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No. On the real vector space $V={\mathbb{R}}^2$, use the symmetric bilinear formIts matrix of a bilinear form is $\mathrm{diag}(1,-1)$, whose determinant is $-1\ne 0$, so it is a nondegenerate bilinear form. LetA vector $(s,t)$ belongs to the bilinear orthogonal complement $U^{\perp }$ exactly when $s-t=0$. HenceIn particular, the claimed direct sum fails. The isotropic vector $(1,1)$ has zero self-pairing, although it pairs nontrivially with other vectors of $V$.

The dimension identity in part (i) counts both vector subspaces but does not force their intersection of vector subspaces to be zero. For a symmetric bilinear form, $U\cap U^{\perp }$ is the radical of a bilinear form restricted to $U$. Thus the precise criterion isIndeed, trivial intersection of vector subspaces and part (i) give a sum of vector subspaces of dimension $\dim V$. A positive-definite bilinear form would guarantee this for every $U$; symmetry alone does not.

A vector space over the real numbers is a set $V$ with an addition making it an abelian group and a scalar multiplication by $\mathbb{R}$ satisfying, for scalars $a,b$ and vectors $u,v$, the identities $a(u+v)=au+av$, $(a+b)v=av+bv$, $a(bv)=(ab)v$ and $1v=v$. A vector subspace is a nonempty subset closed under addition and scalar multiplication; a proper vector subspace is one unequal to $V$. A spanning set is a set whose finite linear combinations give every vector of $V$. A basis is a linearly independent spanning set. The dimension is the number of vectors in a basis, or its cardinality when the basis is infinite.

The sum of vector subspaces and their intersection of vector subspaces areBoth are vector subspaces: addition and scalar multiplication preserve the displayed forms or simultaneous membership. The intersection of vector subspaces is never empty because $0\in U\cap W$.

For the two given hyperplanes, adding and subtracting their defining equations gives $x_1=x_2$ and $x_3=x_4$. HenceThese two nonzero vectors are orthogonal, and consequently form an orthogonal basis of the intersection of vector subspaces. Put $u=(1,-1,-1,1)$ and $w=(1,-1,1,-1)$. Direct substitution gives $u\in U$, $w\in W$. Their dot products with $b_1,b_2$ vanish, and $u\cdot w=0$. Each defining equation of $U$ or $W$ is one nonzero linear constraint in four variables, so each vector subspace has dimension three. Thus each space has the indicated orthogonal basis:The last four vectors are nonzero and mutually orthogonal, so they are a basis of ${\mathbb{R}}^4$. They lie in $U+W$, proving $U+W=V$. The dimension identity is therefore verified numerically:

False. In ${\mathbb{R}}^2$ let $U=\mathrm{span}(1,1)$, $W=\mathrm{span}(1,0)$ and $X=\mathrm{span}(0,1)$. Then $W+X={\mathbb{R}}^2$, so $U\cap (W+X)=U$, but $U\cap W=U\cap X=\{0\}$ and their sum is zero. Thus intersection need not distribute over the sum of vector subspaces. The right side is always contained in the left; this example makes the inclusion strict.

The external direct sum of vector spaces consists of pairs $(v_1,v_2)$ with componentwise operations. For subspaces of one ambient space, the sum of vector subspaces is $V_1+V_2=\{v_1+v_2:v_i\in V_i\}$. The addition map from the external direct sum onto this subspace is surjective, with kernel $\{(w,-w):w\in V_1\cap V_2\}$. This kernel is isomorphic to the intersection. The rank-nullity theorem, together with $\dim (V_1\oplus V_2)=\dim V_1+\dim V_2$, proves the dimension formula for a sum of subspaces.

For the supplied bases, solving for annihilating linear forms givesEach basis has rank three, so these hyperplanes are the whole subspaces. The two independent equations make the intersection two-dimensional. Imposing the requested zero components yieldsBoth satisfy the two equations and are independent, since their first components differ while $v_1\ne 0$. Hence they form the required basis.

Let $r=dim(U\cap W)$. Choose a basis $e_1,\dots ,e_r$ of the intersection of vector subspaces, extend it to a basis $e_1,\dots ,e_r,u_{r+1},\dots ,u_m$ of $U$, and independently extend it to a basis $e_1,\dots ,e_r,w_{r+1},\dots ,w_n$ of $W$. The combined listspans the sum of vector subspaces $U+W$. It is also linearly independent: a vanishing linear combination says that a combination of the $u_i$ belongs to $U\cap W$, and its expression in the chosen basis of $U$ forces all its coefficients to vanish; the coefficients of the $w_i$ then vanish in the same way. This proves the dimension formula for a sum of subspaces

Parametrize the two linear subspaces asPutting a vector of $U$ in $W$ first forces its fourth coordinate to vanish; the remaining defining equation is then automatic. Thusand the dimension formula gives $\dim (U+W)=2+2-1=3$.

The linear functionalvanishes on each displayed generator of $U$ and $W$. Its kernel is three-dimensional because $\ell \ne 0$, so the inclusion $U+W\subseteq \ker \ell$ between two three-dimensional subspaces is equality.

Choose a basis $u_1,\dots ,u_r$ of $U$, extend it to a basis $u_1,\dots ,u_r,w_1,\dots ,w_s$ of $W$, and extend again to a basis of $V$ by adding $v_1,\dots ,v_t$. DefineEvery member of the full basis lies in $W$ or $W^{\prime }$, so the sum of vector subspaces is $W+W^{\prime }=V$. If $x\in W\cap W^{\prime }$, express it in both displayed bases. Uniqueness of the coordinates in the full basis forces its $w_j$ and $v_j$ coefficients to vanish. Thus $x\in U$. The reverse inclusion is built into the construction, proving $W\cap W^{\prime }=U$. This is a subspace supplement with a prescribed intersection.

For the matrix example, letThe skew-symmetric matrices are exactly $U=\mathrm{span}\{J\}$, and the traceless matrices form $W$. A suitable answer isThe matrix trace of $aI+bJ$ is $2a$, so $W\cap W^{\prime }=\mathrm{span}\{J\}=U$. Every $2\times 2$ real matrix has the decompositionwhose first term lies in $W$ and second in $W^{\prime }$. Therefore $W+W^{\prime }=M_2(\mathbb{R})$, as required.