Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 41

Let

${\displaystyle \varphi \colon V\longrightarrow V}$

be given, and let ${\displaystyle {}w\in V}$ be fixed. Then, the mapping

${\displaystyle V\longrightarrow {\mathbb {K} },v\longmapsto \left\langle \varphi (v),w\right\rangle ,}$

is a linear form on ${\displaystyle {}V}$. Therefore, there exists (due to Corollary 38.6 in the real case; for the complex case see Exercise 41.16 ) a right gradient ${\displaystyle {}r=:{\hat {\varphi }}(w)}$ in ${\displaystyle {}V}$ (uniquely determined by ${\displaystyle {}\varphi }$ and ${\displaystyle {}w}$) fulfilling

${\displaystyle {}\left\langle v,{\hat {\varphi }}(w)\right\rangle =\left\langle \varphi (v),w\right\rangle \,.}$

We have to show that the assignment

${\displaystyle w\longmapsto {\hat {\varphi }}(w)}$

is linear. We have

${\displaystyle {}{\begin{aligned}\left\langle v,{\hat {\varphi }}(w_{1}+w_{2})\right\rangle &=\left\langle \varphi (v),w_{1}+w_{2}\right\rangle \\&=\left\langle \varphi (v),w_{1}\right\rangle +\left\langle \varphi (v),w_{2}\right\rangle \\&=\left\langle v,{\hat {\varphi }}(w_{1})\right\rangle +\left\langle v,{\hat {\varphi }}(w_{2})\right\rangle \\&=\left\langle v,{\hat {\varphi }}(w_{1})+{\hat {\varphi }}(w_{2})\right\rangle .\end{aligned}}}$

As this holds for all ${\displaystyle {}v\in V}$, we have

${\displaystyle {}{\hat {\varphi }}(w_{1}+w_{2})={\hat {\varphi }}(w_{1})+{\hat {\varphi }}(w_{2})\,.}$

Moreover,

${\displaystyle {}{\begin{aligned}\left\langle v,{\hat {\varphi }}(sw)\right\rangle &=\left\langle \varphi (v),sw\right\rangle \\&={\overline {s}}\left\langle \varphi (v),w\right\rangle \\&={\overline {s}}\left\langle v,{\hat {\varphi }}(w)\right\rangle \\&=\left\langle v,s{\hat {\varphi }}(w)\right\rangle .\end{aligned}}}$

As this holds for all ${\displaystyle {}v\in V}$, we get

${\displaystyle {}{\hat {\varphi }}(sw)=s{\hat {\varphi }}(w)\,.}$

Let ${\displaystyle {}u_{1},\ldots ,u_{n}}$ be the orthonormal basis, and let

${\displaystyle {}M={\left(a_{ij}\right)}_{ij}\,}$

and

${\displaystyle {}N={\left(b_{ij}\right)}_{ij}\,}$

be the matrices of ${\displaystyle {}\varphi }$ and of ${\displaystyle {}{\hat {\varphi }}}$ with respect to this basis. This means that we have in particular

${\displaystyle {}\varphi (u_{i})=\sum _{k=1}^{n}a_{ki}u_{k}\,}$

and

${\displaystyle {}{\hat {\varphi }}(u_{i})=\sum _{k=1}^{n}b_{ki}u_{k}\,.}$

Due to adjointness, the relation

${\displaystyle {}{\begin{aligned}a_{ji}&=\sum _{k=1}^{n}a_{ki}\left\langle u_{k},u_{j}\right\rangle \\&=\left\langle \sum _{k=1}^{n}a_{ki}u_{k},u_{j}\right\rangle \\&=\left\langle \varphi (u_{i}),u_{j}\right\rangle \\&=\left\langle u_{i},{\hat {\varphi }}(u_{j})\right\rangle \\&=\left\langle u_{i},\sum _{k=1}^{n}b_{kj}u_{k}\right\rangle \\&=\left\langle u_{i},b_{ij}u_{i}\right\rangle \\&={\overline {b_{ij}}}\end{aligned}}}$

holds. That is

${\displaystyle {}{{\overline {N}}^{\text{tr}}}=M\,,}$

and vice versa.

If ${\displaystyle {}\varphi }$ is self-adjoint, then the statement follows from Lemma 41.6 . If ${\displaystyle {}\varphi }$ is described by an Hermitian matrix ${\displaystyle {}M}$ with respect to a orthonormal basis, then, again by Lemma 41.6 , the adjoint endomorphism ${\displaystyle {}{\hat {\varphi }}}$ is described by

${\displaystyle {}{{\overline {M}}^{\text{tr}}}=M\,}$

with respect to this basis. Therefore, it coincides with ${\displaystyle {}\varphi }$.

1. Let ${\displaystyle {}v\in U^{\perp }}$, and ${\displaystyle {}u\in U}$. Because of the invariance of ${\displaystyle {}U}$, we have ${\displaystyle {}\varphi (u)\in U}$. Therefore,

   ${\displaystyle {}\left\langle \varphi (v),u\right\rangle =\left\langle v,\varphi (u)\right\rangle =0\,.}$

   This means that ${\displaystyle {}\varphi (v)}$ is orthogonal to ${\displaystyle {}U}$; thus, it belongs to ${\displaystyle {}U^{\perp }}$, and this means the invariance.

2. This is only relevant in case ${\displaystyle {}{\mathbb {K} }=\mathbb {C} }$. Let ${\displaystyle {}\lambda \in \mathbb {C} }$ be an eigenvalue, and let ${\displaystyle {}v\in V}$ be an eigenvector, that is,

   ${\displaystyle {}\varphi (v)=\lambda v\,.}$

   We may assume that this eigenvector is normed. We have

   ${\displaystyle {}\lambda =\left\langle \lambda v,v\right\rangle =\left\langle \varphi (v),v\right\rangle =\left\langle v,\varphi (v)\right\rangle =\left\langle v,\lambda v\right\rangle ={\overline {\lambda }}\,;}$

   therefore, ${\displaystyle {}\lambda }$ is real.

3. Let ${\displaystyle {}v_{1}}$ be an eigenvector for the eigenvalue ${\displaystyle {}\lambda _{1}}$, and let ${\displaystyle {}v_{2}}$ be an eigenvector for the eigenvalue ${\displaystyle {}\lambda _{2}\neq \lambda _{1}}$. Then

   ${\displaystyle {}\lambda _{1}\left\langle v_{1},v_{2}\right\rangle =\left\langle \lambda _{1}v_{1},v_{2}\right\rangle =\left\langle \varphi (v_{1}),v_{2}\right\rangle =\left\langle v_{1},\varphi (v_{2})\right\rangle =\left\langle v_{1},\lambda _{2}(v_{2})\right\rangle ={\overline {\lambda _{2}}}\left\langle v_{1},v_{2}\right\rangle =\lambda _{2}\left\langle v_{1},v_{2}\right\rangle \,.}$

   This is only possible for

   ${\displaystyle {}\left\langle v_{1},v_{2}\right\rangle =0\,.}$

4. We may assume that ${\displaystyle {}V={\mathbb {K} }^{n}}$, endowed with the standard inner product. In case ${\displaystyle {}{\mathbb {K} }=\mathbb {C} }$, the statement is known; so suppose that ${\displaystyle {}{\mathbb {K} }=\mathbb {R} }$. We may consider the mapping also as a mapping from ${\displaystyle {}\mathbb {C} ^{n}}$ to ${\displaystyle {}\mathbb {C} ^{n}}$. This map is again self-adjoint, and the characteristic polynomial does not change. Therefore, it splits into linear factors, and the zeroes are real by (2).

We do induction over the dimension of ${\displaystyle {}V}$. Because of Lemma 41.10   (4), ${\displaystyle {}\varphi }$ has an eigenvector ${\displaystyle {}v}$; we may assume that this vector is normed. Due to Lemma 41.10   (1), the orthogonal complement

${\displaystyle {}W={\mathbb {K} }v^{\perp }\,}$

is also invariant. Hence, we have a direct sum decomposition

${\displaystyle {}V={\mathbb {K} }v\oplus W\,.}$

The restriction of ${\displaystyle {}\varphi }$ to ${\displaystyle {}W}$ is also self-adjoint. Therefore, the induction hypothesis yields the claim.

1. We have

   ${\displaystyle {}{\begin{aligned}\psi _{\varphi }(av_{1}+bv_{2},w)&=\left\langle \varphi (av_{1}+bv_{2}),w\right\rangle \\&=\left\langle a\varphi (v_{1})+b\varphi (v_{2}),w\right\rangle \\&=a\left\langle \varphi (v_{1}),w\right\rangle +b\left\langle \varphi (v_{2}),w\right\rangle \\&=a\psi _{\varphi }(v_{1},w)+b\psi _{\varphi }(v_{2},w)\end{aligned}}}$

   and

   ${\displaystyle {}{\begin{aligned}\psi _{\varphi }(v,aw_{1}+bw_{2})&=\left\langle \varphi (v),aw_{1}+bw_{2}\right\rangle \\&={\overline {a}}\left\langle \varphi (v),w_{1}\right\rangle +{\overline {b}}\left\langle \varphi (v),w_{2}\right\rangle \\&={\overline {a}}\psi _{\varphi }(v,w_{1})+{\overline {b}}\psi _{\varphi }(v,w_{2}),\end{aligned}}}$

   that is, the assignment is linear in the first component, and antilinear in the second component. Therefore, ${\displaystyle {}\Psi _{\varphi }}$ is a sesquilinear form.

2. The linearity follows from the linearity of the inner product in the first component. In the finite-dimensional case, we have on the left-hand side and on the right-hand side vector spaces of the dimension ${\displaystyle {}{\left(\dim _{K}{\left(V\right)}\right)}^{2}}$; therefore, it is enough to show injectivity. If ${\displaystyle {}\Psi _{\varphi }=0}$ is the zero form, then ${\displaystyle {}\left\langle \varphi (v),w\right\rangle =0}$ for all ${\displaystyle {}v,w}$. In particular, ${\displaystyle {}\left\langle \varphi (v),\varphi (v)\right\rangle =0}$, which implies ${\displaystyle {}\varphi (v)=0}$.
3. If ${\displaystyle {}\varphi }$ is not bijective, then let ${\displaystyle {}v\in \operatorname {kern} \varphi }$, ${\displaystyle {}v\neq 0}$. Then, ${\displaystyle {}\Psi _{\varphi }(v,-)}$ is the zero mapping in the second component, and the form is degenerate. To prove the converse, suppose that ${\displaystyle {}\Psi _{\varphi }(-,-)}$ is degenerate. Then there exists a vector ${\displaystyle {}v\in V}$, ${\displaystyle {}v\neq 0}$, such that ${\displaystyle {}\left\langle \varphi (v),-\right\rangle }$ is the zero-mapping. Since an inner product is nondegenerate, this implies ${\displaystyle {}\varphi (v)=0}$, and ${\displaystyle {}\varphi }$ is not bijective.
4. In the self-adjoint case, we have

   ${\displaystyle {}\Psi _{\varphi }(v,w)=\left\langle \varphi (v),w\right\rangle =\left\langle v,\varphi (w)\right\rangle ={\overline {\left\langle \varphi (w),v\right\rangle }}={\overline {\Psi _{\varphi }(w,v)}}\,.}$

   The converse follows from

   ${\displaystyle {}\left\langle \varphi (v),w\right\rangle =\Psi _{\varphi }(v,w)={\overline {\Psi _{\varphi }(w,v)}}={\overline {\left\langle \varphi (w),v\right\rangle }}=\left\langle v,\varphi (w)\right\rangle \,.}$