Linear-Algebra-Done-Right · Unit 18 · Video 3 · Interactive Practice

Self-Adjoint Operators: Real Eigenvalues and Real ⟨Tv,v⟩\langle Tv, v \rangle

IKey Formulas

FormulaNameWhen it holds
T=T∗  ⟺  ⟨Tv,w⟩=⟨v,Tw⟩T = T^* \iff \langle Tv, w\rangle = \langle v, Tw\rangle for all v,wv, wSelf-adjointT∈L(V)T \in \mathcal{L}(V)
M(T)=M(T)∗\mathcal{M}(T) = \mathcal{M}(T)^*Matrix test (conjugate transpose)Orthonormal basis
λ∥v∥2=⟨Tv,v⟩=⟨v,Tv⟩=λ‾∥v∥2\lambda\|v\|^2 = \langle Tv, v\rangle = \langle v, Tv\rangle = \overline{\lambda}\|v\|^2Eigenvalues are realT=T∗T = T^*, Tv=λvTv = \lambda v, v≠0v \neq 0
T=T∗  ⟺  ⟨Tv,v⟩∈RT = T^* \iff \langle Tv, v\rangle \in \mathbf{R} for every vvReal-value testF=C\mathbf{F} = \mathbf{C} only

Key Insight: T↦T∗T \mapsto T^* plays the role of z↦z‾z \mapsto \overline{z}: a self-adjoint operator (T=T∗T = T^*) is the counterpart of a real number (z=z‾z = \overline{z}).

IIVisualization 1 — Where Does ⟨Tv,v⟩\langle Tv, v\rangle Land?

On C2\mathbf{C}^2, every value ⟨Tv,v⟩\langle Tv, v\rangle lies on the real axis exactly when T=T∗T = T^*.

IIIVisualization 2 — Why Eigenvalues Are Real

The chain λ∥v∥2=⟨Tv,v⟩=⟨v,Tv⟩=λ‾∥v∥2\lambda\|v\|^2 = \langle Tv, v\rangle = \langle v, Tv\rangle = \overline{\lambda}\|v\|^2 forces λ\lambda to equal its mirror image λ‾\overline{\lambda}.

IVVisualization 3 — The Real Identity Needs T=T∗T = T^*

Over R\mathbf{R}, ⟨Tu,w⟩=14(⟨T(u+w),u+w⟩−⟨T(u−w),u−w⟩)\langle Tu, w\rangle = \tfrac{1}{4}\big(\langle T(u+w), u+w\rangle - \langle T(u-w), u-w\rangle\big) holds for self-adjoint TT.

💡 For self-adjoint TT this identity turns ⟨Tv,v⟩=0\langle Tv, v\rangle = 0 for all vv into ⟨Tu,w⟩=0\langle Tu, w\rangle = 0 for all u,wu, w, hence T=0T = 0; the operator (02−20)\begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} has ⟨Tv,v⟩=0\langle Tv, v\rangle = 0 for every v∈R2v \in \mathbf{R}^2 yet T≠0T \neq 0.

VQuiz Questions

Problem 1 · The Matrix Test

Given: C2\mathbf{C}^2 with its standard inner product, and operators given by their matrices with respect to the standard basis. Which operator is self-adjoint?

✅ Correct! Its conjugate transpose swaps 1−3i1-3i and 1+3i1+3i across the diagonal and conjugates each into the other, so the matrix equals its own conjugate transpose.
❌ Not quite. This matrix equals its transpose, but the conjugate transpose has bottom-left entry 1−3i‾=1+3i≠1−3i\overline{1-3i} = 1+3i \neq 1-3i. Over C\mathbf{C} the test needs conjugation, not just symmetry.
❌ Not quite. The off-diagonal entries pair up correctly, but a diagonal entry must equal its own conjugate: 4i‾=−4i≠4i\overline{4i} = -4i \neq 4i. Diagonal entries of a self-adjoint matrix are real.
❌ Not quite. The bottom-left entry should be the conjugate of the top-right one, 1−3i‾=1+3i\overline{1-3i} = 1+3i, not its negative-conjugate −1+3i-1+3i. Here the conjugate transpose has top-right entry −1+3i‾=−1−3i≠1−3i\overline{-1+3i} = -1-3i \neq 1-3i.
Show solution

The standard basis is orthonormal, so TT is self-adjoint exactly when M(T)=M(T)∗\mathcal{M}(T) = \mathcal{M}(T)^*: swap rows and columns, then conjugate every entry.

(41−3i1+3i−2)∗=(4‾1+3i‾1−3i‾−2‾)=(41−3i1+3i−2).\begin{pmatrix} 4 & 1-3i \\ 1+3i & -2 \end{pmatrix}^* = \begin{pmatrix} \overline{4} & \overline{1+3i} \\ \overline{1-3i} & \overline{-2} \end{pmatrix} = \begin{pmatrix} 4 & 1-3i \\ 1+3i & -2 \end{pmatrix}.

The other three fail: (a) has conjugate transpose with bottom-left entry 1+3i1+3i; (b) has 4i‾=−4i\overline{4i} = -4i on the diagonal; (d) has conjugate transpose with top-right entry −1−3i-1-3i.

So the self-adjoint operator is (c). In general: real diagonal entries, and entry (k,j)(k, j) equal to entry (j,k)‾\overline{\text{entry } (j, k)}.

Problem 2 · Eigenvalues of a Self-Adjoint Operator

Given: T∈L(C2)T \in \mathcal{L}(\mathbf{C}^2) with M(T)=(12+i2−i3)\mathcal{M}(T) = \begin{pmatrix} 1 & 2+i \\ 2-i & 3 \end{pmatrix} (standard basis). Find the eigenvalues of TT.

✅ Correct! λ2−4λ−2=0\lambda^2 - 4\lambda - 2 = 0 gives λ=2±6\lambda = 2 \pm \sqrt{6}, both real, as they must be for a self-adjoint operator.
❌ Not quite. Reading eigenvalues off the diagonal needs an upper-triangular matrix, and here the bottom-left entry 2−i2 - i is nonzero. Indeed det⁡(M(T)−I)=0⋅2−5=−5≠0\det\big(\mathcal{M}(T) - I\big) = 0 \cdot 2 - 5 = -5 \neq 0, so 11 is not an eigenvalue.
❌ Not quite. The matrix equals its conjugate transpose, so TT is self-adjoint and every eigenvalue is real; 2±i2 \pm i is ruled out before any computation.
❌ Not quite. This comes from (2+i)(2−i)=3(2+i)(2-i) = 3. In fact (2+i)(2−i)=4−i2=5(2+i)(2-i) = 4 - i^2 = 5, so the determinant is 3−5=−23 - 5 = -2, not 00.
Show solution

M(T)∗=M(T)\mathcal{M}(T)^* = \mathcal{M}(T), so TT is self-adjoint and its eigenvalues are real. The eigenvalues are the roots of

det⁡(1−λ2+i2−i3−λ)=(1−λ)(3−λ)−(2+i)(2−i)=λ2−4λ+3−5=λ2−4λ−2.\det\begin{pmatrix} 1-\lambda & 2+i \\ 2-i & 3-\lambda \end{pmatrix} = (1-\lambda)(3-\lambda) - (2+i)(2-i) = \lambda^2 - 4\lambda + 3 - 5 = \lambda^2 - 4\lambda - 2. λ=4±16+82=2±6.\lambda = \frac{4 \pm \sqrt{16 + 8}}{2} = 2 \pm \sqrt{6}.

Check: the sum is 4=1+34 = 1 + 3 (the trace) and the product is 4−6=−24 - 6 = -2 (the determinant). Both eigenvalues, 2+6≈4.4492 + \sqrt{6} \approx 4.449 and 2−6≈−0.4492 - \sqrt{6} \approx -0.449, are real.

Problem 3 · Computing ⟨Tv,v⟩\langle Tv, v\rangle on C2\mathbf{C}^2

Given: T∈L(C2)T \in \mathcal{L}(\mathbf{C}^2) with M(T)=(3i−i1)\mathcal{M}(T) = \begin{pmatrix} 3 & i \\ -i & 1 \end{pmatrix} (standard basis), the inner product ⟨(z1,z2),(w1,w2)⟩=z1w1‾+z2w2‾\langle (z_1, z_2), (w_1, w_2)\rangle = z_1\overline{w_1} + z_2\overline{w_2}, and v=(1+i,2)v = (1+i, 2). Find ⟨Tv,v⟩\langle Tv, v\rangle.

✅ Correct! Tv=(3+5i,3−i)Tv = (3+5i, 3-i) and ⟨Tv,v⟩=(3+5i)(1−i)+(3−i)⋅2=14\langle Tv, v\rangle = (3+5i)(1-i) + (3-i)\cdot 2 = 14, a real number, as the real-value test predicts for this self-adjoint TT.
❌ Not quite. This multiplies by 1+i1+i instead of 1+i‾=1−i\overline{1+i} = 1 - i: the second slot must be conjugated. A non-real value is also a warning sign here, since M(T)=M(T)∗\mathcal{M}(T) = \mathcal{M}(T)^* makes ⟨Tv,v⟩\langle Tv, v\rangle real for every vv.
❌ Not quite. This multiplies vv by the transpose (3−ii1)\begin{pmatrix} 3 & -i \\ i & 1 \end{pmatrix} (rows and columns swapped, nothing conjugated), a different operator. Multiply the rows of M(T)\mathcal{M}(T) by vv: Tv=(3+5i,3−i)Tv = (3+5i, 3-i).
❌ Not quite. 10=3∣1+i∣2+1⋅∣2∣210 = 3|1+i|^2 + 1\cdot|2|^2 keeps only the diagonal terms. The off-diagonal entries contribute i⋅2⋅(1+i)‾+(−i)(1+i)⋅2‾=(2+2i)+(2−2i)=4i\cdot 2\cdot\overline{(1+i)} + (-i)(1+i)\cdot \overline{2} = (2 + 2i) + (2 - 2i) = 4.
Show solution

Step 1: apply TT.

Tv=(3i−i1)(1+i2)=(3+3i+2i−i+1+2)=(3+5i3−i).Tv = \begin{pmatrix} 3 & i \\ -i & 1 \end{pmatrix}\begin{pmatrix} 1+i \\ 2 \end{pmatrix} = \begin{pmatrix} 3 + 3i + 2i \\ -i + 1 + 2 \end{pmatrix} = \begin{pmatrix} 3+5i \\ 3-i \end{pmatrix}.

Step 2: take the inner product with vv, conjugating the second slot:

⟨Tv,v⟩=(3+5i) (1+i)‾+(3−i) 2‾=(3+5i)(1−i)+(6−2i).\langle Tv, v\rangle = (3+5i)\,\overline{(1+i)} + (3-i)\,\overline{2} = (3+5i)(1-i) + (6 - 2i). (3+5i)(1−i)=3−3i+5i−5i2=8+2i,⟨Tv,v⟩=8+2i+6−2i=14.(3+5i)(1-i) = 3 - 3i + 5i - 5i^2 = 8 + 2i, \qquad \langle Tv, v\rangle = 8 + 2i + 6 - 2i = 14.

Check: M(T)∗=(3−i‾i‾1)=(3i−i1)\mathcal{M}(T)^* = \begin{pmatrix} 3 & \overline{-i} \\ \overline{i} & 1 \end{pmatrix} = \begin{pmatrix} 3 & i \\ -i & 1 \end{pmatrix}, so TT is self-adjoint and ⟨Tv,v⟩\langle Tv, v\rangle must be real. The imaginary parts +2i+2i and −2i-2i cancel, as they must.

Problem 4 · Building Self-Adjoint Operators

Given: SS and TT are nonzero self-adjoint operators on a complex inner product space VV. Which operator must be self-adjoint?

✅ Correct! (S+T)∗=S∗+T∗=S+T(S+T)^* = S^* + T^* = S + T; equivalently, ⟨(S+T)v,v⟩=⟨Sv,v⟩+⟨Tv,v⟩\langle (S+T)v, v\rangle = \langle Sv, v\rangle + \langle Tv, v\rangle is a sum of two real numbers.
❌ Not quite. (iT)∗=i‾ T∗=−iT(iT)^* = \overline{i}\,T^* = -iT, which equals iTiT only if T=0T = 0. In the real-value test, ⟨iTv,v⟩=i⟨Tv,v⟩\langle iTv, v\rangle = i\langle Tv, v\rangle is purely imaginary, not real.
❌ Not quite. (ST)∗=T∗S∗=TS(ST)^* = T^*S^* = TS, so STST is self-adjoint only when ST=TSST = TS. With M(S)=(1000)\mathcal{M}(S) = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} and M(T)=(0110)\mathcal{M}(T) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, M(ST)=(0100)\mathcal{M}(ST) = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} is not self-adjoint.
❌ Not quite. ⟨(T+iI)v,v⟩=⟨Tv,v⟩+i∥v∥2\langle (T + iI)v, v\rangle = \langle Tv, v\rangle + i\|v\|^2 has imaginary part ∥v∥2≠0\|v\|^2 \neq 0 for v≠0v \neq 0, so the real-value test fails. Directly: (T+iI)∗=T∗+i‾ I=T−iI(T + iI)^* = T^* + \overline{i}\,I = T - iI.
Show solution

Use the adjoint rules (A+B)∗=A∗+B∗(A + B)^* = A^* + B^*, (λA)∗=λ‾A∗(\lambda A)^* = \overline{\lambda} A^*, (AB)∗=B∗A∗(AB)^* = B^*A^*, I∗=II^* = I:

  • (S+T)∗=S+T(S + T)^* = S + T: always self-adjoint.
  • (iT)∗=−iT≠iT(iT)^* = -iT \neq iT since T≠0T \neq 0.
  • (ST)∗=TS(ST)^* = TS, which differs from STST in general: for M(S)=(1000)\mathcal{M}(S) = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, M(T)=(0110)\mathcal{M}(T) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, M(ST)=(0100)\mathcal{M}(ST) = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} is not self-adjoint.
  • (T+iI)∗=T−iI≠T+iI(T + iI)^* = T - iI \neq T + iI, since their difference is 2iI≠02iI \neq 0.

The real-value test gives the same verdicts: ⟨Tv,v⟩\langle Tv, v\rangle and ⟨Sv,v⟩\langle Sv, v\rangle are real for every vv, so their sum is real, while multiplying by ii or adding i∥v∥2i\|v\|^2 moves the value off the real axis.

So the answer is (b) S+TS + T.

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