9 Bilinear forms
9.1 Definitions and basic properties
So far in Linear Algebra we have dealt with vector spaces without thinking much about geometric aspects. For example, for an abstract vector space we cannot say what the angle between two vectors is. Likewise, we are not able to talk about the distance between elements of a vector space. To make sense of these notions, the vector space needs further structure given by an inner product.
Example 5.3 • Bilinear maps ➔
, the standard scalar product is an example of a \(2\)-multilinear map.
The first one is the scalar product of two vectors in \(\mathbb{R}^3\) (or more generally \(\mathbb{R}^n\)). So \(V=\mathbb{R}^3\) and \(W=\mathbb{R}.\) Recall that the scalar product is the mapping \[V^2=\mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}, \quad (\vec{x},\vec{y})\mapsto \vec{x}\cdot \vec{y}=x_1y_1+x_2y_2+x_3y_3,\] where we write \(\vec{x}=(x_i)_{1\leqslant i\leqslant 3}\) and \(\vec{y}=(y_i)_{1\leqslant i\leqslant 3}.\) Notice that for all \(s_1,s_2 \in \mathbb{R}\) and all \(\vec{x}_1,\vec{x}_2,\vec{y}\in \mathbb{R}^3\) we have \[(s_1\vec{x}_1+s_2\vec{x}_2)\cdot \vec{y}=s_1(\vec{x}_1\cdot\vec{y})+s_2(\vec{x}_2\cdot\vec{y}),\] so that the scalar product is linear in the first variable. Furthermore, the scalar product is symmetric, \(\vec{x}\cdot \vec{y}=\vec{y}\cdot \vec{x}.\) It follows that the scalar product is also linear in the second variable, hence it is bilinear or \(2\)-multilinear.
The second one is the cross product of two vectors in \(\mathbb{R}^3.\) Here \(V=\mathbb{R}^3\) and \(W=\mathbb{R}^3.\) Recall that the cross product is the mapping \[V^2=\mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3, \quad (\vec{x},\vec{y})\mapsto \vec{x}\times \vec{y}=\begin{pmatrix} x_2y_3-x_3y_2 \\ x_3y_1-x_1y_3 \\ x_1y_2-x_2y_1 \end{pmatrix}.\] Notice that for all \(s_1,s_2 \in \mathbb{R}\) and all \(\vec{x}_1,\vec{x}_2,\vec{y}\in \mathbb{R}^3\) we have \[(s_1\vec{x}_1+s_2\vec{x}_2)\times \vec{y}=s_1(\vec{x}_1\times\vec{y})+s_2(\vec{x}_2\times\vec{y}),\] so that the cross product is linear in the first variable. Likewise, we can check that the cross product is also linear in the second variable, hence it is bilinear or \(2\)-multilinear. Observe that the cross product is alternating.
Definition 9.1 • Bilinear form
Let \(V\) be a \(\mathbb{K}\)-vector space. A bilinear form on \(V\) is a \(2\)-multilinear map with values in \(\mathbb{K}\) \[\langle\cdot{,}\cdot\rangle: V\times V \to \mathbb{K}, \qquad (v_1,v_2) \mapsto \langle v_1,v_2\rangle.\] That is, for all \(s_1,s_2 \in \mathbb{K}\) and all \(v_1,v_2,v_3 \in V\) we have \[\langle s_1 v_1+s_2 v_2, v_3 \rangle=s_1\langle v_1,v_3\rangle+s_2 \langle v_2,v_3\rangle\] as well as \[\langle v_3,s_1v_1+s_2 v_2 \rangle=s_1\langle v_3,v_1\rangle+s_2\langle v_3, v_2 \rangle.\] We say that \(\langle\cdot{,}\cdot\rangle\) is symmetric if \(\langle v_1,v_2\rangle=\langle v_2,v_1\rangle\) for all \(v_1,v_2 \in V\) and alternating if \(\langle v,v\rangle=0\) for all \(v \in V.\)
Example 9.2 • Bilinear forms
The standard scalar product defined by the rule (9.1) is a bilinear form on \(\mathbb{R}^n.\)
Let \(n \in \mathbb{N}\) and \(\mathbf{A}\in M_{n,n}(\mathbb{K})\) be a matrix. Using matrix multiplication, we define a mapping \[\tag{9.2} \langle\cdot{,}\cdot\rangle_\mathbf{A}: \mathbb{K}^n \times \mathbb{K}^n \to \mathbb{K}, \qquad (\vec{x}_1,\vec{x}_2)\mapsto \langle\vec{x}_1,\vec{x}_2\rangle_\mathbf{A}=\vec{x}_1^{T}\mathbf{A}\vec{x}_2.\] Notice that \(\mathbf{A}\vec{x}_2 \in M_{n,1}(\mathbb{K})\) and \(\vec{x}_1^T \in M_{1,n}(\mathbb{K})\) so that \(\vec{x}_1^{T}\mathbf{A}\vec{x}_2 \in M_{1,1}(\mathbb{K})=\mathbb{K}.\) The properties of the transpose and matrix multiplication imply that \(\langle\cdot{,}\cdot\rangle_\mathbf{A}\) is indeed a bilinear form on \(\mathbb{K}^n.\) Also, observe that the standard scalar product on \(\mathbb{R}^n\) arises by taking \(\mathbf{A}\) to be the identity matrix \(\mathbf{1}_{n}\in M_{n,n}(\mathbb{R}).\) That is, for all \(\vec{x},\vec{y} \in \mathbb{R}^n,\) we have \(\vec{x}\cdot \vec{y}=\langle \vec{x},\vec{y}\rangle_{\mathbf{1}_{n}}.\)
Using the determinant of a \(2\times 2\)-matrix, we obtain a map \[\langle\cdot{,}\cdot\rangle: \mathbb{K}_2 \times \mathbb{K}_2 \to \mathbb{K}\qquad (\vec{\xi}_1,\vec{\xi}_2) \mapsto \langle \vec{\xi}_1,\vec{\xi}_2\rangle=\det \begin{pmatrix} \vec{\xi}_1 \\ \vec{\xi}_2\end{pmatrix}.\] The properties of the determinant then imply that \(\langle\cdot{,}\cdot\rangle\) is an alternating bilinear form on the \(\mathbb{K}\)-vector space \(\mathbb{K}_2.\)
For \(n \in \mathbb{N}\) we consider \(V=M_{n,n}(\mathbb{K}),\) the \(\mathbb{K}\)-vector space of \(n\times n\)-matrices with entries in \(\mathbb{K}.\) We define \(\langle\cdot{,}\cdot\rangle: M_{n,n}(\mathbb{K}) \times M_{n,n}(\mathbb{K}) \to \mathbb{K}\) by the rule \[\tag{9.3} (\mathbf{A},\mathbf{B})\mapsto \langle \mathbf{A},\mathbf{B}\rangle=\operatorname{Tr}(\mathbf{A}\mathbf{B}).\] Definition 6.17 Definition 6.17 • Trace of a matrix ➔implies that \[\operatorname{Tr}(s_1 \mathbf{A}_1+s_2 \mathbf{A}_2)=s_1\operatorname{Tr}(\mathbf{A}_1)+s_2\operatorname{Tr}(\mathbf{A}_2)\] for all \(s_1,s_2 \in \mathbb{K}\) and all \(\mathbf{A}_1,\mathbf{A}_2 \in M_{n,n}(\mathbb{K}),\) that is, the trace is a linear map from \(M_{n,n}(\mathbb{K})\) into \(\mathbb{K}.\) Hence we obtain for all \(s_1,s_2 \in \mathbb{K}\) and all \(\mathbf{A}_1,\mathbf{A}_2,\mathbf{B}\in M_{n,n}(\mathbb{K})\) \[\begin{aligned} \langle s_1 \mathbf{A}_1+s_2 \mathbf{A}_2,B\rangle&=\operatorname{Tr}((s_1\mathbf{A}_1+s_2\mathbf{A}_2)B)=s_1\operatorname{Tr}(\mathbf{A}_1\mathbf{B})+s_2\operatorname{Tr}(\mathbf{A}_2\mathbf{B})\\ &=s_1\langle \mathbf{A}_1,\mathbf{B}\rangle+s_2\langle \mathbf{A}_2,\mathbf{B}\rangle. \end{aligned}\] showing that \(\langle\cdot{,}\cdot\rangle\) is linear in the first argument. Proposition 6.19Let \(n \in \mathbb{N}\) and \(\mathbf{A}\in M_{n,n}(\mathbb{K}).\) The sum \(\sum_{i=1}^n [\mathbf{A}]_{ii}\) of its diagonal entries is called the trace of \(\mathbf{A}\) and denoted by \(\operatorname{Tr}(\mathbf{A})\) or \(\operatorname{Tr}\mathbf{A}.\)
Proposition 6.19 ➔implies that \(\langle \mathbf{A},\mathbf{B}\rangle=\langle \mathbf{B},\mathbf{A}\rangle\) for all \(\mathbf{A},\mathbf{B}\in M_{n,n}(\mathbb{K}),\) hence \(\langle\cdot{,}\cdot\rangle\) is symmetric and therefore also linear in the second variable. We conclude that (9.3) defines a symmetric bilinear form on the vector space \(M_{n,n}(\mathbb{K}).\)Let \(n \in \mathbb{N}\) and \(\mathbf{A},\mathbf{B}\in M_{n,n}(\mathbb{K}).\) Then we have \[\operatorname{Tr}(\mathbf{A}\mathbf{B})=\operatorname{Tr}(\mathbf{B}\mathbf{A}).\]
We consider \(V=\mathsf{P}(\mathbb{K}),\) the \(\mathbb{K}\)-vector space of polynomials. For some fixed scalar \(x_0 \in \mathbb{K}\) we may define \[\langle\cdot{,}\cdot\rangle: \mathsf{P}(\mathbb{K}) \times \mathsf{P}(\mathbb{K}) \to \mathbb{K}, \quad (p,q)\mapsto \langle p,q\rangle=p(x_0)q(x_0).\] Then we have for all \(s_1,s_2 \in \mathbb{K}\) and polynomials \(p_1,p_2,q \in \mathsf{P}(\mathbb{K})\) \[\begin{aligned} \langle s_1\cdot_{\mathsf{P}(\mathbb{K})}p_1+_{\mathsf{P}(\mathbb{K})}s_2\cdot_{\mathsf{P}(\mathbb{K})}p_2,q\rangle &=\left(s_1\cdot_{\mathsf{P}(\mathbb{K})}p_1+_{\mathsf{P}(\mathbb{K})}s_2\cdot_{\mathsf{P}(\mathbb{K})}p_2\right)(x_0)q(x_0)\\ &=(s_1p_1(x_0)+s_2p_2(x_0))q(x_0)\\ &=s_1p_1(x_0)q(x_0)+s_2p_2(x_0)q(x_0)\\ &=s_1\langle p_1,q\rangle+s_2\langle p_2,q\rangle. \end{aligned}\] Hence \(\langle\cdot{,}\cdot\rangle\) is linear in the first variable. Clearly \(\langle\cdot{,}\cdot\rangle\) is also symmetric and therefore defines a symmetric bilinear form on \(V=\mathsf{P}(\mathbb{K}).\)
We consider \(V=\mathsf{C}([-1,1],\mathbb{R}),\) the \(\mathbb{R}\)-vector space of continuous real-valued functions defined on the interval \([-1,1].\) Recall from M03 Analysis I that continuous functions are integrable, hence we can define \[\langle\cdot{,}\cdot\rangle: V \times V \to \mathbb{R}, \qquad (f,g) \mapsto \langle f,g\rangle=\int_{-1}^1f(x)g(x)\mathrm{d}x.\] The properties of integration imply that this defines a symmetric bilinear form on \(\mathsf{C}([-1,1],\mathbb{R}).\)
Recall that the choice of an ordered basis of a finite dimensional \(\mathbb{K}\)-vector space \(V\) allowed to associate a matrix to every endomorphism \(f : V \to V.\) Similarly, an ordered basis also allows to associate a matrix to a bilinear form \(\langle\cdot{,}\cdot\rangle\) on \(V.\)
Definition 9.3 • Matrix representation of a bilinear form
Let \(V\) be a finite dimensional \(\mathbb{K}\)-vector space, \(\mathbf{b}=(v_1,\ldots,v_n)\) an ordered basis of \(V\) and \(\langle\cdot{,}\cdot\rangle\) a bilinear form on \(V.\) The matrix representation of \(\langle\cdot{,}\cdot\rangle\) with respect to \(\mathbf{b}\) is the matrix \(\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\) satisfying \[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})=(\langle v_i,v_j\rangle)_{1\leqslant i,j\leqslant n}.\]
Remark 9.4(\(\heartsuit\) – not examinable). Let \(\operatorname{Bil}(V)\) denote the set of bilinear forms on some \(\mathbb{K}\)-vector space \(V.\) By definition, \(\operatorname{Bil}(V)\) is a subset of the vector space of functions from \(V\times V\) into \(\mathbb{K}.\) By Definition 3.21
Definition 3.21 • Vector subspace ➔
, it follows that \(\operatorname{Bil}(V)\) is itself a \(\mathbb{K}\)-vector space. Moreover, if \(\dim V=n \in \mathbb{N}\) and \(V\) is equipped with an ordered basis \(\mathbf{b},\) the mapping from \(\operatorname{Bil}(V)\) into \(M_{n,n}(\mathbb{K})\) which sends a bilinear form to its matrix representation with respect to \(\mathbf{b}\) \[\langle\cdot{,}\cdot\rangle\mapsto \mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\] is an isomorphism. In particular, \(\dim \operatorname{Bil}(V)=n^2.\) The proof is left to the interested reader.Let \(V\) be a \(\mathbb{K}\)-vector space. A subset \(U\subset V\) is called a vector subspace of \(V\) if \(U\) is non-empty and if \[\tag{3.8} s_1\cdot_Vv_1+_Vs_2\cdot_V v_2 \in U\quad \text{for all}\; s_1,s_2 \in \mathbb{K}\; \text{and all}\; v_1,v_2 \in U.\]
Example 9.5
Let \(\langle\cdot{,}\cdot\rangle\) denote the standard scalar product on \(\mathbb{R}^n\) and \(\mathbf{e}=(\vec{e}_1,\ldots,\vec{e}_n)\) the standard basis of \(\mathbb{K}^n.\) Then, one easily computes that \[\langle \vec{e}_i,\vec{e}_j\rangle=\vec{e}_i^T\vec{e}_j=\delta_{ij}\] and hence \(\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{e})=(\delta_{ij})_{1\leqslant i,j\leqslant n}=\mathbf{1}_{n}.\)
Likewise, if \(\mathbf{A}\in M_{n,n}(\mathbb{K}),\) then \(\mathbf{M}(\langle\cdot{,}\cdot\rangle_\mathbf{A},\mathbf{e})=\mathbf{A}.\) Indeed, writing \(\mathbf{A}=(A_{ij})_{1\leqslant i,j\leqslant n},\) we have \[\mathbf{A}\vec{e}_j=\sum_{k=1}^n A_{kj}\vec{e}_k\] and thus \[\langle \vec{e}_i,\vec{e}_j\rangle_\mathbf{A}=\vec{e}_i^T\mathbf{A}\vec{e}_j=\vec{e}_i^T\sum_{k=1}^n A_{kj}\vec{e}_k=\sum_{k=1}^nA_{kj}\vec{e}_i^T\vec{e}_k=\sum_{k=1}^nA_{kj}\delta_{ik}=A_{ij}.\]
Let \(\langle\cdot{,}\cdot\rangle\) denote the alternating bilinear form on \(\mathbb{K}_2\) from Example 9.2 Example 9.2 • Bilinear forms ➔above and \[\mathbf{b}=\left(\begin{pmatrix} 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \end{pmatrix}\right).\] The alternating property of \(\langle\cdot{,}\cdot\rangle\) implies that the diagonal entries of \(\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\) vanish. Hence we obtain \[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})=\begin{pmatrix} 0 & \det\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ \det\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & 0 \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.\]The standard scalar product defined by the rule (9.1) is a bilinear form on \(\mathbb{R}^n.\)
Let \(n \in \mathbb{N}\) and \(\mathbf{A}\in M_{n,n}(\mathbb{K})\) be a matrix. Using matrix multiplication, we define a mapping \[\tag{9.2} \langle\cdot{,}\cdot\rangle_\mathbf{A}: \mathbb{K}^n \times \mathbb{K}^n \to \mathbb{K}, \qquad (\vec{x}_1,\vec{x}_2)\mapsto \langle\vec{x}_1,\vec{x}_2\rangle_\mathbf{A}=\vec{x}_1^{T}\mathbf{A}\vec{x}_2.\] Notice that \(\mathbf{A}\vec{x}_2 \in M_{n,1}(\mathbb{K})\) and \(\vec{x}_1^T \in M_{1,n}(\mathbb{K})\) so that \(\vec{x}_1^{T}\mathbf{A}\vec{x}_2 \in M_{1,1}(\mathbb{K})=\mathbb{K}.\) The properties of the transpose and matrix multiplication imply that \(\langle\cdot{,}\cdot\rangle_\mathbf{A}\) is indeed a bilinear form on \(\mathbb{K}^n.\) Also, observe that the standard scalar product on \(\mathbb{R}^n\) arises by taking \(\mathbf{A}\) to be the identity matrix \(\mathbf{1}_{n}\in M_{n,n}(\mathbb{R}).\) That is, for all \(\vec{x},\vec{y} \in \mathbb{R}^n,\) we have \(\vec{x}\cdot \vec{y}=\langle \vec{x},\vec{y}\rangle_{\mathbf{1}_{n}}.\)
Using the determinant of a \(2\times 2\)-matrix, we obtain a map \[\langle\cdot{,}\cdot\rangle: \mathbb{K}_2 \times \mathbb{K}_2 \to \mathbb{K}\qquad (\vec{\xi}_1,\vec{\xi}_2) \mapsto \langle \vec{\xi}_1,\vec{\xi}_2\rangle=\det \begin{pmatrix} \vec{\xi}_1 \\ \vec{\xi}_2\end{pmatrix}.\] The properties of the determinant then imply that \(\langle\cdot{,}\cdot\rangle\) is an alternating bilinear form on the \(\mathbb{K}\)-vector space \(\mathbb{K}_2.\)
For \(n \in \mathbb{N}\) we consider \(V=M_{n,n}(\mathbb{K}),\) the \(\mathbb{K}\)-vector space of \(n\times n\)-matrices with entries in \(\mathbb{K}.\) We define \(\langle\cdot{,}\cdot\rangle: M_{n,n}(\mathbb{K}) \times M_{n,n}(\mathbb{K}) \to \mathbb{K}\) by the rule \[\tag{9.3} (\mathbf{A},\mathbf{B})\mapsto \langle \mathbf{A},\mathbf{B}\rangle=\operatorname{Tr}(\mathbf{A}\mathbf{B}).\] Definition 6.17 implies that \[\operatorname{Tr}(s_1 \mathbf{A}_1+s_2 \mathbf{A}_2)=s_1\operatorname{Tr}(\mathbf{A}_1)+s_2\operatorname{Tr}(\mathbf{A}_2)\] for all \(s_1,s_2 \in \mathbb{K}\) and all \(\mathbf{A}_1,\mathbf{A}_2 \in M_{n,n}(\mathbb{K}),\) that is, the trace is a linear map from \(M_{n,n}(\mathbb{K})\) into \(\mathbb{K}.\) Hence we obtain for all \(s_1,s_2 \in \mathbb{K}\) and all \(\mathbf{A}_1,\mathbf{A}_2,\mathbf{B}\in M_{n,n}(\mathbb{K})\) \[\begin{aligned} \langle s_1 \mathbf{A}_1+s_2 \mathbf{A}_2,B\rangle&=\operatorname{Tr}((s_1\mathbf{A}_1+s_2\mathbf{A}_2)B)=s_1\operatorname{Tr}(\mathbf{A}_1\mathbf{B})+s_2\operatorname{Tr}(\mathbf{A}_2\mathbf{B})\\ &=s_1\langle \mathbf{A}_1,\mathbf{B}\rangle+s_2\langle \mathbf{A}_2,\mathbf{B}\rangle. \end{aligned}\] showing that \(\langle\cdot{,}\cdot\rangle\) is linear in the first argument. Proposition 6.19 implies that \(\langle \mathbf{A},\mathbf{B}\rangle=\langle \mathbf{B},\mathbf{A}\rangle\) for all \(\mathbf{A},\mathbf{B}\in M_{n,n}(\mathbb{K}),\) hence \(\langle\cdot{,}\cdot\rangle\) is symmetric and therefore also linear in the second variable. We conclude that (9.3) defines a symmetric bilinear form on the vector space \(M_{n,n}(\mathbb{K}).\)
We consider \(V=\mathsf{P}(\mathbb{K}),\) the \(\mathbb{K}\)-vector space of polynomials. For some fixed scalar \(x_0 \in \mathbb{K}\) we may define \[\langle\cdot{,}\cdot\rangle: \mathsf{P}(\mathbb{K}) \times \mathsf{P}(\mathbb{K}) \to \mathbb{K}, \quad (p,q)\mapsto \langle p,q\rangle=p(x_0)q(x_0).\] Then we have for all \(s_1,s_2 \in \mathbb{K}\) and polynomials \(p_1,p_2,q \in \mathsf{P}(\mathbb{K})\) \[\begin{aligned} \langle s_1\cdot_{\mathsf{P}(\mathbb{K})}p_1+_{\mathsf{P}(\mathbb{K})}s_2\cdot_{\mathsf{P}(\mathbb{K})}p_2,q\rangle &=\left(s_1\cdot_{\mathsf{P}(\mathbb{K})}p_1+_{\mathsf{P}(\mathbb{K})}s_2\cdot_{\mathsf{P}(\mathbb{K})}p_2\right)(x_0)q(x_0)\\ &=(s_1p_1(x_0)+s_2p_2(x_0))q(x_0)\\ &=s_1p_1(x_0)q(x_0)+s_2p_2(x_0)q(x_0)\\ &=s_1\langle p_1,q\rangle+s_2\langle p_2,q\rangle. \end{aligned}\] Hence \(\langle\cdot{,}\cdot\rangle\) is linear in the first variable. Clearly \(\langle\cdot{,}\cdot\rangle\) is also symmetric and therefore defines a symmetric bilinear form on \(V=\mathsf{P}(\mathbb{K}).\)
We consider \(V=\mathsf{C}([-1,1],\mathbb{R}),\) the \(\mathbb{R}\)-vector space of continuous real-valued functions defined on the interval \([-1,1].\) Recall from M03 Analysis I that continuous functions are integrable, hence we can define \[\langle\cdot{,}\cdot\rangle: V \times V \to \mathbb{R}, \qquad (f,g) \mapsto \langle f,g\rangle=\int_{-1}^1f(x)g(x)\mathrm{d}x.\] The properties of integration imply that this defines a symmetric bilinear form on \(\mathsf{C}([-1,1],\mathbb{R}).\)
Proposition 9.6
Let \(V\) be a finite dimensional \(\mathbb{K}\)-vector space, \(\mathbf{b}=(v_1,\ldots,v_n)\) an ordered basis of \(V\) with associated linear coordinate system \(\boldsymbol{\beta}: V \to \mathbb{K}^n\) and \(\langle\cdot{,}\cdot\rangle\) a bilinear form on \(V.\) Then
for all \(w_1,w_2 \in V\) we have \[\langle w_1,w_2\rangle=(\boldsymbol{\beta}(w_1))^T\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\boldsymbol{\beta}(w_2).\]
\(\langle\cdot{,}\cdot\rangle\) is symmetric if and only if \(\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\) is symmetric;
if \(\mathbf{b}^{\prime}\) is another ordered basis of \(V,\) then \[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b}^{\prime})=\mathbf{C}^T\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\mathbf{C},\] where \(\mathbf{C}=\mathbf{C}(\mathbf{b}^{\prime},\mathbf{b})\) denotes the change of basis matrix, see Definition 3.103 Definition 3.103 • Change of basis matrix ➔.Let \(V\) be a finite dimensional \(\mathbb{K}\)-vector space and \(\mathbf{b}, \mathbf{b}^{\prime}\) be ordered bases of \(V\) with corresponding linear coordinate systems \(\boldsymbol{\beta},\boldsymbol{\beta}^{\prime}.\) The change of basis matrix from \(\mathbf{b}\) to \(\mathbf{b}^{\prime}\) is the matrix \(\mathbf{C}\in M_{n,n}(\mathbb{K})\) satisfying \[f_\mathbf{C}=\boldsymbol{\beta}^{\prime}\circ \boldsymbol{\beta}^{-1}\] We will write \(\mathbf{C}(\mathbf{b},\mathbf{b}^{\prime})\) for the change of basis matrix from \(\mathbf{b}\) to \(\mathbf{b}^{\prime}.\)
Proof. (i) Since \(\mathbf{b}\) is a basis of \(V,\) it follows that for all \(w_1,w_2 \in V\) there exist unique scalars \(s_1,\ldots,s_n\) and \(t_1,\ldots,t_n\) so that \[w_1=\sum_{i=1}^n s_i v_i \qquad \text{and} \qquad w_2=\sum_{i=1}^n t_i v_i.\] Recall that this means that \[\boldsymbol{\beta}(w_1)=\begin{pmatrix} s_ 1 \\ \vdots \\s_n\end{pmatrix}\qquad \text{and}\qquad \boldsymbol{\beta}(w_2)=\begin{pmatrix} t_ 1 \\ \vdots \\t_n\end{pmatrix}.\] Using the bilinearity of \(\langle\cdot{,}\cdot\rangle,\) this gives \[\begin{aligned} \langle w_1,w_2\rangle&=\left\langle \sum_{i=1}^n s_i v_i,\sum_{j=1}^n t_j v_j\right\rangle=\sum_{i=1}^ns_i\sum_{j=1}^nt_j\langle v_i,v_j\rangle\\ &=\sum_{i=1}^ns_i\sum_{j=1}^n\,[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})]_{ij}t_j=(\boldsymbol{\beta}(w_1))^T\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\boldsymbol{\beta}(w_2). \end{aligned}\]
(ii) Suppose \(\langle\cdot{,}\cdot\rangle\) is symmetric. Then for all \(1\leqslant i,j\leqslant n,\) we have \[[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})]_{ij}=\langle v_i,v_j\rangle=\langle v_j,v_i\rangle=[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})]_{ji}\] so that \(\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\) is symmetric. Conversely, suppose \(\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\) is symmetric. Using notation as in (i), we obtain for all \(w_1,w_2 \in V\) \[\begin{aligned} \langle w_1,w_2\rangle&=\sum_{i=1}^n\sum_{j=1}^ns_i[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})]_{ij}t_j=\sum_{j=1}^n\sum_{i=1}^nt_j[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})]_{ji}s_i\\ &=\langle w_2,w_1\rangle \end{aligned}\] so that \(\langle\cdot{,}\cdot\rangle\) is symmetric as well.
Definition 3.103 • Change of basis matrix ➔
, we have \(\mathbf{C}(\mathbf{b}^{\prime},\mathbf{b})=(C_{ij})_{1\leqslant i,j\leqslant n}.\) Writing \(\mathbf{C}=(C_{ij})_{1\leqslant i,j\leqslant n}\) and using the bilinearity of \(\langle\cdot{,}\cdot\rangle,\) we calculate \[\begin{aligned}
[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b}^{\prime})]_{ij}&=\langle v^{\prime}_i,v^{\prime}_j\rangle=\left\langle \sum_{k=1}^n C_{ki}v_k,\sum_{l=1}^n C_{lj}v_l\right\rangle=\sum_{k=1}^n\sum_{l=1}^n C_{ki}C_{lj}\langle v_k,v_l\rangle \\
&=\sum_{k=1}^nC_{ki}\sum_{l=1}^n\langle v_k,v_l\rangle C_{lj}=\sum_{k=1}^n C_{ki}[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})C]_{kj}\\
&=\sum_{k=1}^n [C^T]_{ik}[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})C]_{kj}=[C^T\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})C]_{ij}
\end{aligned}\] as claimed.Let \(V\) be a finite dimensional \(\mathbb{K}\)-vector space and \(\mathbf{b}, \mathbf{b}^{\prime}\) be ordered bases of \(V\) with corresponding linear coordinate systems \(\boldsymbol{\beta},\boldsymbol{\beta}^{\prime}.\) The change of basis matrix from \(\mathbf{b}\) to \(\mathbf{b}^{\prime}\) is the matrix \(\mathbf{C}\in M_{n,n}(\mathbb{K})\) satisfying \[f_\mathbf{C}=\boldsymbol{\beta}^{\prime}\circ \boldsymbol{\beta}^{-1}\] We will write \(\mathbf{C}(\mathbf{b},\mathbf{b}^{\prime})\) for the change of basis matrix from \(\mathbf{b}\) to \(\mathbf{b}^{\prime}.\)
Example 9.7 We consider the symmetric bilinear form \(\langle\cdot{,}\cdot\rangle_\mathbf{A}\) on \(\mathbb{R}^2\) arising from the matrix \[\mathbf{A}=\begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix}.\] via the rule (9.2). Let \(\mathbf{e}=(\vec{e}_1,\vec{e}_2)\) denote the ordered standard basis of \(\mathbb{R}^2\) and \(\mathbf{b}=(\vec{e}_1+\vec{e}_2,\vec{e}_2-\vec{e}_1).\) In Example 9.5
Example 9.5 ➔
we have seen that \(\mathbf{M}(\langle\cdot{,}\cdot\rangle_\mathbf{A},\mathbf{e})=\mathbf{A}.\) In Example 3.105
Let \(\langle\cdot{,}\cdot\rangle\) denote the standard scalar product on \(\mathbb{R}^n\) and \(\mathbf{e}=(\vec{e}_1,\ldots,\vec{e}_n)\) the standard basis of \(\mathbb{K}^n.\) Then, one easily computes that \[\langle \vec{e}_i,\vec{e}_j\rangle=\vec{e}_i^T\vec{e}_j=\delta_{ij}\] and hence \(\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{e})=(\delta_{ij})_{1\leqslant i,j\leqslant n}=\mathbf{1}_{n}.\)
Likewise, if \(\mathbf{A}\in M_{n,n}(\mathbb{K}),\) then \(\mathbf{M}(\langle\cdot{,}\cdot\rangle_\mathbf{A},\mathbf{e})=\mathbf{A}.\) Indeed, writing \(\mathbf{A}=(A_{ij})_{1\leqslant i,j\leqslant n},\) we have \[\mathbf{A}\vec{e}_j=\sum_{k=1}^n A_{kj}\vec{e}_k\] and thus \[\langle \vec{e}_i,\vec{e}_j\rangle_\mathbf{A}=\vec{e}_i^T\mathbf{A}\vec{e}_j=\vec{e}_i^T\sum_{k=1}^n A_{kj}\vec{e}_k=\sum_{k=1}^nA_{kj}\vec{e}_i^T\vec{e}_k=\sum_{k=1}^nA_{kj}\delta_{ik}=A_{ij}.\]
Let \(\langle\cdot{,}\cdot\rangle\) denote the alternating bilinear form on \(\mathbb{K}_2\) from Example 9.2 above and \[\mathbf{b}=\left(\begin{pmatrix} 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \end{pmatrix}\right).\] The alternating property of \(\langle\cdot{,}\cdot\rangle\) implies that the diagonal entries of \(\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\) vanish. Hence we obtain \[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})=\begin{pmatrix} 0 & \det\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ \det\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & 0 \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.\]
Example 3.105 ➔
we computed that \[\mathbf{C}(\mathbf{b},\mathbf{e})=\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}.\] By definition, we have \[\begin{aligned}
\phantom{mathjax}[\mathbf{M}(\langle\cdot{,}\cdot\rangle_A,\mathbf{b})]_{11}&=\begin{pmatrix} 1 & 1 \end{pmatrix}\begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}=12,\\
[\mathbf{M}(\langle\cdot{,}\cdot\rangle_A,\mathbf{b})]_{12}&=\begin{pmatrix} 1 & 1 \end{pmatrix}\begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix}\begin{pmatrix} -1 \\ 1 \end{pmatrix}=0,\\
[\mathbf{M}(\langle\cdot{,}\cdot\rangle_A,\mathbf{b})]_{22}&=\begin{pmatrix} -1 & 1 \end{pmatrix}\begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix}\begin{pmatrix} -1 \\ 1 \end{pmatrix}=8,
\end{aligned}\] so that \[\mathbf{M}(\langle\cdot{,}\cdot\rangle_\mathbf{A},\mathbf{b})=\begin{pmatrix} 12 & 0 \\ 0 & 8 \end{pmatrix}.\] Indeed, writing \(\mathbf{C}=\mathbf{C}(\mathbf{b},\mathbf{e}),\) we have \[\mathbf{C}^T\mathbf{M}(\langle\cdot{,}\cdot\rangle_\mathbf{A},\mathbf{e})\mathbf{C}=\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}=\begin{pmatrix} 12 & 0 \\ 0 & 8 \end{pmatrix}=\mathbf{M}(\langle\cdot{,}\cdot\rangle_\mathbf{A},\mathbf{b}),\] in agreement with Proposition 9.6Let \(V=\mathbb{R}^2\) and \(\mathbf{e}=(\vec{e}_1,\vec{e}_2)\) be the ordered standard basis and \(\mathbf{b}=(\vec{v}_1,\vec{v}_2)=(\vec{e}_1+\vec{e}_2,\vec{e}_2-\vec{e}_1)\) another ordered basis. According to the recipe mentioned in Example 3.94, if we want to compute \(\mathbf{C}(\mathbf{e},\mathbf{b})\) we simply need to write each vector of \(\mathbf{e}\) as a linear combination of the elements of \(\mathbf{b}.\) The transpose of the resulting coefficient matrix is then \(\mathbf{C}(\mathbf{e},\mathbf{b}).\) We obtain \[\begin{aligned} \vec{e}_1&=\frac{1}{2}\vec{v}_1-\frac{1}{2}\vec{v}_2,\\ \vec{e}_2&=\frac{1}{2}\vec{v}_1+\frac{1}{2}\vec{v}_2, \end{aligned}\] so that \[\mathbf{C}(\mathbf{e},\mathbf{b})=\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\end{pmatrix}.\] Reversing the role of \(\mathbf{e}\) and \(\mathbf{b}\) gives \(\mathbf{C}(\mathbf{b},\mathbf{e})\) \[\begin{aligned} \vec{v}_1&=1\vec{e}_1+1\vec{e}_2,\\ \vec{v}_2&=-1 \vec{e}_1+1\vec{e}_2, \end{aligned}\] so that \[\mathbf{C}(\mathbf{b},\mathbf{e})=\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}.\] Notice that indeed we have \[\mathbf{C}(\mathbf{e},\mathbf{b})\mathbf{C}(\mathbf{b},\mathbf{e})=\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\] so that \(\mathbf{C}(\mathbf{e},\mathbf{b})^{-1}=\mathbf{C}(\mathbf{b},\mathbf{e}).\)
Proposition 9.6 ➔
.Let \(V\) be a finite dimensional \(\mathbb{K}\)-vector space, \(\mathbf{b}=(v_1,\ldots,v_n)\) an ordered basis of \(V\) with associated linear coordinate system \(\boldsymbol{\beta}: V \to \mathbb{K}^n\) and \(\langle\cdot{,}\cdot\rangle\) a bilinear form on \(V.\) Then
for all \(w_1,w_2 \in V\) we have \[\langle w_1,w_2\rangle=(\boldsymbol{\beta}(w_1))^T\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\boldsymbol{\beta}(w_2).\]
\(\langle\cdot{,}\cdot\rangle\) is symmetric if and only if \(\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\) is symmetric;
if \(\mathbf{b}^{\prime}\) is another ordered basis of \(V,\) then \[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b}^{\prime})=\mathbf{C}^T\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\mathbf{C},\] where \(\mathbf{C}=\mathbf{C}(\mathbf{b}^{\prime},\mathbf{b})\) denotes the change of basis matrix, see Definition 3.103.
Remark 9.8
You may remember from school that two non-zero vectors \(\vec{x}_1,\vec{x}_2 \in \mathbb{R}^n\) are perpendicular if and only if \(\vec{x}_1\cdot \vec{x}_2=0.\) In particular, no non-zero vector in \(\mathbb{R}^n\) is perpendicular to all vectors, or phrased differently, if \(\vec{x}\cdot\vec{x}_0=0\) for all vectors \(\vec{x},\) then \(\vec{x}_0=0_{\mathbb{R}^n}.\)
This condition also makes sense for a bilinear form:
Definition 9.9 • Non-degenerate bilinear form
Let \(\langle\cdot{,}\cdot\rangle\) be a bilinear form on a finite dimensional \(\mathbb{K}\)-vector space \(V.\) Then \(\langle\cdot{,}\cdot\rangle\) is called non-degenerate, if whenever a vector \(v_0\in V\) satisfies \(\langle v,v_0\rangle=0\) for all vectors \(v \in V,\) then we must have \(v_0=0_V.\)
Non-degeneracy of a bilinear form \(\langle\cdot{,}\cdot\rangle\) can be characterized in terms of its matrix representation, more precisely:
Proposition 9.10
Let \(\langle\cdot{,}\cdot\rangle\) be a bilinear form on a finite dimensional \(\mathbb{K}\)-vector space \(V\) and \(\mathbf{b}\) an ordered basis of \(V.\) Then \(\langle\cdot{,}\cdot\rangle\) is non-degenerate if and only if \(\det \mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\neq 0.\)
Proof. Let \(n=\dim V.\) First observe that a vector \(\vec{y}\in \mathbb{K}^n\) satisfies \(\vec{x}^T\vec{y}=0\) for all \(\vec{x} \in \mathbb{K}^n\) if and only if \(\vec{y}=0_{\mathbb{K}^n}.\)
Proposition 6.22 ➔
, \(\det \mathbf{A}=0\) is equivalent to the mapping \(f_\mathbf{A}: \mathbb{K}^n \to \mathbb{K}^n\) not being injective and hence by Lemma 3.31Let \(V\) be a finite dimensional \(\mathbb{K}\)-vector space and \(g : V \to V\) an endomorphism. Then the following statements are equivalent:
\(g\) is injective;
\(g\) is surjective;
\(g\) is bijective;
\(\det(g) \neq 0.\)
Lemma 3.31 ➔
equivalent to the existence of a non-zero vector \(\vec{x}_0 \in \mathbb{K}^n\) with \(\mathbf{A}\vec{x}_0=0_{\mathbb{K}^n}.\) Let \(v_0 \in V\) be the non-zero vector whose coordinate representation is \(\vec{x}_0,\) that is, \(\boldsymbol{\beta}(v_0)=\vec{x}_0,\) where \(\boldsymbol{\beta}: V \to \mathbb{K}^n\) denotes the linear coordinate system associated to \(\mathbf{b}.\) By Proposition 9.6A linear map \(f : V \to W\) is injective if and only if \(\operatorname{Ker}(f)=\{0_V\}.\)
Proposition 9.6 ➔
we have for all \(v \in V\) \[\tag{9.4}
\langle v,v_0\rangle=(\boldsymbol{\beta}(v))^T\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\boldsymbol{\beta}(v_0)=(\boldsymbol{\beta}(v))^T\mathbf{A}\vec{x}_0.\] Writing \(\vec{y}=\mathbf{A}\vec{x}_0,\) the observation at the beginning of the proof shows that (9.4) is \(0\) for all \(v \in V\) if and only if \(\mathbf{A}\vec{x}_0=0_{\mathbb{K}^n}.\)Let \(V\) be a finite dimensional \(\mathbb{K}\)-vector space, \(\mathbf{b}=(v_1,\ldots,v_n)\) an ordered basis of \(V\) with associated linear coordinate system \(\boldsymbol{\beta}: V \to \mathbb{K}^n\) and \(\langle\cdot{,}\cdot\rangle\) a bilinear form on \(V.\) Then
for all \(w_1,w_2 \in V\) we have \[\langle w_1,w_2\rangle=(\boldsymbol{\beta}(w_1))^T\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\boldsymbol{\beta}(w_2).\]
\(\langle\cdot{,}\cdot\rangle\) is symmetric if and only if \(\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\) is symmetric;
if \(\mathbf{b}^{\prime}\) is another ordered basis of \(V,\) then \[\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b}^{\prime})=\mathbf{C}^T\mathbf{M}(\langle\cdot{,}\cdot\rangle,\mathbf{b})\mathbf{C},\] where \(\mathbf{C}=\mathbf{C}(\mathbf{b}^{\prime},\mathbf{b})\) denotes the change of basis matrix, see Definition 3.103.
Exercises
Exercise 9.11
We consider \(V=M_{2,2}(\mathbb{R})\) and define \[\langle\cdot{,}\cdot\rangle: V \times V \to \mathbb{R}, \qquad (\mathbf{A},\mathbf{B}) \mapsto \langle \mathbf{A},\mathbf{B}\rangle=\frac{1}{4}\left(\det(\mathbf{A}+\mathbf{B})-\det(\mathbf{A}-\mathbf{B})\right).\] Show that \(\langle\cdot{,}\cdot\rangle\) defines a symmetric bilinear form on \(V=M_{2,2}(\mathbb{R}).\)
Solution
We first show that \(\langle \mathbf{A},\mathbf{B}\rangle= \langle \mathbf{B},\mathbf{A}\rangle\): Since \(\mathbf{A},\mathbf{B}\in M_{2,2}(\mathbb{R}),\) it holds that \(\det(\mathbf{B}-\mathbf{A}) = \det(-(\mathbf{A}-\mathbf{B})) = (-1)^2\det(\mathbf{A}-\mathbf{B})=\det(\mathbf{A}-\mathbf{B})\) and we find \[\langle \mathbf{B},\mathbf{A}\rangle = \frac14(\det(\mathbf{B}+\mathbf{A}) - \det(\mathbf{B}-\mathbf{A})) = \frac14(\det(\mathbf{A}+\mathbf{B})-\det(\mathbf{A}-\mathbf{B}))=\langle\mathbf{A},\mathbf{B}\rangle,\] which establishes the symmetry of \(\langle\cdot{,}\cdot\rangle.\) We are left to show the linearity of \(\langle\cdot{,}\cdot\rangle\) in the first slot, i.e. we need to show that the map \(\mathbf{A}\mapsto \langle \mathbf{A},\mathbf{B}\rangle\) is linear. Let \(\mathbf{A}= (A_{ij})_{1\leqslant i,j\leqslant 2}\) and \(\mathbf{B}= (B_{ij})_{1\leqslant i,j\leqslant 2}.\) Then \(\langle \mathbf{A},\mathbf{B}\rangle\) is given by \[\begin{aligned} &\frac14((A_{11}+B_{11})(A_{22}+B_{22})-(A_{12}+B_{12})(A_{21}+B_{21})-((A_{11}-B_{11})(A_{22}-B_{22})+\\ &-(A_{12}-B_{12})(A_{21}-B_{21})))\\ =~& \frac14(2A_{11}B_{22} + 2B_{11}A_{22}-2A_{12}B_{21}-2B_{12}A_{21})\\ =~& \frac12(A_{11}B_{22} + B_{11}A_{22}-A_{12}B_{21}-B_{12}A_{21}). \end{aligned}\] This expression is linear in the components of \(\mathbf{A}\) and the claim follows. Note that this could also be expressed as follows: We can identify \(M_{2,2}(\mathbb{R})\) and \(\mathbb{R}^4\) via the linear isomorphism \(\Psi:M_{2,2}(\mathbb{R})\to\mathbb{R}^4\) given by \[\mathbf{A}\mapsto \begin{pmatrix}A_{11} \\ A_{12} \\ A_{21} \\ A_{22}\end{pmatrix}.\] If we further define \(f:\mathbb{R}^4\to\mathbb{R}\) by \[\begin{pmatrix}A_{11} \\ A_{12} \\ A_{21} \\ A_{22}\end{pmatrix} \mapsto \frac12 \begin{pmatrix}B_{22} & -B_{21} & -B_{12} & B_{11}\end{pmatrix}\begin{pmatrix}A_{11} \\ A_{12} \\ A_{21} \\ A_{22}\end{pmatrix} ,\] we might write the map \(\mathbf{A}\mapsto\langle \mathbf{A},\mathbf{B}\rangle\) as \(f\circ\Psi.\) Since both \(f\) and \(\Psi\) are linear, so is the map \(\mathbf{A}\mapsto\langle \mathbf{A},\mathbf{B}\rangle\) as a composition of linear maps.