LINEAR-ALGEBRA · Interactive Practice | Unit 24 · Video 2

Why Symmetric Matrices Always Have Real Eigenvalues

IKey Formulas

Formula	Name	Description
$A\mathbf{x} = \lambda \mathbf{x}$	Eigenvalue equation	$\lambda$ is an eigenvalue, $\mathbf{x}$ its eigenvector
$A^{T} = A$	Symmetric condition	The hinge of the entire proof
$\bar{\mathbf{x}}^{T}\mathbf{x} = \sum_k (a_k^2 + b_k^2) > 0$	Squared complex length	Why we can divide by it when $\mathbf{x} \neq 0$
$\lambda = \bar{\lambda} \iff \lambda \in \mathbb{R}$	Reality test	Conjugate-equals-self means imaginary part is zero
$A = \bar{A}^{T}$	Hermitian condition	Complex generalization; same conclusion

IIVisualization 1 — Symmetry Keeps Eigenvalues Real

With the diagonal fixed at $2$ and $3$ , equal off-diagonals ( $b=c$ ) pin both eigenvalues to the real axis — unequal ones can push them off.

b (top-right) 1 c (bottom-left) 1

💡 The eigenvalues leave the real axis only once $bc < -\tfrac14$ . Symmetry forces $b=c$ , hence $bc = b^2 \ge 0$ — safely on the real side of that threshold.

IIIVisualization 2 — The Symmetric Family Stays Real

Sweep the entries of any symmetric matrix $A = A^T$ : the discriminant is a sum of squares, so the eigenvalues can never leave the axis.

a 2 b 1 c 3

IVVisualization 3 — Why the Conjugate Matters

For $\mathbf{x} = (1,\ \alpha + i\beta)$ the naive $\mathbf{x}^T\mathbf{x}$ can collapse to $0$ on a nonzero vector; the conjugate form $\bar{\mathbf{x}}^T\mathbf{x}$ never does.

α = Re(x₂) 0 β = Im(x₂) 1

VQuiz Questions

Question 1

Consider the symmetric matrix $A = \begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix}.$

Without computing them numerically, what can you say about the eigenvalues of $A$ ?

✅ Correct! Symmetry alone guarantees the eigenvalues are real — no computation needed.

❌ Not quite. The matrix is real and symmetric ( $A^T = A$ ), so the theorem applies and the eigenvalues are guaranteed to be real.

Show solution

Solution:

The matrix is real and satisfies $A^T = A$ . The theorem proved in the video states:

Every real symmetric matrix has only real eigenvalues.

So without any computation we know both eigenvalues lie on the real axis. (For the curious: the characteristic polynomial is $\lambda^2 - 5\lambda = \lambda(\lambda - 5)$ , giving $\lambda = 0$ and $\lambda = 5$ — both real.)

The proof in one breath: conjugate the eigenvalue equation, transpose it, use $A^T = A$ to swap, and combine with the original to get $\lambda\,\bar{\mathbf{x}}^T\mathbf{x} = \bar{\lambda}\,\bar{\mathbf{x}}^T\mathbf{x}$ . Since $\bar{\mathbf{x}}^T\mathbf{x} > 0$ , we cancel and get $\lambda = \bar{\lambda}$ , so $\lambda \in \mathbb{R}$ .

Question 2

Consider the complex vector $\mathbf{x} = (1,\ i)$ . A student computes the "squared length" as $\mathbf{x}^T \mathbf{x} = 1 \cdot 1 + i \cdot i = 0.$

The proof in the video relies on the squared length being strictly positive for a nonzero vector. Which expression actually delivers a positive real number for the same $\mathbf{x}$ ?

✅ Correct! Conjugating either factor produces the same positive real number — that's the point of the Hermitian inner product.

❌ Not quite. The fix is to conjugate one of the two factors. Either $\bar{\mathbf{x}}^T \mathbf{x}$ or $\mathbf{x}^T \bar{\mathbf{x}}$ gives 2, and they're equal because each is the conjugate of the other.

Show solution

Solution:

For a complex vector $\mathbf{x}$ with components $x_k = a_k + i b_k$ , the correct squared-length is $\bar{\mathbf{x}}^T \mathbf{x} \;=\; \sum_k \bar{x}_k\, x_k \;=\; \sum_k (a_k - i b_k)(a_k + i b_k) \;=\; \sum_k (a_k^2 + b_k^2),$ which is real and strictly positive whenever $\mathbf{x} \neq \mathbf{0}$ .

For $\mathbf{x} = (1, i)$ : $\bar{\mathbf{x}}^T \mathbf{x} = (1)(1) + (-i)(i) = 1 + 1 = 2.$

Both forms $\bar{\mathbf{x}}^T \mathbf{x}$ and $\mathbf{x}^T \bar{\mathbf{x}}$ evaluate to the same positive real number — they are conjugates of each other, but a real number equals its own conjugate. The crucial move is conjugating one of the two factors, which turns each $i \cdot i = -1$ into $-i \cdot i = +1$ .

The naive $\mathbf{x}^T \mathbf{x} = 0$ is the disaster the conjugate is designed to prevent.

Question 3

The proof has the following structure (steps in order):

Start with $A\mathbf{x} = \lambda \mathbf{x}$ .
Conjugate both sides: $A\bar{\mathbf{x}} = \bar{\lambda}\bar{\mathbf{x}}$ .
Transpose: $\bar{\mathbf{x}}^T A^T = \bar{\lambda}\,\bar{\mathbf{x}}^T$ .
Replace $A^T$ with $A$ : $\bar{\mathbf{x}}^T A = \bar{\lambda}\,\bar{\mathbf{x}}^T$ .
Combine with step 1 to get $\lambda\,\bar{\mathbf{x}}^T \mathbf{x} = \bar{\lambda}\,\bar{\mathbf{x}}^T \mathbf{x}$ .
Cancel $\bar{\mathbf{x}}^T \mathbf{x}$ to conclude $\lambda = \bar{\lambda}$ .

Which step is the one place where the symmetry hypothesis $A^T = A$ is actually used?

✅ Correct! Step 4 is the hinge — every other step works for arbitrary matrices.

❌ Not quite. Look for the step that wouldn't be valid without $A^T = A$ . Conjugation, transposition, and cancellation each work without symmetry; only the substitution $A^T \to A$ requires it.

Show solution

Solution:

The hypothesis $A^T = A$ is invoked in exactly one spot — Step 4 — when we replace $A^T$ with $A$ . The other steps work for any matrix:

Step 2 uses only that $A$ is real ( $\bar{A} = A$ ) — that gives the bonus result that complex eigenvalues of real matrices come in conjugate pairs.
Step 3 is just the algebraic rule $(MN)^T = N^T M^T$ .
Step 6 uses the fact that $\bar{\mathbf{x}}^T\mathbf{x} > 0$ for any nonzero complex vector.

Knock out Step 4 and the proof collapses. That single substitution is what funnels $\bar{\lambda}$ to the same combination $\bar{\mathbf{x}}^T A \mathbf{x}$ that $\lambda$ already controls, forcing $\lambda = \bar{\lambda}$ .

Generalization: for complex matrices, the analogous condition is $A = \bar{A}^T$ (Hermitian). It plays exactly the same role in Step 4, and the conclusion is identical — real eigenvalues.

Question 4

Consider the complex matrix $A = \begin{bmatrix} 2 & 1 + i \\ 1 - i & 3 \end{bmatrix}.$

Note that $A^T \neq A$ (the off-diagonal entries are $1+i$ and $1-i$ , which differ). Nevertheless, all eigenvalues of $A$ turn out to be real.

True or False: This is a counterexample to the theorem from the video — a non-symmetric matrix with real eigenvalues shows that symmetry is not actually required.

✅ Correct! The matrix is Hermitian — the complex generalization of symmetric — so the theorem applies and predicts real eigenvalues.

❌ Not quite. Check whether $A = \bar{A}^T$ . Take the conjugate of every entry, then transpose — you'll find you recover $A$ exactly. That's the Hermitian condition, the correct hypothesis for complex matrices.

Show solution

Solution:

This is False — and the reason is precisely the Hermitian generalization mentioned at the end of the video.

Compute the conjugate transpose: $\bar{A}^T = \overline{\begin{bmatrix} 2 & 1+i \\ 1-i & 3 \end{bmatrix}}^{T} = \begin{bmatrix} 2 & 1-i \\ 1+i & 3 \end{bmatrix}^{T} = \begin{bmatrix} 2 & 1+i \\ 1-i & 3 \end{bmatrix} = A.$

So $A = \bar{A}^T$ — the matrix is Hermitian. For complex matrices, "symmetric" ( $A^T = A$ ) is the wrong condition; the correct condition is "Hermitian" ( $A = \bar{A}^T$ ). Replace $A^T$ with $\bar{A}^T$ throughout the proof and the conclusion is identical: real eigenvalues.

Real symmetric matrices are simply the real special case of Hermitian matrices — when all entries are real, $\bar{A}^T$ collapses to $A^T$ , and Hermitian collapses to symmetric.

(For verification: the eigenvalues of this $A$ are $(5 \pm \sqrt{9})/2 \cdot \ldots$ well, $\lambda^2 - 5\lambda + 4 = 0$ , giving $\lambda = 1$ and $\lambda = 4$ — both real, as the theorem predicts.)

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