Symmetric matrices — spectral theorem (real eigenvalues, orthogonal eigenvectors): Worked examples

This page is the "worked-out drill" child of the parent topic. The parent proved why real symmetric matrices behave. Here we hit every kind of case a problem can throw at you and grind each one to a numeric answer.

Before we begin, one plain-words reminder so nothing is used before it is built.

The scenario matrix

Every symmetric-matrix problem falls into one of these cells. The examples below are labelled with the cell they cover, so together they leave no gap.

#	Case class	What is unusual	Example
C1	Both eigenvalues positive	ordinary "stretch/stretch"	Ex 1
C2	Mixed signs (one $+$ , one $-$ )	a reflection hides inside	Ex 2
C3	A zero eigenvalue (singular $A$ )	$A$ squashes a whole direction to $0$	Ex 3
C4	Repeated eigenvalue (multiplicity 2)	eigenvectors not forced — you choose them	Ex 4
C5	Diagonal $A$ (degenerate/limiting)	eigenvectors are the axes themselves	Ex 5
C6	$3 \times 3$ with a clean block	full spectral decomposition in 3D	Ex 6
C7	Word problem (real-world)	you must build the symmetric matrix first	Ex 7
C8	Exam twist — rebuild $A$ from $Q, Λ$	run the theorem backwards	Ex 8

Example 1 — C1: both eigenvalues positive

Worked example Statement

Orthogonally diagonalize $A = (4114)$ .

Forecast: trace $= 8$ , det $= 16 - 1 = 15$ . Which two numbers add to $8$ and multiply to $15$ ? Guess before reading on.

Steps:

Solve $λ^{2} - 8 λ + 15 = 0 = (λ - 5) (λ - 3)$ , so $λ = 5, 3$ . Why this step? The characteristic quadratic is built straight from trace and det — no need to expand $det (A - λ I)$ by hand.
$λ = 5$ : solve $(A - 5 I) x = (- 1 1 1 - 1) x = 0$ . The two rows both say $- x_{1} + x_{2} = 0$ , so $x = (1, 1)$ . Why this step? An eigenvector is any nonzero vector the shifted matrix sends to $0$ ; both rows agree, confirming the eigenvalue is right.
$λ = 3$ : $(A - 3 I) x = (1111) x = 0 \Rightarrow x = (1, - 1)$ .
Normalize (divide each by its length $2$ ): $q_{1} = \frac{1}{2} (1, 1)$ , $q_{2} = \frac{1}{2} (1, - 1)$ . Why this step? Only unit-length columns make $Q^{⊤} = Q^{- 1}$ , which is what turns $P Λ P^{- 1}$ into the tidy $Q Λ Q^{⊤}$ . See Diagonalization and Orthogonal Matrices. $Q = \frac{1}{2} (11 1 - 1), Λ = (5003) .$

Verify: dot product $(1, 1) \cdot (1, - 1) = 0$ ✓ (orthogonal, as promised for distinct eigenvalues). Rebuild: $Q Λ Q^{⊤} = (4114) = A$ ✓. Both $λ > 0$ , so $A$ stretches (never flips) — it is positive definite, see Positive Definite Matrices.

Both eigenvalues positive means every direction gets pushed outward. The picture below shows the eigen-axes (the only two directions $A$ leaves pointing the same way) and how a circle balloons into an ellipse.

Figure — Symmetric matrices — spectral theorem (real eigenvalues, orthogonal eigenvectors)

Example 2 — C2: mixed signs (a reflection inside)

The magenta arrow below is the reflected eigen-direction: same line, reversed arrow.

Example 3 — C3: a zero eigenvalue (singular matrix)

Worked example Statement

Diagonalize $A = (1111)$ .

Forecast: trace $= 2$ , det $= 1 - 1 = 0$ . A zero determinant means one eigenvalue is $0$ (product of eigenvalues is $0$ ). The other must be the whole trace. Guess them.

Steps:

$λ^{2} - 2 λ + 0 = 0 = λ (λ - 2)$ , so $λ = 2, 0$ . Why this step? A zero eigenvalue means $A$ flattens an entire direction onto the origin — that direction is the null space.
$λ = 2$ : $(- 1 1 1 - 1) x = 0 \Rightarrow x = (1, 1)$ .
$λ = 0$ : $A x = 0$ directly $\Rightarrow (1, 1) \cdot (x_{1}, x_{2})$ ... solve $x_{1} + x_{2} = 0 \Rightarrow x = (1, - 1)$ . Why this step? The eigenvector for $λ = 0$ is the null space: everything on the line $x = (t, - t)$ is sent to $0$ .
$Q = \frac{1}{2} (11 1 - 1)$ , $Λ = (2000)$ .

Verify: $A (1, - 1) = (0, 0)$ ✓ (squashed). $A (1, 1) = (2, 2) = 2 (1, 1)$ ✓. Spectral sum: $A = 2 q_{1} q_{1}^{⊤} = 2 \cdot \frac{1}{2} (1111) = A$ ✓ (the $λ = 0$ piece contributes nothing — this is exactly why singular matrices lose rank).

The orange line below is the squashed direction — every point on it lands at the origin.

Example 4 — C4: a repeated eigenvalue

Worked example Statement

Diagonalize $A = (3003)$ ... wait, that is too easy. Take the disguised version $A = (3003)$ but prove the eigenvalue repeats and any orthonormal pair works.

Forecast: trace $= 6$ , det $= 9$ . Two numbers summing to $6$ , multiplying to $9$ : they must both be $3$ . What are the eigenvectors then?

Steps:

$λ^{2} - 6 λ + 9 = (λ - 3)^{2}$ , so $λ = 3$ twice. Why this step? A repeated root warns you the eigenvector is not forced to a single line.
Solve $(A - 3 I) x = (0000) x = 0$ . Every vector solves this. Why this step? The shifted matrix is all zeros, so the eigenspace is the entire plane — a 2D eigenspace, never defective (the parent's promise for symmetric matrices).
Since any two perpendicular unit vectors form a valid $Q$ , pick the axes $q_{1} = (1, 0)$ , $q_{2} = (0, 1)$ , OR the rotated pair $\frac{1}{2} (1, 1), \frac{1}{2} (1, - 1)$ . Both give $Λ = (3003)$ . Why this step? When you can't distinguish eigen-directions, you choose an orthonormal basis (use Gram-Schmidt Process in higher dimensions).

Verify: For the rotated pair, $Q Λ Q^{⊤} = 3 Q Q^{⊤} = 3 I = A$ ✓ because $Q Q^{⊤} = I$ for any orthogonal $Q$ . A scalar-multiple-of-identity is invariant to your choice of axes — that is the geometric meaning of "repeated eigenvalue."

Example 5 — C5: diagonal matrix (limiting / degenerate case)

Example 6 — C6: a $3 \times 3$ block matrix (full 3D decomposition)

Worked example Statement

Diagonalize $A = 500021012$ .

Forecast: The top-left is decoupled from the $2 \times 2$ bottom block. What are the three eigenvalues?

Steps:

The isolated $5$ gives $λ = 5$ with eigenvector $(1, 0, 0)$ . Why this step? Row/column 1 has no coupling, so $e_{1}$ passes straight through: $A e_{1} = 5 e_{1}$ .
Bottom block $(2112)$ : trace $4$ , det $3$ , so $λ = 3, 1$ (eigenvectors $(1, 1)$ and $(1, - 1)$ in the $y, z$ slots). Why this step? Reuse the exact $2 \times 2$ trick from Ex 1 on the sub-block; block-diagonal problems split into independent smaller ones.
Full eigenvectors: $λ = 5 : (1, 0, 0)$ ; $λ = 3 : (0, 1, 1)$ ; $λ = 1 : (0, 1, - 1)$ . Normalize the last two by $2$ . $Q = 100 0 \frac{1}{2} \frac{1}{2} 0 \frac{1}{2} - \frac{1}{2}, Λ = 500030001 .$

Verify: All three eigenvectors pairwise orthogonal: $(1, 0, 0) \cdot (0, 1, 1) = 0$ , $(1, 0, 0) \cdot (0, 1, - 1) = 0$ , $(0, 1, 1) \cdot (0, 1, - 1) = 1 - 1 = 0$ ✓. Rebuild $Q Λ Q^{⊤} = A$ ✓. This links to Singular Value Decomposition: for symmetric positive-definite $A$ , the SVD equals the spectral decomposition.

Example 7 — C7: word problem (build the matrix first)

Worked example Statement

A stretchy sheet deforms so that a point at position $(x, y)$ moves to $(x + 0.5 y, 0.5 x + y)$ . The strain is symmetric. Find the two perpendicular directions of pure stretch and their stretch factors.

Forecast: The transformation matrix is $A = (1 0.5 0.5 1)$ . Trace $= 2$ , det $= 1 - 0.25 = 0.75$ . Guess the stretch factors.

Steps:

Confirm $A$ is symmetric: off-diagonals both $0.5$ ✓, so the Spectral Theorem applies and pure-stretch axes exist. Why this step? Physically, a symmetric strain has principal axes with no shear — the theorem guarantees them.
$λ^{2} - 2 λ + 0.75 = 0$ . Using $λ = \frac{2 \pm 4 - 3}{2} = \frac{2 \pm 1}{2}$ , so $λ = 1.5, 0.5$ . Why this step? These are the stretch factors: $1.5$ (stretch by 50%) and $0.5$ (compress to half).
$λ = 1.5$ : direction $(1, 1)$ (the $4 5^{\circ}$ diagonal). $λ = 0.5$ : direction $(1, - 1)$ (the anti-diagonal). Why this step? Along $(1, 1)$ the sheet stretches most; along the perpendicular $(1, - 1)$ it compresses. This is exactly the Quadratic Forms geometry of an ellipse's axes.

Verify: $A (1, 1) = (1.5, 1.5) = 1.5 (1, 1)$ ✓; $A (1, - 1) = (0.5, - 0.5) = 0.5 (1, - 1)$ ✓. Directions perpendicular: $(1, 1) \cdot (1, - 1) = 0$ ✓. Units: stretch factors are dimensionless ratios (new length / old length) ✓.

Example 8 — C8: exam twist (run the theorem backwards)

Worked example Statement

You are given $Q = \frac{1}{2} (11 - 1 1)$ and $Λ = (6002)$ . Reconstruct the symmetric matrix $A = Q Λ Q^{⊤}$ and confirm it is symmetric.

Forecast: With eigenvalues $6$ and $2$ (both positive) and $4 5^{\circ}$ -rotated eigen-axes, guess whether $A$ has equal diagonal entries and a nonzero off-diagonal.

Steps:

First $Λ Q^{⊤}$ : $Λ Q^{⊤} = (6002) \cdot \frac{1}{2} (1 - 1 11) = \frac{1}{2} (6 - 2 62)$ . Why this step? Multiplying right-to-left keeps each matrix small; $Q^{⊤}$ is $Q$ with rows and columns swapped.
Then $Q (Λ Q^{⊤}) = \frac{1}{2} (11 - 1 1) \cdot \frac{1}{2} (6 - 2 62) = \frac{1}{2} (8448) = (4224)$ . Why this step? This is the spectral theorem used constructively — you can manufacture any symmetric matrix from chosen orthonormal axes and chosen eigenvalues.

Verify: $A = (4224)$ has $A = A^{⊤}$ ✓ (off-diagonals both $2$ ). Trace $= 8 = 6 + 2$ ✓, det $= 16 - 4 = 12 = 6 \times 2$ ✓. Eigenvalues recovered match $Λ$ .

Recall Quick self-test

Given $det A < 0$ for a symmetric $2 \times 2$ matrix, what can you say about its eigenvalues? ::: They have opposite signs (product is negative), so $A$ reflects along one eigen-axis (Case C2). A symmetric matrix has $det = 0$ . What does one eigenvalue equal, and what happens geometrically? ::: One eigenvalue is $0$ ; that eigen-direction is squashed onto the origin — $A$ is singular (Case C3). For a repeated eigenvalue on a symmetric matrix, are eigenvectors uniquely determined? ::: No — the eigenspace is full-dimensional, so you choose an orthonormal basis (Gram–Schmidt) (Case C4).