\) are not equal for all \(k\).

VIII. Distributivity (Scalar Addition)

\((k + \ell)(x, y) = ((k+\ell)x, (k+\ell)y)\)
\(k(x, y) + \ell(x, y) = (kx + \ell x + 1, ky + \ell y + 1)\)
✘ Fails: The results differ by the constant \(1\).

Conclusion: Since the distributive laws fail to hold, the set \(V\) with the defined operations is NOT a vector space.