Math 303: Section 22

Dr. Janssen

\[ \def\R{{\mathbb R}} \def\P{{\mathbb P}} \def\b{{\mathbf{b}}} \def\x{{\mathbf{x}}} \def\y{{\mathbf{y}}} \def\u{{\mathbf{u}}} \def\v{{\mathbf{v}}} \def\w{{\mathbf{w}}} \def\z{{\mathbf{z}}} \DeclareMathOperator{\null}{Nul} \DeclareMathOperator{\span}{Span} \]

Some mathematical structures behave like others!

Generally, *abstraction* is the process of deciding what structure/properties we care about. In linear algebra, we care about **vector spaces**.

A set \(V\) on which an operation of addition and a multiplication by scalars is defined is a **vector space** if for all \(\u,\v,\w\) in \(V\) and all scalars \(a\) and \(b\):

- \(\u + \v\) is an element of \(V\)
- \(\u + \v = \v + \u\)
- \((\u + \v) + \w = \u + (\v + \w)\)
- There is a zero vector \(\mathbf{0}\) in \(V\) so that \(\u + \mathbf{0} = \u\)
- For each \(\x\) in \(V\) there is a \(\y\) in \(V\) so that \(\x + \y = \mathbf{0}\)
- \(a\u\) is an element of \(V\)
- \((a+b)\u = a\u + b\u\)
- \(a(\u + \v) = a\u + b\v\)
- \((ab)\u = a(b\u)\)
- \(1\u = \u\)

- \(\R^n\) with the usual vector addition and scalar multiplication
- \(\P_n\)
^{1} - The eigenspace of an \(n\times n\) matrix corresponding to an eigenvalue \(\lambda\) is a subspace of \(\R^n\), and therefore a vector space.
- The null space of a matrix
- The column space of a matrix
- The span of
*any set of vectors in \(\R^n\)* - The space \(\mathcal{M}_{m\times n}(\R)\) of \(m\times n\) matrices with real entries.
- The space \(\mathcal{F}\) of functions from \(\R\) to \(\R\)
- The space \(\R^\infty\) of infinite real sequences!
- The set of solutions to a second order homogeneous differential equation

**Theorem 1 (Multiplying by 0) **Let \(\v\) be a vector in a vector space \(V\). Then \(0\v = \mathbf{0}\).

Let \(\u,\v,\w\) be vectors in a vector space \(V\), and suppose that

\[ \u + \w = \v + \w \qquad(1)\]

- Why does \(V\) contain an additive inverse \(\mathbf{z}\) of \(\w\)?
- Add \(\z\) to both sides of Equation 1 to obtain

\[ (\u + \w) + \z = (\v + \w) + \z. \]

Which property of a vector space allows us to state the following?

\[ \u + (\w + \mathbf{z}) = \v + (\w + \mathbf{z}) \]

- Now use the properties of additive inverses and the additive identity to explain why \(\u = \v\). Conclude that we have a cancellation law for addition in
*any*vector space.

Let \(V\) be any vector space with identity \(\mathbf{0}\).

- \(0\v = \mathbf{0}\) for any \(\v\) in \(V\)
- The vector \(\mathbf{0}\) is unique.
- \(c\mathbf{0} = \mathbf{0}\) for any scalar \(c\).
- For any \(\v\) in \(V\), the additive inverse of \(\v\) is unique.
- The additive inverse of a vector \(\v\) in \(V\) is the vector \((-1)\v\).
- If \(\u, \v\), and \(\w\) are in \(V\) and \(\u + \w = \v + \w\), then \(\u = \v\).

Often a vector space \(V\) contains a vector space \(W\). In this case, we say that \(W\) is a *subspace* of \(V\).

Let \(H = \{at : a\in \R\}\). Notice that \(H\) is a *subset* of \(\P_1\).

- Is \(H\) closed under the addition in \(\P_1\)? Verify.
- Does \(H\) contain the zero vector from \(\P_1\)? Verify.
- Is \(H\) closed under multiplication by scalars? Verify.
- Explain why \(H\) satisfies every other property of the definition of a vector space automatically just by being a subset of \(\P_1\) and using the same operations as in \(\P_1\). Conclude that \(H\) is a vector space.

**Definition 1 (Subspace) **A subset \(H\) of a vector space \(V\) is a **subspace** of \(V\) if

- whenever \(\u,\v\in H\) it is also true that \(\u+\v\in H\)
- whenever \(\u\) is in \(H\) and \(a\) is a scalar, it is also true that \(a\u\in H\)
- \(\mathbf{0}\in H\)

Is the given subset \(H\) a subspace of the indicated vector space \(V\)? Verify your answer.

- \(V\) is any vector space and \(H = \{\mathbf{0}\}\)
- \(V = M_{2\times 2}\), and \(H = \left\{\left[\begin{matrix} 2x & y \\ 0 & x \end{matrix}\right] \ : \ x,y\in\R \right\}\)
- \(V = \P_2\), \(H = \{2at^2 + 1 \ : \ a\in\R\}\)
- \(V = \P_2\), \(H = \{at \ : \ a\in \R \} \cup \{bt^2 \ : \ b\in \R\}\)
- \(V = \mathcal{F}\), \(H = \P_2\)

A given (randomly chosen?) set of vectors \(S\) is probably not a vector (sub)space. But we can build one by considering the *span* of \(S\).

**Definition 2 **Let \(V\) be a vector space. A **linear combination** of vectors \(\v_1, \v_2, \ldots, \v_k\) in \(V\) is a vector of the form

\[ x_1 \v_1 + x_2 \v_2 + \cdots + x_k \v_k, \]

where \(x_1, x_2, \ldots, x_k\) are scalars. The **span** of the vectors \(\v_1, \v_2, \ldots, \v_k\) is the set of *all* linear combinations of \(\v_1, \v_2, \ldots, \v_k\):

\[ \span\{\v_1, \v_2, \ldots, \v_k\} = \{x_1 \v_1 + \cdots + x_k \v_k \ : \ x_i\in\R\}. \]

Spans are subspaces!

Let \(H = \{ a_2 t^2 - a_1 t \ : \ a_1,a_2\in \R\}\). Note that \(H\) is a subset of \(\P_2\). Find two vectors \(\v_1,\v_2\in \P_2\) so that \(H = \Span \{\v_1, \v_2\}\) and hence conclude that \(H\) is a subspace of \(\P_2\).

Let \(p_1(t) = 1 - t^2\) and \(p_2(t) = 1 + t^2\), and let \(S = \{p_1(t), p_2(t)\}\) in \(\P_2\). Is the polynomial \(q(t) = 3 - 2t^2 \in \span S\)?

With \(S\) as in the previous part, describe as best you can the subspace \(\span S\) of \(\P_2\).