Exercise 9.B.5

Proof. Suppose $V$ is a real inner product space and is self-adjoint. Then $T_{}$

C$isaself-adjointoperatoronthecomplexinnerproductspace$V_C$definedinExercise3.BytheComplexSpectralTheorem,thereisanorthonormalbasis$e_1 + if_1, …, e_n + if_n$of$V_C$consistingofeigenvectorsof$T_C$.By7.13,theeigenvaluesof$T_C$areallreal.Thus,thereare$${\lambda }_1,\dots ,{\lambda }_n\in ℝ$ such that

C(e_j + if_j) = ${\lambda }_j{e}_j+i{\lambda }_{j}T{f}_j$$$

for each $j$. The equation above shows that the $e$’s and $f$’s are eigenvectors of $T$. Fix $j\in \{1,\dots ,n\}$. Let $e_1^{\prime }+i{f}_1^{\prime },\dots ,e_{d({\lambda }_j)}^{\prime }+i{f}_{d({\lambda }_j)}^{\prime }$ denote the basis vectors that correspond to ${\lambda }_j$, where $d({\lambda }_j)=\dim <!--FUNCTION\; APPLICATION-->E({\lambda }_j,T_{}$

C)$.Thenthelist$

is in $E({\lambda }_j,T)$. Since $E(\lambda ,T_{}$

C)$iscontainedinthecomplexspanofthelistabove,itfollowsthatthedimensionofthecompelxspanofthelistaboveisatleast$d(${\lambda }_j)$. Thus, by 9.4, the dimension of the real span of the list above is at least $d({\lambda }_j)$. Hence $\dim <!--FUNCTION\; APPLICATION-->E({\lambda }_j,T)\ge d({\lambda }_j)$. Because

and

we must have $\dim <!--FUNCTION\; APPLICATION-->E({\lambda }_j,T)=d({\lambda }_j)$ for each $j$. The sum of eigenspaces corresponding to different eigenvalues is a direct sum. The equation above thus shows that the sum of the eigenspaces of $T$ has the same dimension as $V$. This means that

For each eigenspace of $T$, we can select an orthonormal basis. Putting these bases together, we get a basis of $V$ (by the equation above) consisting of eigenvectors of $T$. By 7.22, this basis is orthonormal. □