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Discovery guide 17.1 Discovery guide

Recall that a vector space is a collection of objects (called vectors) that satisfies all of the axioms in Definition 16.4.1.

Discovery 17.1.

Sometimes you have a subcollection of vectors inside a larger vector space, and would like to know whether the subcollection is also a vector space, all on its own.

(a)

In the large vector space, you would already know (from having checked) that all ten axioms are true. Because all the vectors in the subcollection also “live” in the large vector space, six of the axioms will automatically be true for the subcollection (and the remaining four may or may not be true). Identify these six axioms that are automatically true.

Hint

It is easier to identify the six that are definitely true rather than the four that might be false.

(b)

Using \(\R^2\) as the large vector space, for each of the following subcollections, which of those four remaining axioms are true and which are false? (Consider all vectors as positioned with initial point at the origin.)

(i)

All points on the line \(y=x\text{.}\)

(ii)

All points on the line \(y=x+1\text{.}\)

(iii)

All points on the circle of radius \(1\) centred at the origin.

In Proposition 17.5.1, we will prove that the task of checking the four “possibly false” axioms you identified in Discovery 17.1 for a particular subcollection can be refined to the following test.

The Subspace Test.
  1. Nonempty.

    The subcollection contains at least one vector.

  2. Closed under vector additition.

    The sum of two vectors in the subcollection is always equal to another vector in the subcollection.

  3. Closed under scalar multiplication.

    A scalar multiple of a vector in the subcollection is always equal to another vector in the subcollection.

Discovery 17.2.

In each of the following, check each part of the Subspace Test for subcollection \(W\) inside vector space \(V\text{.}\)

  • If you think a part of the Subspace Test is true, justify it without resorting to examples.
  • If you think a part of the Subspace Test is false, provide an explicit example that demonstrates it.
(a)

\(V = \R^3\text{;}\) \(W =\) the \(xy\)-plane.

(b)

\(V = \R^3\text{;}\) \(W =\) the plane parallel to the \(xy\)-plane at height \(z=1\text{.}\)

(c)

\(V =\) all \(10\times 10\) matrices; \(W =\) diagonal \(10\times 10\) matrices.

(d)

\(V =\) all \(12\times 12\) matrices; \(W =\) those \(12\times 12\) matrices with a \(7\) in the \((1,1)\) entry.

(e)

\(V =\) all \(6\times 4\) matrices; \(W =\) those \(6\times 4\) matrices with \(0\) in each of the four corner entries.

(f)

\(V =\) all polynomials; \(W =\) those polynomials of degree \(2\) or less.

(g)

\(V =\) all polynomials; \(W =\) those polynomials of degree exactly \(2\text{.}\)

(h)

\(V =\) all polynomials; \(W =\) those polynomials with constant term equal to \(0\text{.}\)

(i)

\(V = \R^3\text{;}\) \(W =\) all column vectors \(\uvec{x}\) that satisfy the matrix equation \(A\uvec{x} = \zerovec\text{,}\) where \(A\) is some fixed \(2 \times 3\) matrix.

Hint

You don't need to know the entries of the matrix \(A\) to carry out the Subspace Test — use matrix algebra instead to test a sum or scalar multiple in the equation \(A\uvec{x}=\zerovec\text{.}\)

(j)

\(V =\R^3\text{;}\) \(W =\) all possible linear combinations of vectors \(\uvec{u} = (1,1,1)\) and \(\uvec{v} = (3,2,-1)\text{.}\)

As we will see from Proposition 17.5.5 in Subsection 17.5.2, the pattern in Discovery 17.2.j always works: if \(V\) is a vector space and \(S\) is a set of vectors in \(V\text{,}\) then the subcollection \(W\) of all possible linear combinations of vectors from \(S\) is a subspace of \(V\text{,}\) called the span of \(S\), and we write \(W = \Span S\text{.}\)

Discovery 17.3.

In each of the following, determine if the given vector \(\uvec{v}\) is a member of \(\Span S\text{.}\) That is, determine if \(\uvec{v}\) can be expressed as a linear combination of the vectors in \(S\text{.}\)

Hint.

Don't guess at it, set up equations and solve! The unknowns in your equations will be the scalars in the linear combination of the \(S\)-vectors to try to make the vector \(\uvec{v}\text{.}\) Start with a vector equation

\begin{equation*} \uvec{v} = \text{linear combination of }S\text{-vectors with variables as scalars} \text{.} \end{equation*}

This should somehow lead to a (gasp!) system of linear equations in your unknown scalars.

(a)

\(V = \R^3\text{,}\) \(S = \bigl\{(1,0,1),(2,1,-1)\bigr\}\text{,}\) \(\uvec{v} = (1,-1,4)\text{.}\)

(b)

\(V =\) all \(2\times 3\) matrices, \(S = \left\{ \left[\begin{smallmatrix} 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \end{smallmatrix}\right], \left[\begin{smallmatrix} 0 \amp 0 \amp 0 \\ 1 \amp 1 \amp 0 \end{smallmatrix}\right], \left[\begin{smallmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \end{smallmatrix}\right] \right\}\text{,}\) \(\uvec{v} = \left[\begin{smallmatrix} 0 \amp 2 \amp 2 \\ 3 \amp -3 \amp -3 \end{smallmatrix}\right]\text{.}\)

(c)

\(V = \) all polynomials, \(S = \{1, 1+x, 1+x^2\}\text{,}\) \(\uvec{v} = 2 - x + 3x^2\text{.}\)

Discovery 17.4.

In each of the following, try to convince yourself that \(V = \Span S\text{.}\) That is, convince yourself that every vector in \(V\) can be expressed as a linear combination of the vectors in \(S\text{.}\)

(a)

\(V = \R^5\text{,}\) \(S = \{\uvec{e}_1,\uvec{e}_2,\uvec{e}_3,\uvec{e}_4,\uvec{e}_5\}\text{.}\)

(b)

\(V =\) all \(2\times 2\) matrices, \(S = \left\{ \left[\begin{smallmatrix} 1 \amp 0 \\ 0 \amp 0 \end{smallmatrix}\right], \left[\begin{smallmatrix} 0 \amp 1 \\ 0 \amp 0 \end{smallmatrix}\right], \left[\begin{smallmatrix} 0 \amp 0 \\ 1 \amp 0 \end{smallmatrix}\right], \left[\begin{smallmatrix} 0 \amp 0 \\ 0 \amp 1 \end{smallmatrix}\right] \right\}\text{.}\)

(c)

\(V = \) all polynomials of degree \(3\) or less, \(S = \{1,x,x^2,x^3\}\text{.}\)