R is not a Complex Linear Space

In mathematics an object is a set of properties that define it. In Linear Algebra a set of elements make a vector space if they satisfy a certain set of properties.

Now one of those properties is a scalar multiplication. Mind you that what is considered a vector or scalar depends on the context. In one context \(a+ib\) is a scalar and in another context it is a vector.

Consider the set of real numbers \(\mathbb{R}\) and a set of complex numbers as \(\mathbb{C}\). Then the following holds:

1. \(\mathbb{C}\) is a real linear space over \(\mathbb{R}\).
2. \(\mathbb{C}\) is a complex linear space over \(\mathbb{C}\).
3. \(\mathbb{R}\) is a real linear space over \(\mathbb{R}\).
4. \(\mathbb{R}\) is NOT a complex linear space over \(\mathbb{C}\).

Let’s see each case one by one, specifically why \(\mathbb{R}\) is NOT a complex linear space over \(\mathbb{C}\).

One of the properties of a linear space is that it should be closed under scalar multiplication. That is, if \(\vec{x}\) is a vector in the space and \(a\) is a scalar then \(a\vec{x}\) should also be in the same space.

That is, for \(\mathbb{R}\) to qualify for real or complex linear space the following should hold :

$$a\vec{x}\in\mathbb{R}$$

Similarly for \(\mathbb{C}\) to qualify for real or complex linear space $$a\vec{x}\in\mathbb{C}$$

Both the above conditions assume that the scalar part in the multiplication comes from \(\mathbb{R}\) or \(\mathbb{C}\).

That is scalar \(a \in \mathbb{R}\) or \(a \in \mathbb{C}\).

Now lets see what happens in the cases listed above :

1. \(\mathbb{C}\) is a real linear space over \(\mathbb{R}\) :
   • \(\mathbb{C}\) is a set of complex numbers.
   • \(\mathbb{R}\) is a set of real numbers.
   • If \(\vec{x}\in\mathbb{C}\) and \(a\in\mathbb{R}\) then \(a\vec{x}\in\mathbb{C}\) because \(a\vec{x}\) is a complex number and the operation is closed.
   • Example : \(5 * (4 + 5i) = 20 + 25i \in\mathbb{C}\)
2. \(\mathbb{C}\) is a complex linear space over \(\mathbb{C}\) :
   • If \(\vec{x}\in\mathbb{C}\) and \(a\in\mathbb{C}\) then \(a\vec{x}\in\mathbb{C}\) because \(a\vec{x}\) is a complex number and the operation is closed.
   • Example : \((4 + 5i) * (4 + 5i) = -9 + 40i \in\mathbb{C}\)
3. \(\mathbb{R}\) is a real linear space over \(\mathbb{R}\) :
   • If \(\vec{x}\in\mathbb{R}\) and \(a\in\mathbb{R}\) then \(a\vec{x}\in\mathbb{R}\) because \(a\vec{x}\) is a real number and the operation is closed.
   • Example : \(5 * 4 = 20 \in\mathbb{R}\)
4. \(\mathbb{R}\) is NOT a complex linear space over \(\mathbb{C}\) :
   • If \(\vec{x}\in\mathbb{R}\) and \(a\in\mathbb{C}\) then \(a\vec{x}\in\mathbb{C}\) and \(a\vec{x}\notin\mathbb{R}\) because \(a\vec{x}\) is a complex number and the operation is not closed.
   • Example : \((4 + 5i) * 5 = 20 + 25i \notin\mathbb{R}\)

• Note : The first part of the mutiplication is a scalar and the second part a vector.

In the last case, the operation is not closed because the result of the operation is not in the same space as the operands. This is why \(\mathbb{R}\) is not a complex linear space over \(\mathbb{C}\). The scalar multiplication(where the scalar comes from \(\mathbb{C}\) and vector comes from \(\mathbb{R}\)) result can go out of \(\mathbb{R}\).