Link to Problem: HERE

In \(\mathbb{C}^3\), the vector space of column vectors of size 3, prove that the set \(Z\) is a subspace.

\[ Z = \Bigg\{ \left[\begin{array} {rrr} x_1 \\ x_2 \\ x_3 \\ \end{array}\right] \Bigg \vert 4x_1 - x_2 + 5x_3 = 0 \Bigg\} \]

As per Theorem TSS: Testing Subsets for Subspaces, a vector \(W\) is a subspace if \(W\) is a subset of \(V\) endowed with the same operations as \(V\) and has the following 3 properties:

  1. \(W\) is nonempty, \(W \neq \emptyset\).
  2. If \(x \epsilon W\) and \(y \epsilon W\), then \(x + y \epsilon W\).
  3. If \(\alpha \epsilon \mathbb{C}\) and \(x \epsilon W\), then \(\alpha x \epsilon W\).

I will prove each one of these, demonstrating that \(Z\) is a subspace.

  1. \(W\) is nonempty, \(W \neq \emptyset\).

If \(x_1 = -1\), \(x_2 = 1\), and \(x_3 = 1\) then \(4x_1 -1x_2 + 5x_3 = 4(1) - 1(1) + 5(1) = 0\).

\(\left[\begin{array} {rrr} -1 \\ 1 \\ 1 \\ \end{array}\right]\) is a member of \(Z\) therefor \(Z\) is not empty.

  1. If \(x \epsilon W\) and \(y \epsilon W\), then \(x + y \epsilon W\).

We must prove that \(4x_1 -1x_2 + 5x_3 + 4y_1 -1y_2 + 5y_3 = 0\).

Since x and y are both members of the space, we know that \(4x_1 -1x_2 + 5x_3 = 0\) and \(4x_1 -1x_2 + 5x_3 = 0\) and thus

\((4x_1 -1x_2 + 5x_3) + (4y_1 -1y_2 + 5y_3) = 0\)

\(0 + 0 = 0\)

This statement is true therefore \(x + y \epsilon Z\)

  1. If \(\alpha \epsilon \mathbb{C}\) and \(x \epsilon W\), then \(\alpha x \epsilon W\).

We must prove that \(4(\alpha x_1) -1(\alpha x_2) + 5(\alpha x_3) = 0\).

\(\alpha(4x_1 -1x_2 + 5x_3) = 0\)

\(\alpha(0) = 0\)

\(0 = 0\)

This statement is true there \(\alpha x \epsilon Z\)

All of this is summarized by Theorem NSMS: Null Space of a Matrix is a Subspace. If \(Z\) is an \(m\times n\) matrix then the null space of \(Z\) is a subspace of \(\mathbb{C}^3\).