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For vector subspaces $U,W$ of a common vector space, this is a surjective linear map. Its kernel of a linear map is $\{(x,-x):x\in U\cap W\}$, which has a linear isomorphism to $U\cap W$ by $x\mapsto (x,-x)$. The rank-nullity theorem and the direct sum basis give the dimension formula for a sum of subspaces. The paired domain keeps the two contributions separate even when their underlying vectors overlap.

For a bilinear form $B$ on a finite-dimensional vector space $V$, let $\Phi :V\to V^{\ast }$ send $u$ to $B(u,\cdot )$. Under the bidual evaluation map, $U^{\perp }$ is the annihilator of a vector subspace $\mathrm{im}(\Phi {|}_U)$, so $\dim U^{\perp }=\dim V-\mathrm{rank}(\Phi {|}_U)$. The rank-nullity theorem on $U$ now gives the displayed equality. In particular the sum of dimensions is at least $\dim V$. If $U\cap U^{\perp }=0$, the dimension formula for a sum of subspaces forces $V=U\oplus U^{\perp }$. The exact formula records the extra dimensions contributed by the left radical $\ker \Phi$, and does not assume an inner product.

The dimension formula for a sum of subspaces and the zero intersection implyBut $U+U^{\perp }$ is a vector subspace of the $n$-dimensional space $V$, so its dimension is at most $n$. Equality follows, and the sum is all of $V$. Since the intersection is zero, it is a direct sum:The argument remains valid for a degenerate ambient symmetric bilinear form; it uses the stated intersection hypothesis rather than assuming nondegeneracy everywhere.

The dimension formula for a sum of subspaces givesSince the sum lies in ${\mathbb{R}}^3$ and each plane has dimension two, the possibilities areThe first case means the planes coincide; two distinct planes through the origin intersect in a line.

Choose a basis $e_1,\dots ,e_k$ for $U\cap U^{\prime }$. Extend it to a basis $e_1,\dots ,e_k,u_1,\dots ,u_r$ of $U$, and separately to a basis $e_1,\dots ,e_k,v_1,\dots ,v_s$ of $U^{\prime }$. The combined list spans $U+U^{\prime }$.

To prove linear independence, suppose a linear combination of the combined list is zero. Then the combination of the $v_j$ lies in $U\cap U^{\prime }$, so it can be expressed using the $e_i$. Independence of the chosen basis of $U^{\prime }$ forces every $v_j$ coefficient to vanish. Independence of the chosen basis of $U$ then forces the remaining coefficients to vanish. Thus the combined list is a basis, with $k+r+s$ elements. This proves the dimension formula for a sum of subspaces:

For every $v\in V$, $(\alpha +\beta )v=\alpha v+\beta v$, so $\mathrm{im}(\alpha +\beta )\subseteq \mathrm{im}\alpha +\mathrm{im}\beta$. Taking dimensions and applying the preceding dimension formula for a sum of subspaces givesThis proves the upper rank bounds for a sum of linear maps, without assuming the two images have zero intersection.

Both candidate sets contain zero. The sum of subspaces is closed under linear combinations because $a(u_1+w_1)+b(u_2+w_2)=(a{u}_1+b{u}_2)+(a{w}_1+b{w}_2)$, with the two terms in $U$ and $W$. The intersection is closed because any linear combination of vectors lying in both subspaces still lies in both. Thus each is a vector subspace.

Take a basis $s_1,\dots ,s_r$ of the intersection, extend it by $u_1,\dots ,u_a$ to a basis of $U$, and by $w_1,\dots ,w_b$ to a basis of $W$. The combined list $s,u,w$ spans $U+W$. To prove independence, a zero combination makes the $u$ combination equal a vector in both $U$ and $W$, so it is a combination of the $s$'s. Independence of the basis of $U$ makes all $u$ coefficients zero; independence of the basis of $W$ then makes all remaining coefficients zero. Hence $\dim (U+W)=r+a+b$, giving the dimension formula for a sum of subspaces

For the concrete kernels, the equations $Ax=0$ giveApplying the two rows of $B$ to this expression gives $4(s-t)$ and $5(s-t)/3$. Thus the intersection has basis $v_0=(0,2,1,1)$. Choose $v_1=(5,-1,3,0)$ in $U$ and $v_2=(-4,-2,1,0)$ in $W$. ThenIndeed $U$ has dimension two because $A$ has rank two, and $v_0,v_1$ are independent. Likewise $W$ has dimension two and $v_0,v_2$ are independent, so the last list spans the sum. Its first three coordinates have determinant $-48$, confirming independence and dimension three.

Associate to $Q$ its symmetric bilinear form by the polarization identity,We first construct a diagonal basis by induction on the dimension. A zero quadratic form is already diagonal. Otherwise choose $v$ with $Q(v)\ne 0$. Every $x$ has the decompositionwhose second term lies in $W=\{w:B(w,v)=0\}$. Also $\mathbb{R}v\cap W=\{0\}$, so $V=\mathbb{R}v\oplus W$ and $\dim W=n-1$. Apply the induction hypothesis to the restricted quadratic form on $W$. The cross terms with $v$ vanish, giving a diagonal basis on $V$. Rescale each vector with diagonal coefficient $a\ne 0$ by $1/\sqrt{|a|}$, and reorder positive, negative and zero coefficients. This yieldsNo nondegeneracy assumption was used.

For uniqueness, let $V_{+}$ be the positive coordinate subspace and $V_{\le 0}$ the span of all negative and zero coordinate vectors. Their dimensions are $p$ and $n-p$. Any subspace $W$ on which $Q$ has positive definiteness intersects $V_{\le 0}$ only in zero. The dimension formula for a sum of subspaces therefore gives $\dim W\le p$. Conversely $V_{+}$ itself has dimension $p$ and positive restriction. Thus $p$ is the largest possible dimension of a positive subspace, an intrinsic description independent of the basis. Applying the same argument to $-Q$ shows that $q$ is the largest dimension of a negative subspace. This proves Sylvester's law of inertia, including uniqueness of the zero count $n-p-q$.

The extremal subspaces themselves need not be unique. For $Q(x,y)=x^2-y^2$, the coordinate axes give one positive and one negative line. For any $t\ne 0$, putThen $Q(u)=1$, $Q(v)=-1$, $B(u,v)=0$, and the determinant of the change of basis is one. The distinct lines $\mathbb{R}u$ and $\mathbb{R}v$ give another positive/negative decomposition, with the same intrinsic counts $p=q=1$.

Identify a configuration with its set $S$ of open edges. Let $k(S)$ count the connected components of a graph on the full graph vertex set, including isolated graph vertices. The random-cluster measure isFor $p\in (0,1)$ and finite $q\ge 1$ all its weights are positive. We compare $1\le q_1\le q_2$, taking ${\mu }_1={\mu }_{p,q_1}$ and ${\mu }_2={\mu }_{p,q_2}$.

The key fact is supermodularity of graph component count:To prove it, orient the graph edges arbitrarily and let $U_S$ be the real span of their incidence vectors for edges in $S$. Its dimension of a vector space is $r(S)=|V|-k(S)$: on each connected component of a graph these vectors span the vectors whose coordinates sum to zero. Since $U_{S\cup T}=U_S+U_T$ and $U_{S\cap T}\subset U_S\cap U_T$, the dimension formula for a sum of subspaces giveswhich is exactly the component-count inequality.

Put $a=k(S)-k(S\cup T)\ge 0$ and $b=k(S\cap T)-k(T)\ge 0$. The preceding inequality gives $b\ge a$. In the ratio of the two sides of Holley's condition, the normalization constants and Bernoulli edge factors cancel, because $|S\cup T|+|S\cap T|=|S|+|T|$. The ratio is thereforeBy the Holley inequality, ${\mu }_{p,q_1}$ stochastically dominates ${\mu }_{p,q_2}$. Thus increasing cluster weight suppresses every increasing event:The random-cluster single-edge conditional probability confirms the mechanism: it is $p$ when the endpoints are already connected, and $p/(p+q(1-p))$ otherwise. Larger $q$ favours configurations with more separate components. The condition $q\ge 1$ enters explicitly in the last factor $q_2^{b-a}$.

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.

Row reduction of the matrices whose columns are the displayed spanning vectors givesRow-reducing the matrix formed from all six columns givesThe dimension formula for a sum of subspaces now yields

Let $S$ be the vector subspace of symmetric matrices. A basis consists of $E_{ii}$ for $1\le i\le n$ and $E_{ij}+E_{ji}$ for $i<j$, soThe dimension formula for a sum of subspaces now givesChoose $0\ne A\in V\cap S$. It belongs to $V$ and is a symmetric matrix, so part (a)(i) makes it a diagonalisable matrix. The preceding example shows that the strict threshold is sharp.

A projective conic in ${\mathbb{P}}^2(\mathbb{C})$ is the zero locus of a nonzero homogeneous quadratic polynomial, up to nonzero scaling of that polynomial. Under an invertible linear change of homogeneous coordinates, its symmetric matrix has rank $3$, $2$, or $1$. Over $\mathbb{C}$, the corresponding forms are a nonsingular conic such as $XZ-Y^2=0$, two distinct lines such as $XY=0$, or a double line $X^2=0$. In the last case the repeated factor records multiplicity, even though the underlying point set is one line.

A projective line is the projectivization of a two-dimensional linear subspace of ${\mathbb{C}}^3$, equivalently the zero locus of one nonzero linear form. For two such subspaces $U,W$, the dimension formula for a sum of subspaces gives $\dim (U\cap W)\ge 2+2-3=1$. A nonzero vector in the intersection therefore supplies a common projective point. Distinct lines have exactly one such point.

The kernel of a linear map $g$ is precisely $U$, sincePart (c) gives $\mathrm{im}g=W_1+W_2$. Apply the rank-nullity theorem to the finite-dimensional vector space $W_1\times W_2$, and then parts (a) and (b):This proves the dimension formula for a sum of subspaces. The kernel of a linear map records exactly the freedom to transfer a shared vector from one summand to the other without changing their sum.

The image of $A+B$ lies in $\mathrm{im}A+\mathrm{im}B$. The dimension formula for a sum of subspaces gives the upper bound. Applying it to $A=(A+B)-B$ and then interchanging $A,B$ gives the lower bound. Both extremes are sharp: cancelling diagonal blocks gives the difference of ranks; diagonal maps with supports covering the whole space and positive coefficients give full rank when the sum of prescribed ranks is at least the dimension.

For open-edge sets $S,T$ in a fixed finite graph, including isolated graph vertices in $k$, one has $k(S\cap T)+k(S\cup T)\ge k(S)+k(T)$. Incidence-vector spans have rank $r(S)=|V|-k(S)$; their union span is the sum of the two spans, while the intersection-edge span is contained in the intersection of spans. The vector-space dimension formula for a sum of subspaces proves submodularity of $r$, equivalently supermodularity of $k$.