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5.28 Theorems of self-adjoint map

\(f: V \rightarrow V / f^{*} \circ f = f \circ f^{*}\) normal operators

Aim: \(\dim_k V < \infty\), if \(f\) is normal, then \(f\) is diagonalizable and \(\exists\) an orthonormal basis of eigenvectors.

Theorem 1

\((V,\lang,\rang)\) be k-IPVS, let \(f:V\to V\) be a self-adjoint and k-linear map. Suppose \(\lambda\in k\) is an eigenvalue of \(f\)

Then

  1. \(\lambda \in \R\)​

  2. If \(v\) is an eigenvector of eigenvalue \(\lambda\) and \(\mu\) is another eigenvalue of \(f\) s.t. \(\lambda \neq u\) and \(w\) is an eigenvector with eigenvalue \(\mu\), then \(v\perp w\)​

Proof

  1. Let \(\lambda\) be an eigenvalue of \(f\), then suppose \(v\neq 0\), \(v\) is eigenvector of eigenvalue \(\lambda\)​

    Then \(\lambda \cdot ||v||^{2}=\lambda \lang v,v\rang=\lang \lambda v,v\rang=\lang f(v),v\rang\) and since \(f\) is self-adjoint, then \(f(v)=f^{*}(v)\), then \(=\lang v,\lambda v\rang=\bar \lambda\lang v,v\rang =\bar\lambda ||v||^2\)​

    Since \(||v||\neq 0\), we have \(\lambda =\bar\lambda\) then \(\lambda\in \R\)​

  2. \(\lambda ,\mu\in \R\) and \(\lambda\neq \mu\) where \(f(v)=\lambda v\) and \(f(w)=\mu w\) where \(v,w\neq 0_V\), then \(\lambda \lang v,w\rang=\lang \lambda v,w\rang =\lang f(v),w\rang=\lang v,f^*(w)\rang=\lang v,f(w)\rang=\lang v,\mu w\rang=\mu \lang v,w\rang\) since \(\bar u=u\)​

    Then \((\lambda -\mu)\lang v,w\rang=0\), since \(\lambda \neq \mu\), then \(\lang v,w\rang=0\)​

Theorem 2

Suppose \((V,\lang,\rang)\) is a finite dimension k-IPVS. Let \(f:V\to V\) be a self-adjoint linear map.

Then \(\exists v\in V:v\neq 0_{V}:v\) is an eigenvector of \(f\) corresponding to an eigenvalue \(\lambda\)

Proof

Denote \(n=\dim V\), \(f:V\to V,f=f^{*}\), let \(B\) be an orthonormal basis of \(V\)

Denote \(A=[f]_B\), since \(f=f^*\), we have \(A=[f]_{B}=[f^{*}]_{B}=A^*\), so \(A^*=A\)

Let \(W=\mathbb{C}^{n\times1}\) with inner product \(\lang X,Y\rang =\underbrace{Y^{*}}_{\in \mathbb{C}^{1\times n}}\underbrace{X}_{\in\mathbb{C}^{n\times 1}}\in \mathbb{C}\)

Define a \(\mathbb{C}\)-linear map, \(U:W\to W\) by \(U(X)=AX\), we know that \(U=U^{*}\) since the following

\(\forall X,Y\in W\), \(\lang U(X),Y\rang =\lang AX,Y\rang=Y^*AX=Y^*A^*X=(AY)^*X=\lang X,AY\rang =\lang X,U(Y)\rang\)

Then \(\chi_{U}(x)\in\mathbb{C}[x]\) and \(\deg=n>0\), so it has at least one root \(\lambda\in\mathbb{C}\), that is \(\chi_{U}(\lambda)=0=\det(\lambda\cdot Id_{n}-A)=\det(\lambda\cdot Id_{n}-[U]_{E} )\)

Thus \(\lambda Id_n-A\) is not invertible, that is \(\exists X\in \mathbb{C}^{n\times 1}\) s.t. \((\lambda Id_{n}-A)\cdot X=0_{W}\iff \lambda Id_n\cdot X=\lambda X=AX\)

Thus \(\lambda\) is an eigenvalue of \(U\) with eigenvector \(X\), since we know \(U\) is self-adjoint, by theorem 1, we conclude that \(\lambda\in\R\)

Moreover, \(\chi_U(x)=\chi_A(x)\), now we have \(\chi_{A}(x)=\chi_{f}(x)\), so \(\lambda\) is a root of \(\chi_f(x)\), that is \(\lambda\) is an eigenvalue of \(f\), so there exists \(v\neq 0_V\) s.t. \(f(v)=\lambda v\)

  1. why \(A=[U]_{E}\). Write \(X\) is basis

  2. why \(\lambda Id_n-A\) is not invertible, that is \(\exists X\in \mathbb{C}^{n\times 1}\) s.t. \((\lambda Id_{n}-A)\cdot X=0_{W}\iff \lambda Id_n\cdot X=\lambda X=AX\).

    Because this is says that \(X\) is in the kernel

  3. why \(\chi_U(x)=\chi_A(x)\). Since \(\det(\lambda\cdot Id_{n}-A)=\det(\lambda\cdot Id_{n}-[U]_{E})\)

Remark

If \(V\) is not finite dimension, then \(f\)(self-adjoint) may have eigenvalue

Example

\(V=\{f:[0,1]\to \R:f\text{ is continuous}\}\), \(\displaystyle\lang f,g\rang=\int^{1}_{0}f(t)g(t)dt\) where \(V\) is infinite dimension

Consider \(\phi:V\to V\) defined by \(\phi(f)(t)=t\cdot f(t)\)

  • \(\phi\) is self-adjoint: Given \(f,g\in V\), \(\displaystyle\lang \phi(f),g\rang=\int^1_0\phi(f)(t)\cdot g(t)dt=\int^1_0f(t)\cdot t\cdot g(t)dt=\lang f,\phi(g)\rang\)​

Suppose \(\lambda\in \R\) s.t. \(\lambda\) is an eigenvalue of \(\phi\), let \(f\in V\) be an eigenvector with eigenvalue \(\lambda\) : \(t\cdot f(t)=\phi(f)(t)=\lambda f(t)\)

Then \((\lambda -t)\cdot f(t)=0\), then since \(f\) is continuous, then \(f(t)=0\), contradiction

Theorem 3

\((V,\lang,\rang)\) k-IPVS, finite dimensional. \(f:V\to V\) be \(k\)-linear. Suppose \(W\subseteq V\) is a subspace s.t. \(W\) is \(f\)-invariant. Then \(W^\perp\) is \(f^*\)-invariant.

Proof

Let \(v\in W^{\perp}\), we want to prove that \(f^*(v)\in W^\perp\). Let \(w\in W\), \(\lang w,f^{*}(v)\rang=\lang \underbrace{f(w)}_{\in W},\underbrace{v}_{\in W^\perp}\rang\) because \(w\in W\) and \(W\) is \(f\)-invariant, then \(=0\)

Remark

If \(V\) is not finite dimension but \(f^*\) exists, then the result is also true.

Theorem 4

Let \((V,\lang,\rang)\) a k-IPVS, finite dimension. Let \(f:V\to V\) be k-linear and self-adjoint.

Then there exists an orthonormal basis \(B\) of \(V\), \(B=\{v_1,...,v_n\}\) s.t. \(v_i\) is eigenvector of \(f\), \(\forall i,1\leq i\leq n\)

Proof

Let \(n=\dim V\). By theorem 2: \(\exists \lambda\in \R\) : \(\lambda\) is an eigenvector of \(f\).

Let \(0_{V}\neq v\) be an eigenvector of eigenvalue \(\lambda\). Let \(v_1=\frac{v}{||v||}\), then \(v_1\) is an eigenvector of \(f\) with eigenvalue \(\lambda\) and \(||v_1||=1\)

If \(n=1\), there is nothing to do.

If not, we will proceed by induction \(n\). Let \(W=\lang v_1\rang\). \(W\) is \(f\)-invariant since it's an eigenvector

Here \(W^\perp\) is \(f^*=f\) invariant by theorem 3 and \(\dim W^\perp=n-1<n\). Restrict the inner product of \(V\) to \(W^\perp\).

So \((W^{\perp},\lang,\rang_{V}\Big|_{W^\perp})\) is a k-IPVS, let \(\tilde f=f\Big|_{W^\perp}: W^{\perp}\to W^{\perp}\) and \(\tilde f\) is also self-adjoint.

Then use inductive hypothesis, which says that there is an orthonormal basis \(\tilde B\) of \(W^\perp\) where all the vectors are eigenvectors of \(\tilde f\)

\(\tilde B=\{v_2,...,v_n\}\), now \(B=\{v_1\}\cup \tilde B\) is such that \(||v_i||=1,\forall i\) and \(v_i\) is an eigenvector of \(f\), \(\forall i\)

It is also orthogonal since \(v_{1}\in W\), \(v_i\in W^\perp,\forall i,2\leq i\leq n\)

Corollary

Let \(A\in M_n(k)\) s.t. \(A=A^*\). Then \(\exists P\in Gl_n(k)\) s.t.

  1. \(P\) is unitary (\(P^*=P^{-1}\)​)

    Change of basis matrix satisfy this exercise

  2. \(P^{-1}AP\) is diagonal

Now let's consider \(f\) normal

Theorem 5

\((V,\lang,\rang)\) be k-IPVS, finite dimension. \(f:V\to V\) be \(k\)-linear normal (\(f^{*}\circ f=f\circ f^*\))

Let \(v\in V\), \(v\) is an eigenvector of \(f\) with eigenvalue \(\lambda\)​\(\iff\)​\(v\) is an eigenvector of \(f^*\) with eigenvalue \(\bar\lambda\)

Proof

Let \(g:V\to V\) be k-linear and normal. Given \(v\in V\), \(||g(v)||^{2}=\lang g(v),g(v)\rang=\lang v,g^*\circ g(v)\rang\)

Since \(g\) is normal, then \(=\langle v,g\circ g^{*}(v)\rangle=\langle g^{*}(v),g^{*}(v)\rangle=||g^{*}(v)||^{2}\)

Let \(\lambda\in k\), let \(g=f-\lambda\cdot Id_{V}\), then \(g^{*}=f^{*}-(\lambda Id_{V})^{*}=f^{*}-\bar\lambda Id_{V}\)

Given \(v_1,v_2\in V\), \(\lang g(v_{1}),v_{2}\rang=\lang v_1,g^*(v_2)\rang\). Also \(g^{*}\circ g=g\circ g^*\) exercise

Then \(g\) is normal, \(||g(v)||^2=||g^*(v)||^2\). So \(v\) is eigenvector of \(f\) with eigenvalue \(\lambda\)​\(\iff\)​\(v\) is an eigenvector of \(f^*\) with eigenvalue \(\bar \lambda\)

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