Subspaces and Spanning Sets

Chapter 6 • Section 6-2

"Spaces within spaces... Inception style! 🏢"

🧪 The Subspace Test

A subset $U$ of a vector space $V$ is a subspace if:

  • 1. Zero Vector: $\vec{0} \in U$.
  • 2. Closed under Addition: If $\vec{u}, \vec{v} \in U$, then $\vec{u} + \vec{v} \in U$.
  • 3. Closed under Scaling: If $\vec{u} \in U$ and $c \in \mathbb{R}$, then $c\vec{u} \in U$.

Example: Symmetric Matrices

The set of symmetric matrices $\{ A \in \M_{nn} \mid A^T = A \}$ is a subspace of $\M_{nn}$.

Zero matrix is symmetric. Sum of symmetric matrices is symmetric. Scaled symmetric matrix is symmetric.

Non-Example: Degree exactly 2

Polynomials with degree exactly 2 is NOT a subspace.

Fails closure under addition: $(x^2 + x) + (-x^2) = x$ (degree 1).

Subspace Checker

Determine if the following sets are subspaces.

1. The set of all polynomials $p(x)$ such that $p(0) = 0$.

2. The set of all polynomials $p(x)$ such that $p(0) = 1$.

🔗 Linear Combinations

A vector $\vec{w}$ is a linear combination of vectors $\vec{v}_1, \dots, \vec{v}_k$ if there exist scalars $c_1, \dots, c_k$ such that:

$$ \vec{w} = c_1\vec{v}_1 + c_2\vec{v}_2 + \dots + c_k\vec{v}_k $$

Example: Is $x^2+1$ a combination?

Is $x^2+1$ a linear combination of $\{x+1, x^2+x, x^2+2\}$?

We set up the equation:

$$ x^2+1 = a(x+1) + b(x^2+x) + c(x^2+2) $$

Comparing coefficients leads to a system. If consistent, the answer is Yes!

Spanning Sets

The span of a set of vectors $\{\vec{v}_1, \dots, \vec{v}_k\}$ is the set of all possible linear combinations:

$$ \span\{\vec{v}_1, \dots, \vec{v}_k\} = \{ c_1\vec{v}_1 + \dots + c_k\vec{v}_k \mid c_i \in \mathbb{R} \} $$

The span is always a subspace!

Example: Spanning $\M_{22}$

The set of matrices with a single 1 and zeros elsewhere spans the entire space.

$$ \left\{ \begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\1&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix} \right\} $$

Theorem: Smallest Subspace

Let $U = \span\{\vec{v}_1, \dots, \vec{v}_k\}$. Then:

  • $U$ is a subspace of $V$ containing each $\vec{v}_i$.
  • If $W$ is any subspace containing each $\vec{v}_i$, then $U \subseteq W$.

In other words, the span is the "smallest" subspace containing the vectors.

🧠 Knowledge Check Win $15

Does the set $\{1, x, x^2\}$ span $\calP_2$?