<Home>Teaching><MATH 115A (Spring 2023)>Worksheet 9

Worksheet 10

Problem 1

Let VV be a finite-dimensional inner product space, let T∈L(V)T \in \mathcal{L}\p{V}. Prove that λ\lambda is an eigenvalue of TT if and only if λ‾\conj{\lambda} is an eigenvalue of T∗T^*.

Solution.

Recall that a linear map S∈L(V)S \in \mathcal{L}\p{V} is invertible if and only if S∗S^* is invertible. Applying this to S=T−λid⁡VS = T - \lambda \operatorname{id}_V, we get

λ is an eigenvalue of T  ⟺  T−λid⁡V is not invertible  ⟺  (T−λid⁡V)∗ is not invertible  ⟺  T∗−λ‾id⁡V is not invertible  ⟺  λ‾ is an eigenvalue of T∗.\begin{aligned} \lambda\text{ is an eigenvalue of }T &\iff T - \lambda \operatorname{id}_V\text{ is not invertible} \\ &\iff \p{T - \lambda \operatorname{id}_V}^*\text{ is not invertible} \\ &\iff T^* - \conj{\lambda} \operatorname{id}_V\text{ is not invertible} \\ &\iff \conj{\lambda}\text{ is an eigenvalue of }T^*. \end{aligned}

Problem 7

Let VV be a complex inner product space, let T∈L(V)T \in \mathcal{L}\p{V}. Define T1=T+T∗2T_1 = \frac{T + T^*}{2} and T2=iT∗−T2T_2 = i\frac{T^* - T}{2}.

  1. Prove that T1T_1 and T2T_2 are self-adjoint and that T=T1+iT2T = T_1 + iT_2.
  2. Suppose that T=U1+iU2T = U_1 + iU_2 for U1,U2∈L(V)U_1, U_2 \in \mathcal{L}\p{V} self-adjoint. Prove that U1=T1U_1 = T_1 and U2=T2U_2 = T_2.
  3. Prove that TT is normal if and only if T1T2=T2T1T_1T_2 = T_2T_1.
Solution.

  1. Recall that T∗∗=TT^{**} = T. Then we check

    (T1)∗=(T+T∗2)∗=T∗+T2=T1(T2)∗=(iT∗−T2)∗=−iT−T∗2=T2.\begin{gathered} \p{T_1}^* = \p{\frac{T + T^*}{2}}^* = \frac{T^* + T}{2} = T_1 \\ \p{T_2}^* = \p{i\frac{T^* - T}{2}}^* = -i \frac{T - T^*}{2} = T_2. \end{gathered}

    Note that for T2T_2, we used the fact that i‾=−i\conj{i} = -i. Lastly, because i2=−1i^2 = -1,

    T1+iT2=T+T∗2−T∗−T2=T.T_1 + iT_2 = \frac{T + T^*}{2} - \frac{T^* - T}{2} = T.

  2. Note that since (U1)∗=U1\p{U_1}^* = U_1 and (U2)∗=U2\p{U_2}^* = U_2, we have

    {T=U1+iU2,T∗=U1−iU2.\begin{cases} T = U_1 + iU_2, \\ T^* = U_1 - iU_2. \end{cases}

    Adding the equations, we get

    T+T∗=2U1  ⟹  U1=T+T∗2=T1.T + T^* = 2U_1 \implies U_1 = \frac{T + T^*}{2} = T_1.

    Similarly, subtracting them gives

    T−T∗=2iU2  ⟹  U2=T−T∗2i⋅ii=iT∗−T2=T2.T - T^* = 2iU_2 \implies U_2 = \frac{T - T^*}{2i} \cdot \frac{i}{i} = i\frac{T^* - T}{2} = T_2.

  3. We need to show that TT∗=T∗TTT^* = T^*T. We can just expand

    TT∗=(T1+iT2)(T1−iT2)=T12+iT2T1−iT1T2+T22=T12+iT1T2−iT2T1+T22=(T1−iT2)(T1+iT2)=T∗T.\begin{aligned} TT^* &= \p{T_1 + iT_2}\p{T_1 - iT_2} \\ &= T_1^2 + iT_2T_1 - iT_1T_2 + T_2^2 \\ &= T_1^2 + iT_1T_2 - iT_2T_1 + T_2^2 \\ &= \p{T_1 - iT_2}\p{T_1 + iT_2} \\ &= T^*T. \end{aligned}

    Note that in the third-to-last equality, we used T1T2=T2T1T_1T_2 = T_2T_1.