Simply speaking, a subset [math]H[/math] of a vector space [math]V[/math] is a [b]subspace[/b] if [math]H[/math] is a vector space itself when inheriting the addition and scalar multiplication defined in [math]V[/math]. [br][br][u]Definition[/u]: [math]H[/math] is a subspace of [math]V[/math] if [math]H[/math] is a [u]non-empty[/u] subset of [math]V[/math] that satisfies the following conditions:[br][list=1][*]For any [math]u[/math] and [math]v[/math] in [math]H[/math], [math]u+v[/math] is in [math]H[/math].[/*][*]For any [math]u[/math] in [math]H[/math] and real number [math]c[/math], [math]cu[/math] is in [math]H[/math]. (Note: In particular, we can set [math]c=0[/math] and it implies that the zero vector is in [math]H[/math].)[br][/*][/list][br][br][u]Example 1[/u]: [math]\left\{0\right\}[/math], the set containing only the zero vector, is a subspace of any vector space. It is called the [b]zero subspace[/b].[br][br][br][u]Example 2[/u]: For any non-negative integer [math]n[/math], [math]\mathbb{P}_n[/math], the vector space of all polynomial of degree [math]\le n[/math] with real coefficients, is a subspace of [math]\mathbb{P}[/math], the vector space of all polynomials with real coefficients.[br][br][br][u]Example 3[/u]: Any plane in [math]\mathbb{R}^3[/math] containing the origin is a subspace of [math]\mathbb{R}^3[/math]. For example, [math]H=\left\{x\hat{\mathbf{i}}+y\hat{\mathbf{j}}+z\hat{\mathbf{k}}\in \mathbb{R}^3 \ | \ 2x+3y-4z=0\right\}[/math] is a subspace of [math]\mathbb{R}^3[/math] because [br][list][*]Zero vector is obviously in [math]H[/math] so [math]H[/math] is non-empty.[br][/*][*]For any [math]u=u_1\hat{\mathbf{i}}+u_2\hat{\mathbf{j}}+u_3\hat{\mathbf{k}},v=v_1\hat{\mathbf{i}}+v_2\hat{\mathbf{j}}+v_3\hat{\mathbf{k}}\in H[/math], [math]2u_1+3u_2 -4u_3=0[/math] and [math]2v_1+3v_2-4v_3=0[/math]. Then [math]u+v=(u_1+v_1)\hat{\mathbf{i}}+(u_2+v_2)\hat{\mathbf{j}}+(u_3+v_3)\hat{\mathbf{k}}[/math] and [math] 2(u_1+v_1)+3(u_2+v_2)-4(u_3+v_3)=(2u_1+3u_2 -4u_3)+(2v_1+3v_2-4v_3)=0[/math]. Therefore, [math]u+v[/math] is also in [math]H[/math].[/*][*]For any real number [math]c[/math] and [math]u=u_1\hat{\mathbf{i}}+u_2\hat{\mathbf{j}}+u_3\hat{\mathbf{k}}[/math] in [math]H[/math] i.e. [math]2u_1+3u_2 -4u_3=0[/math]. [math]cu=cu_1\hat{\mathbf{i}}+cu_2\hat{\mathbf{j}}+cu_3\hat{\mathbf{k}}[/math]. Hence [math]2(cu_1)+3(cu_2)-4(cu_3)=c(2u_1+3u_2 -4u_3)=0[/math], which means [math]cu[/math] is also in [math]H[/math].[br][/*][/list][br][br][u]Example 4[/u]: Let [math]\mathbb{S}[/math] be the vector space of all real number sequences and [math]F=\left\{(a_n)\in \mathbb{S} \ | \ a_n=0 \ \text{for all even} \ n\right\}[/math]. Then [math]F[/math] is a subspace of [math]\mathbb{S}[/math] because[br][list][*]The sequence of all zeros is in [math]F[/math] so [math]F[/math] is non-empty.[br][/*][*]If [math]\left(a_n\right),\left(b_n\right)[/math] are in [math]F[/math], then [math]a_n=b_n=0[/math] for all even [math]n[/math]. Therefore, [math]a_n+b_n=0[/math] for all even [math]n[/math] and [math]\left(a_n\right)+\left(b_n\right)=\left(a_n+b_n\right)[/math] is also in [math]F[/math].[/*][*]For any real number [math]c[/math] and any [math]\left(a_n\right)[/math] in [math]F[/math], [math]a_n=0[/math] for all even [math]n[/math]. Therefore, [math]ca_n=0[/math] for all even [math]n[/math] and [math]c\left(a_n\right)=\left(ca_n\right)[/math] is also in [math]F[/math]. [br][/*][/list][br][br][u]Example 5[/u]: Let [math]M_{n\times n}[/math] be the vector space of all n x n matrices. Let [math]U=\left\{A\in M_{n\times n} \ | \ A \ \text{is an upper triangular matrix}\right\}[/math]. Then [math]U[/math] is a subspace of [math]M_{n\times n}[/math] because[br][list][*]Zero matrix is in [math]U[/math] so [math]U[/math] is non-empty.[br][/*][*]For any two n x n matrices [math]A,B[/math] in [math]U[/math], they are upper triangular. Hence [math]A+B[/math] is obviously upper triangular and thus in [math]U[/math].[/*][*]For any real number [math]c[/math] and any n x n matrix [math]A[/math] in [math]U[/math], [math]A[/math] is upper triangular. Therefore, [math]cA[/math] is also upper triangular and thus in [math]U[/math].[br][/*][/list]