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Choose an ordered basis $(e_1,\dots ,e_n)$ of a finite-dimensional vector space. The matrix of a bilinear form records all pairings of basis vectors, with $A_{ij}=B(e_i,e_j)$. Writing $v={\sum }_i{v}_i{e}_i$ and $w={\sum }_j{w}_j{e}_j$, separate linearity giveswhere the last expression uses the coordinate vectors of $v,w$. Conversely, every matrix $A$ defines a bilinear form by this formula.^[ref]

For a concrete bilinear form on ${\mathbb{R}}^2$, takeThen $B((1,0),(0,1))=2$ but $B((0,1),(1,0))=0$: a bilinear form need not be a symmetric bilinear form. For $v=(2,1{)}^T$ and $w=(-1,3{)}^T$, first compute $Aw=(5,3{)}^T$, then $v^{T}Aw=2\cdot 5+1\cdot 3=13$. Reversing the inputs gives $Av=(4,1{)}^T$ and $w^{T}Av=-4+3=-1$.

Under change of basis, let the columns of an invertible matrix $P$ be the new basis vectors in old coordinates. Thus $v=P{v}^{\prime }$ and $w=P{w}^{\prime }$, soThis is matrix congruence, rather than the matrix similarity $P^{-1}AP$ for an endomorphism. Both inputs have changed coordinates, which explains the two factors of $P$. For the displayed $A$ and $P=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, direct multiplication gives $A^{\prime }=\left(\begin{array}{cc}1& 3\\ 1& 4\end{array}\right)$; its $(1,2)$ entry is $B(e_1,e_1+e_2)=3$.^[ref]

Square matrices $A$ and $B$ are similar when $B=P^{-1}AP$ for some invertible matrix $P$. They represent the same linear operator in different bases and therefore have the same Jordan normal form.

The nullity of a linear map $T:V\to W$ is $n(T)=\dim \ker T$, the dimension of its kernel. It counts independent input directions that can change without changing the output. For a finite-dimensional vector space $V$, the rank-nullity theorem givesFor an $m\times n$ matrix $A$ of matrix rank $r$, this is $n-r$, using the number of columns, not the number of rows. This dimension identity is ^[ref].

The nullity answers a practical question about a linear system: once one input is known to produce the requested output, how much freedom remains? If $T(x_0)=b$, its fiber of a function over $b$ isIndeed, $T(x)=T(x_0)$ exactly when $T(x-x_0)=0$. Thus every nonempty fiber is an affine subspace of dimension $n(T)$, with the same direction space as the kernel. The kernel itself is the fiber over zero. Outputs outside the image of a linear map have empty fibers; the nullity does not guarantee existence, and we assign no solution-space dimension to an empty fiber here.

In the diagram, $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ has matrix $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$, matrix rank one and nullity one. Its kernel has basis $(1,-1)$, and the full solution of $F(x,y)=(c,0)$ is $(x,y)=(c,0)+s(1,-1)$. One free scalar moves the solution along a straight line; the output $(c,1)$ is impossible.

The existing three-input example $T(x,y,z)=(x+y,z)$ instead has rank two and nullity one. Its image of a linear map is all of ${\mathbb{R}}^2$, andEvery output has a solution straight line. In contrast, $U(x,y,z)=(x+y,0)$ has rank one and nullity two, with kernel basis $(1,-1,0),(0,0,1)$. Its reachable output $(a,0)$ has the solution affine subspacea plane, while $(a,b)$ with $b\ne 0$ has no solution. These two linear maps have the same input and output dimensions but different freedom and reachability. Computing the kernel separates what an output determines from what an extra constraint would have to fix. For either example, a basis of the kernel gives one independent free scalar per basis vector, which is why nullity measures freedom rather than the number of solutions.

An ordered basis is a basis equipped with an ordering of its basis vectors. The order fixes which position of a coordinate vector corresponds to each basis vector.

Rotate an orthonormal basis counterclockwise by $\theta$ while keeping the vector $v$ fixed. The new basis vectors, written in the old basis, are the columns of $P=R_{\theta }$. The change-of-basis matrix convention therefore givesFor example, the fixed vector with old coordinate vector $(2,0{)}^T$ has new coordinate vector $(1,-\sqrt{3}{)}^T$ after the basis turns by $\pi /3$ radians. Its negative second coordinate means that the arrow lies below the new first axis. By contrast, an active planar rotation through that same angle moves the arrow to old coordinates $(1,\sqrt{3}{)}^T$. The sign difference follows from which object moves. The passive component formulas are $x^{\prime }=x\cos \theta +y\sin \theta$, $y^{\prime }=-x\sin \theta +y\cos \theta$.^[ref]

If coordinate vectors are instead written as rows, transposing the active equation gives $(v^{\prime }{)}^T=v^T{R}_{\theta }^T$. Switching to rows changes the side and the transpose; it does not by itself turn an active planar rotation into passive rotation of coordinates.