KindiakMath

Create Your Own Vector Space

Problem 1. Let V be a vector space over a field \mathbb K. For any set K, suppose there exists a bijection T : K \to V. Prove that K can be equipped with addition and scalar multiplication operations such that K becomes a vector space over \mathbb K.

(Click for Solution)

Solution. The key idea is to define the addition and scalar multiplication operators in V, and then transport them back to K. Define

x + y := T^{-1}(T(x) + T(y)),\quad c \cdot x := T^{-1}(cT(x)).

Since T is bijective, the operations are well-defined. The additive identity will be T^{-1}(\mathbf 0), since

\begin{aligned} x + T^{-1}(\mathbf 0) &= T^{-1}(T(x) + T(T^{-1}(\mathbf 0))) \\ &= T^{-1}(T(x) + \mathbf 0) \\ &= T^{-1}(T(x)) = x.\end{aligned}

For any x \in K, and the additive inverse will be T^{-1}(-T(x)), since

\begin{aligned}x + T^{-1}(-T(x)) &= T^{-1}(T(x) + T(T^{-1}(-T(x)))\\ &= T^{-1}(T(x) + (-T(x)) \\ &= T^{-1}(\mathbf 0). \end{aligned}

We leave it as an exercise to verify the rest of the vector space axioms.

Remark. These definitions turn T into a linear transformation from K to V.

Problem 2. Let V be a vector space over \mathbb K. Prove the following properties:

  • Suppose \mathbf u \in V has the following property: For any \mathbf v, \mathbf v + \mathbf u = \mathbf v. Then \mathbf u = \mathbf 0.
  • Fix \mathbf v \in V. If \mathbf u \in V has the property that \mathbf v + \mathbf u = \mathbf 0, then \mathbf u = -\mathbf v.
  • For any \mathbf v \in V, 0\mathbf v = \mathbf 0.
  • For any \mathbf v \in V, (-1)\mathbf v = -\mathbf v.
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Solution. For the first property, consider \mathbf v = \mathbf 0. Then

\mathbf u = \mathbf 0 + \mathbf u = \mathbf 0.

For the second property, adding -\mathbf v on both sides yields

\begin{aligned}-\mathbf v + (\mathbf v + \mathbf u) &= -\mathbf v + \mathbf 0 \\ (-\mathbf v + \mathbf v) + \mathbf u &= -\mathbf v \\ \mathbf 0 + \mathbf u &= -\mathbf v \\ \mathbf u &= -\mathbf v,\end{aligned}

where each equation implies the next. For the third property, use the property 0 + 0 = 0 to obtain

\begin{aligned}(0 + 0) \mathbf v &= 0\mathbf v \\ 0 \mathbf v + 0 \mathbf v &= 0 \mathbf v \\ -0 \mathbf v + (0 \mathbf v + 0 \mathbf v ) &= -0 \mathbf v + 0 \mathbf v \\ (-0 \mathbf v + 0 \mathbf v) + 0 \mathbf v &= -0 \mathbf v + 0 \mathbf v \\ \mathbf 0 + 0 \mathbf v &= \mathbf 0 \\ 0 \mathbf v &= \mathbf 0,\end{aligned}

where each equation implies the next. For the fourth property, use the property 1 + (-1) = 0 to obtain

\begin{aligned} \mathbf 0 &= 0 \mathbf v \\ \mathbf 0 &= (1 + (-1))\mathbf v \\ \mathbf 0 &= \mathbf v + (-1)\mathbf v \\ -\mathbf v + \mathbf 0 &= -\mathbf v + (\mathbf v + (-1)\mathbf v ) \\ -\mathbf v + \mathbf 0 &= (-\mathbf v + \mathbf v) + (-1)\mathbf v \\ -\mathbf v &= \mathbf 0 + (-1)\mathbf v \\ -\mathbf v &= (-1)\mathbf v,\end{aligned}

where each equation implies the next.

—Joel Kindiak, 27 Feb 25, 2355H

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