5.21 Preserves the inner products

Let \((V,\lang,\rang)\) be k-IPVS, \(\dim V=n,B=\{v_1,...,v_n\}\) and \(X\in V,X=\sum_{i=1}^{n}x_{i}v_{i},x_{i}\in k\) and \(Y\in V,Y=\sum_{i=1}^{n}y_{j}v_{j},y_{j}\in k\)

Then \(\langle X,Y\rangle=\langle\sum_{i=1}^{n}x_{i}v_{i},\sum_{j=1}^{n}y_{j}v_{j}\rangle =\sum_{i=1}^{n}\sum_{j=1}^{n}x_{i}\bar y_{j}\langle v_{i},v_{j}\rangle\)

Let \(A\) be \([\lang,\rang]_B\) s.t. \(\underbrace{\bar Y^{t}}_{\in k^{1\times n}}\underbrace{A}_{\in k^{n\times n}}\underbrace{X} _{\in k^{n\times 1}}=(\bar Y^{t} AX)_{11}=(\bar y_{1},...,\bar y_{n})(A_{ij}) \begin{pmatrix} x_{1} \\ ... \\ x_{n} \end{pmatrix}=\sum^n_{r=1}[(\bar y_1,...,\bar y_n)A]_{1r}\begin{pmatrix} x_1\\...\\x_n \end{pmatrix}_{r1}=\sum_{r=1}^{n}\sum_{s=1}^{n}(\bar y_1,...,\bar y_n)_{1s}A_{sr}x_r=\sum_{r=1}^{n}\sum_{s=1}^{n} x_r\bar y_s A_{sr}\)

Let \(r\to i,s\to j\), then \(=\sum^n_{r=1}\sum^n_{s=1}x_i\bar y_jA_{ji}\)

Thus define \(A_{ji}=\lang v_i,v_j\rang\), then \(\lang X,Y\rang =\bar Y^tAX\) where \(X,Y\) are columns

Definition

Given \(C\in M_{n}(k)=k^{n\times n}\), the adjoint of \(C\) is \(\bar C^t\). Notation: \(\bar C^t=C^*\)

Remark

\(\lang V,\lang,\rang)\) finite dimensional k-IPVS and orthonormal basis, \(f:V\to V\), we've proved that \(f\) has an adjoint \(f^*\) and \([f^*]_B=\overline{[f]_B}^t =([f]_B)^*\)

Lemma

\(C,D\in M_n(\mathbb{C})\), then \((C\cdot D)^{*}=\overline{(C\cdot D)}^{t}=(\bar C\cdot\bar D)^t=\bar D^t\cdot \bar C^t=D^*\cdot C^*\)

\(C^{**}=C\)

Example of adjoint linear map

Let \(V=\mathbb{C}^{n\times n}\) and \(\lang A,B\rang=\text{tr}(\bar B^{t}\cdot A)\in k\) where \(A,B\in \mathbb{C}^{n\times n}\) note that inner product should be a number

Let \(M\in k^{n\times n}\) be a fixed matrix. Consider the k-linear map: \(f_M:V\to V\) s.t. \(f_M(A)=M\cdot A\)

Solve \(f_M^*\)

Given \(A,B\in \mathbb{C}^{n\times n}\), \(\lang f_{M}(A),B\rang =\lang M\cdot A,B\rang =\text{tr}(B^{*}\cdot M\cdot A)\), using lemma \(=\text{tr}((M^{*}\cdot B)^{*} \cdot A)=\text{tr}((f_{M^*}(B))^* \cdot A)\)

\(\lang A,(f_{M})^{*}(B)\rang=\text{tr}((f^*_M(B))^*\cdot A)\), thus take \((f_{M})^{*}=f_{M^*}:V\to V\) s.t. \(V\to V\) where \(B\mapsto M^*\cdot B\)

In conclusion: The adjoint linear map is charactered by Theorem
Let \(V=\mathbb{C}^{n\times 1}\). Given \(X,Y\in \mathbb{C}^{n\times 1}\), define \(\lang X,Y\rang=Y^{*}\cdot X\in \mathbb{C}^{1\times 1}\)

Let \(A\in \mathbb{C}^{n\times n}\) be a fixed matrix. Consider \(f_A:V\to V\) where \(f_A(X)=AX\)

Is it true that \((f_{A})^{*}=f_{A*}\) ?

\(\lang f_{A}(X),Y\rang=\lang AX,Y\rang =Y^*AX\)

\(\lang X,f_{A^*}(Y)\rang=(f_{A^*}(Y))^*\cdot X=(A^*\cdot Y)^*\cdot X=Y^*\cdot A\cdot X\)

Thus it's true
\(V=\R[x]\) be \(\R\)-vs infinite dimensional

\(\displaystyle\lang f(x),g(x)\rang =\int^{1}_{0}f(t)g(t)dt\) is inner product

Fix \(p(x)\in V\), \(\gamma_{P}:\R[x]\to \R[x]\) s.t. \(\gamma_P(f(x))=p(x)f(x)\)

Solve \((\gamma_P)^*\)

\(\displaystyle \lang \gamma_{P}(f(x)),g(x)\rang =\lang p(x)f(x),g(x)\rang =\int^{1} _{0}p(t)f(t)g(t)dt=\int^{1}_{0}f(t)p(t)g(t)dt=\lang f(x),p(x)\cdot g(x)\rang=\lang f(x),\gamma_P(g(x))\rang\)
\(V=\R[x]\) be \(\R\)-vs infinite dimensional

\(\displaystyle\lang f(x),g(x)\rang =\int^{1}_{0}f(t)g(t)dt\) is inner product
Let \(D:V\to V\) where \(D(p(x))=p'(x)\).

Given \(f(x),g(x)\in V\), \(\displaystyle\lang D(f(x)),g(x)\rang =\int^{1}_{0}f'(t)g(t)dt=f(t)g(t)\Big|^{1}_{0} -\int^{1}_{0}f(t)g'(t)dt\)

\({\displaystyle\langle f(x),D^{*}(g(x))\rangle=\int_0^1f(t)D^*(g(t))dt}\)

If exists, then \(\displaystyle f(t)g(t)\Big|_{0}^{1}-\int_{0}^{1}f(t)g^{\prime}(t)=\int_{0}^{1}f( t)D^{*}(g(t) ) dt\)

Then \(\displaystyle f(t)g(t)\Big|_{0}^{1}=\int_{0}^{1}f(t)D^{*}(g(t))dt+\int_{0}^{1}f( t)g^{\prime}(t)dt\)

Then \(\displaystyle \int^{1}_{0}f'(t)g(t)+f(t)g'(t)dt=\int_{0}^{1}f(t)D^{*}(g(t))dt+\int _{0}^{1}f( t)g^{\prime}(t)dt\)

Then \(\displaystyle \int^{1}_{0}f'(t)g(t)dt=\int_{0}^{1}f(t)D^{*}(g(t))dt\), then \(f'(t)g(t)=f(t)D^*(g(t))\)

Then \(D^{*}(g(t))=\frac{f'(t)}{f(t)}g(t)\) which is a contradiction since \(f(t)\) can be any

Theorem

Let \((V,\lang,\rang)\) be k-IPVS finite dimensional. Let \(f,g:V\to V\) be k-linear maps \(\alpha\in k\). Then

\((f+g)^{*}=f^{*}+g^{*}\)
\((\alpha f)^*=\bar \alpha f^*\)
\((f\circ g)^*=g^*\circ f^*\)
\((f^*)^*=f\)

Proof

\(\forall v,w\in V\), examine \(\lang(f+g)(v),w\rang =\lang v,(f^{*}+g^{*})(w)\rang\)
\(\forall v,w\in V\), examine \(\lang(\alpha\cdot f)(v),w\rang =\lang v,\bar \alpha\cdot f^*(w)\rang\)
....
....

Definition

Let \((V,\lang,\rang_{V})\) and \((W,\lang ,\rang_{W})\) be two k-IPVS, we will say that \(f:V\to W\) be k-linear preserves the inner products if \(\forall v_{1},v_{2}\in V\) s.t. \(\lang v_{1},v_{2}\rang_{V}=\lang f(v_{1}),f(v_{2})\rang_{W}\)
\(f:V\to W\) is an isomorphism of k-IPVS if \(f\) is a k-linear isomorphism and \(f\) preserves the inner products

Remarks

If \(f:V\to W\) preserves the inner product, then \(\forall v\in V,||v||_{V}^{2}=\langle v,v\rangle_{V}=\langle f(v),f(v)\rangle_{W} =||f(v)||_{W}^{2}\)

Thus \(||v||_{V}=||f(v)||_W\)

As a consequence, if \(f(v)=0_W\), then \(v=0_V\), so \(f\) is a monomorphism(injective and k-linear, \(\ker f=\{0\}\))
More generally, \(f:V\to W\) is an epimorphism( \(\text{Im}f=W\) ) and \(f\) preserves inner product, then \(f:V\to W\) is an isomorphism of k-IPVS

Moreover, \(f^{-1}:W\to V\) is such that, given \(w_1,w_2\in W\), then \(\lang f^{-1}(w_{1}),f^{-1}(w_{2})\rang_V=\lang f(f^{-1}(w_1),f(f^{-1}(w_2))\rang_W=\lang w_1,w_2\rang_W\)

Thus \(f^{-1}\) also preserves inner product

Theorem

Let \((V,\lang,\rang)\) and \((W,\lang,\rang)\) be k-IPVS s.t. \(\dim V=\dim W\) finite dimensional, \(f:V\to W\) be a k-linear map. The following statement are equivalent

\(f\) preserves inner product
\(f\) is an isomorphism of inner product vector space
\(f\) carries every orthonormal basis of \(V\) into an orthonormal basis of \(W\)
\(\exists\) an orthonormal basis of \(V\) s.t. \(f(B)\) is an orthonormal basis of \(W\)

Proof

\(1\Rightarrow 2\)) Since \(\dim V=\dim W\), then \(f\) is k-linear isomorphism. And since 1, then 2

\(2\Rightarrow 3\)) Given a basis \(B\) of \(V\) which is orthonormal, then \(\forall v_i,v_j\in B\) with \(i\neq j\), \(0=\lang v_{i},v_{j}\rang_{V}=\lang f(v_i),f(v_j)\rang_W\Rightarrow f(v_i)\perp f(v_j)\)

Also \(1=||v_{i}||=||f(v_i)||\Rightarrow f(B)\) is an orthonormal basis of \(W\)

\(3\Rightarrow4\)) Since \(V,W\) are finite dimension, then we know that there exists an orthonormal basis \(B\) of \(V\)

By 3), we know that \(f(B)\) is an orthonormal basis of \(W\)

\(4\Rightarrow 1\)) Let \(B\) be an orthonormal basis of \(V\) s.t. \(f(B)\) is an orthonormal basis of \(W\).

Given \(v_i,v_j\in B,i\neq j\) where \(\lang f(v_{i}),f(v_{j})\rang_{W}=\begin{cases} 0&\text{if }i\neq j\\ 1&\text{if }i=j \end{cases}\)

Given \(v\in V\), \(\exists v_1,...,v_n\in B\) and \(a_{1},...,a_{n}\in k,b_1,...,b_n\in k\) s.t. \(v=\sum^{n}_{i=1}a_{i}v_{i},\,\,\,v'=\sum^{n}_{j=1}b_{j}v_{j }\)

\(\lang v,v'\rang_{V}=\lang \sum^n_{i=1}a_iv_i,\sum^n_{j=1}b_jv_j\rang=\sum^n_{i,j=1}a_i\bar b_j\lang v_i,v_j\rang\)

Also \(\lang v_{i},v_{j}\rang_{V}= \begin{cases} 0 & \text{if }i\neq j \\ 1 & \text{if }i=j \end{cases}\), then \(=\sum^{n}_{i=1}a_{i}\cdot \bar b_{i}\)

And \(\langle f(v),f(v^{\prime})\rangle_{W}=\left\langle\sum_{i=1}^{n}a_{i}f(v_{i}),\sum _{j=1}^{n}b_{j}f(w_{j})\right\rangle_{W}=\sum_{i,j=1}^{n}a_{i}\bar{b_{j}}\langle f(v_{i}),f(w_{j})\rangle_{W}=\sum_{i=1}^{n}a_{i}\bar{b_{i}}\)

Thus \(f\) preserves inner product

Corollary

Let \(V,W\) be finite dimensional k-IPVS, then \(V\cong W(\text{isomor. preseving inner product})\;\char"27FA \;\dim V=\dim W\)

Proof

\(\implies)\) If \(V \cong W\) (as \(I.P.V.S\)), there is an isomorphism \(f: V \to W\), so \(\dim_k V = \dim_k W\).

\(\impliedby\)) Let \(B = \{v_1, ..., v_m\}\) be an O.N. basis of \(V\). Let \(B' = \{w_1, ..., w_m\}\) be an O.N. basis of \(W\).
Let \(f: V \to W\) be the \(k\)-linear map defined by \(f(v_i) = w_i\) for all \(1 \leq i \leq m\).

We know \(f\) sends the \(O.N\) basis \(B\) of \(V\) into the \(O.N\) basis \(B'\) of \(W\).

We use the eq. condition \((4)\) and \((2)\) in the \(TH^M\) to get that \(f\) is an Isomorphism of \(k\)-IPVS.

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