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MATH
[1251]

Examples from Subspaces

All examples from Subspaces.

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A line through the origin

For a fixed non-zero direction vector w∈Rnw\in\mathbb R^n, consider

S={x∈Rn:x=λw, λ∈R}.S=\{x\in\mathbb R^n:x=\lambda w,\ \lambda\in\mathbb R\}.

The zero vector is in SS by taking λ=0\lambda=0. If x1=λ1wx_1=\lambda_1w and x2=λ2wx_2=\lambda_2w, then

x1+x2=(λ1+λ2)w∈S.x_1+x_2=(\lambda_1+\lambda_2)w\in S.

For μ∈R\mu\in\mathbb R,

μx1=(μλ1)w∈S.\mu x_1=(\mu\lambda_1)w\in S.

The subspace test shows that SS is a subspace of Rn\mathbb R^n.

[routine]

  1. [1]:Subspace test

Polynomials satisfying a homogeneous condition

Consider

S={p∈P3(R):p(1)=0}.S=\{p\in P_3(\mathbb R):p(1)=0\}.

The zero polynomial belongs to SS. If p1,p2∈Sp_1,p_2\in S, then

(p1+p2)(1)=p1(1)+p2(1)=0,(p_1+p_2)(1)=p_1(1)+p_2(1)=0,

and if p∈Sp\in S and λ∈R\lambda\in\mathbb R, then

(λp)(1)=λp(1)=0.(\lambda p)(1)=\lambda p(1)=0.

Thus SS is a subspace of P3(R)P_3(\mathbb R). It is not merely {0}\{0\}; for example, x−1∈Sx-1\in S.

[routine]

  1. [1]:Subspace test

Matrices annihilating a fixed vector

For a fixed x∈Rnx\in\mathbb R^n, let

S={A∈Mm×n(R):Ax=0}.S=\{A\in M_{m\times n}(\mathbb R):Ax=0\}.

The zero matrix belongs to SS. If A1,A2∈SA_1,A_2\in S, then

(A1+A2)x=A1x+A2x=0.(A_1+A_2)x=A_1x+A_2x=0.

If A∈SA\in S and λ∈R\lambda\in\mathbb R, then

(λA)x=λ(Ax)=0.(\lambda A)x=\lambda(Ax)=0.

Therefore SS is a subspace of Mm×n(R)M_{m\times n}(\mathbb R).

[routine]

  1. [1]:Subspace test

Sets that are not subspaces

For a fixed non-zero u∈Rnu\in\mathbb R^n, the set

{x∈Rn:x=λu, −10≤λ≤10}\{x\in\mathbb R^n:x=\lambda u,\ -10\leq\lambda\leq10\}

is not closed under addition: 10u10u belongs to it but 10u+10u=20u10u+10u=20u does not.

The affine line

{x∈R2:x1+2x2=1}\{x\in\mathbb R^2:x_1+2x_2=1\}

does not contain the zero vector. Likewise, neither

{p∈P3(R):p(x)=1 for every x∈R}nor{p∈P2(R):p(0)=1}\{p\in P_3(\mathbb R):p(x)=1\text{ for every }x\in\mathbb R\} \qquad\text{nor}\qquad \{p\in P_2(\mathbb R):p(0)=1\}

contains the zero polynomial. Finally, the invertible matrices in Mn×n(R)M_{n\times n}(\mathbb R) do not form a subspace because the zero matrix is not invertible.

Replacing the right-hand side 11 by 00 gives homogeneous conditions such as p(x)=0p(x)=0 for every xx or p(0)=0p(0)=0 and these do define subspaces. The same contrast between homogeneous and non-homogeneous conditions returns with kernels of linear transformations.

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