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A real matrix is orthogonal if \(AA^T=I\) (so \(A^{-1}=A^T\)); a complex matrix is unitary if \(AA^\theta=I\) (so \(A^{-1}=A^\theta\)).

Explained simply

Rotate a photo on your phone. Its size and shape do not change; only its direction does. Matrices that perform such "rigid" turns or reflections are orthogonal. They never stretch lengths or change angles. Their inverse is just the transpose, which makes them very easy to undo. Unitary matrices are the complex-number version and are central to signal processing and quantum computing.

Technically, \(A\) (real, \(n\times n\)) is orthogonal if \(AA^T=A^TA=I\). Equivalently, its rows (and columns) are unit vectors perpendicular to each other. A complex \(A\) is unitary if \(AA^\theta=A^\theta A=I\).

How it works

  1. Compute \(A^T\) (or \(A^\theta\) for complex entries).
  2. Multiply \(AA^T\) (or \(AA^\theta\)).
  3. If the result is \(I\), the matrix is orthogonal (unitary). A quick check: each row has length 1 and different rows have dot product 0.
  4. Then write \(A^{-1}=A^T\) (or \(A^\theta\)) directly.

Key formula

  • Orthogonal: \(AA^T=I\), \(A^{-1}=A^T\), \(|A|=\pm1\).
  • Unitary: \(AA^\theta=I\), \(A^{-1}=A^\theta\), \(|\,|A|\,|=1\).
  • Product of two orthogonal (unitary) matrices is orthogonal (unitary); so is the inverse and the transpose.

Properties are stated without proof in the syllabus.

Quick example

\(A=\dfrac13\begin{bmatrix}1&2&2\\2&1&-2\\2&-2&1\end{bmatrix}\). Row lengths: \(\frac19(1+4+4)=1\) for each row. Dot products: row1·row2 \(=\frac19(2+2-4)=0\), row1·row3 \(=\frac19(2-4+2)=0\), row2·row3 \(=\frac19(4-2-2)=0\). So \(AA^T=I\): orthogonal, and \(A^{-1}=A^T\).

Unitary: \(U=\dfrac1{\sqrt2}\begin{bmatrix}1&i\\i&1\end{bmatrix}\), \(U^\theta=\dfrac1{\sqrt2}\begin{bmatrix}1&-i\\-i&1\end{bmatrix}\). \(UU^\theta=\frac12\begin{bmatrix}1+1&-i+i\\i-i&1+1\end{bmatrix}=I\). Unitary.

Common mistakes

  • Using only \(A^T\) for a complex matrix; unitary needs the conjugate transpose.
  • Thinking \(|A|=\pm1\) is enough to prove orthogonality; it is necessary, not sufficient.
  • Forgetting the scalar factor (like \(\frac13\) or \(\frac1{\sqrt2}\)) when multiplying.

How MU asks it

Frequent 5/6-mark parts: "Show that \(A\) is unitary and hence find \(A^{-1}\)", "Find \(a,b,c\) if \(A=\begin{bmatrix}0&2b&c\\a&b&-c\\a&-b&c\end{bmatrix}\) is orthogonal" (answer \(a=\pm\frac1{\sqrt2}\), \(b=\pm\frac1{\sqrt6}\), \(c=\pm\frac1{\sqrt3}\)). Show the product \(AA^T\) or \(AA^\theta\) entry by entry and state the conclusion and the inverse.

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report
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