For any vector space $V$:

• $\{0\}$ is a subspace (the zero subspace or trivial subspace).
• $V$ itself is a subspace.

These are called the improper subspaces. All other subspaces are proper.

In $\mathbb{R}^3$:

• Any line through the origin is a subspace: $W = \{t(a, b, c) \mid t \in \mathbb{R}\}$ for a fixed nonzero vector $(a, b, c)$.
• Any plane through the origin is a subspace: $W = \{(x, y, z) \mid ax + by + cz = 0\}$ for constants $a, b, c$ (not all zero).
• A line or plane not through the origin is not a subspace (it does not contain $0$).

Let $V = M_{n \times n}(F)$. Then:

• $\text{Sym}_n(F) = \{A \in V \mid A^T = A\}$ is a subspace (symmetric matrices).
• $\text{Skew}_n(F) = \{A \in V \mid A^T = -A\}$ is a subspace (skew-symmetric matrices).

When $\text{char}(F) \neq 2$, we have $V = \text{Sym}_n(F) \oplus \text{Skew}_n(F)$ via $A = \tfrac{1}{2}(A + A^T) + \tfrac{1}{2}(A - A^T)$.

The set of $n \times n$ diagonal matrices $D_n(F) = \{A \in M_{n \times n}(F) \mid a_{ij} = 0 \text{ for } i \neq j\}$ is a subspace of $M_{n \times n}(F)$ of dimension $n$.

The set $U_n(F) = \{A \in M_{n \times n}(F) \mid a_{ij} = 0 \text{ for } i > j\}$ of upper triangular matrices is a subspace of $M_{n \times n}(F)$ of dimension $n(n+1)/2$.

$P_n(F) \subseteq P(F)$ is a subspace for each $n \geq 0$. Note that the set of polynomials of degree exactly $n$ is not a subspace (it does not contain $0$, and it is not closed under addition: $(x^2 + x) + (-x^2) = x \notin$ degree-2 polynomials).

In $F(\mathbb{R}, \mathbb{R})$:

• $W_{\text{even}} = \{f \mid f(-x) = f(x) \text{ for all } x\}$ is a subspace.
• $W_{\text{odd}} = \{f \mid f(-x) = -f(x) \text{ for all } x\}$ is a subspace.

Every function decomposes as $f(x) = \tfrac{f(x)+f(-x)}{2} + \tfrac{f(x)-f(-x)}{2}$ (even + odd).

For $A \in M_{m \times n}(F)$, the null space $N(A) = \{x \in F^n \mid Ax = 0\}$ is a subspace of $F^n$. If $Au = 0$ and $Av = 0$, then $A(au + bv) = aAu + bAv = 0$.

$W = \{(x, y) \in \mathbb{R}^2 \mid x \geq 0, y \geq 0\}$ is not a subspace. It contains $(1, 1)$ but not $(-1)(1, 1) = (-1, -1)$, so it fails closure under scalar multiplication.

$W = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1\}$ is not a subspace: $0 \notin W$, and $(1, 0) + (0, 1) = (1, 1) \notin W$.

$W = \{(x, y) \in \mathbb{R}^2 \mid x + y = 1\}$ is not a subspace because $(0, 0) \notin W$. However, $W$ is an affine subspace (a translate of the subspace $\{(x, y) \mid x + y = 0\}$).

$C^1(\mathbb{R}) \subseteq C(\mathbb{R}) \subseteq F(\mathbb{R}, \mathbb{R})$ are nested subspaces. The differentiable functions form a subspace of the continuous functions, which form a subspace of all functions.

In $\mathbb{R}^3$, let $W_1 = \{(a, 0, 0) \mid a \in \mathbb{R}\}$ (the $x$-axis) and $W_2 = \{(0, b, c) \mid b, c \in \mathbb{R}\}$ (the $yz$-plane). Then $\mathbb{R}^3 = W_1 \oplus W_2$: every $(a, b, c) = (a, 0, 0) + (0, b, c)$ uniquely.