ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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An explicit real symmetric 3x3 matrix is orthogonally diagonalised

Example

For

$A = (\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{array}),$

an orthogonal matrix

$Q = (\begin{array}{ccc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ - \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{array})$

satisfies

$Q^{T} A Q = diag (1, 3, 3) .$

Facts & Assumptions

Given: The real symmetric matrix $A$ above, acting on $R^{3}$ with the standard inner product.

[L1]

A self-adjoint operator on a finite-dimensional real inner product space has an orthonormal eigenbasis (Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis).

Verification

technique · direct

1.1

Direct multiplication gives $A (1, - 1, 0)^{T} = (1, - 1, 0)^{T}$ , $A (1, 1, 0)^{T} = 3 (1, 1, 0)^{T}$ , and $A (0, 0, 1)^{T} = 3 (0, 0, 1)^{T}$ . After normalising the first two vectors, the three displayed columns of $Q$ form an orthonormal eigenbasis, exactly as [L1] predicts.

L1algebra

2.1

Because the columns of $Q$ are orthonormal eigenvectors with eigenvalues $1, 3, 3$ , the conjugated matrix $Q^{T} A Q$ is diagonal with those eigenvalues on the diagonal.

step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)