Gram–Schmidt, orthogonal/unitary/normal matrices, spectral theorem

In one sentence

Normal is exactly the class the spectral theorem applies to, Hermitian and unitary and skew-Hermitian are the three special cases whose eigenvalues sit on the real axis, the unit circle and the imaginary axis, and every question here is asking which of those you are in.

Why the exam asks it

Because "diagonalisable" and "unitarily diagonalisable" are different conditions, and the second is the useful one — it gives an orthonormal eigenbasis, which is what makes the spectral theorem a tool rather than a classification. And because the word "symmetric" carries real weight over $\mathbb{R}$ and almost none over $\mathbb{C}$.

The idea

An inner product gives $\|v\| = \sqrt{\langle v,v\rangle}$, and

$|\langle u,v\rangle| \le \|u\|\,\|v\| \qquad \text{(Cauchy–Schwarz)},$

with equality exactly when $u$ and $v$ are dependent.

Gram–Schmidt turns any independent list into an orthonormal one spanning the same subspaces step by step. Two things worth noting: it needs the inputs to be independent (a dependent vector produces the zero vector at that step), and it gives the $QR$ factorisation directly.

Orthogonal projection onto a subspace $W$ minimises distance: $\|v - P_W v\| \le \|v - w\|$ for every $w \in W$, with equality only at $w = P_W v$. That is the best-approximation property, and it is what least squares is.

$V = W \oplus W^{\perp}$ for finite-dimensional $W$, and $(W^{\perp})^{\perp} = W$.

The adjoint and the three families

$A^*$ is the conjugate transpose. Then:

• Hermitian (self-adjoint): $A^* = A$.
• Unitary: $A^*A = I$; over $\mathbb{R}$, orthogonal, $A^TA = I$.
• Skew-Hermitian: $A^* = -A$.
• Normal: $AA^* = A^*A$.

All three of the first families are normal, and normal is strictly larger than their union.

The spectral theorem

Over $\mathbb{C}$: $A$ is normal $\iff$ $A$ is unitarily diagonalisable, that is $A = U D U^*$ with $U$ unitary.

Over $\mathbb{R}$: $A$ is symmetric $\iff$ $A$ is orthogonally diagonalisable, $A = Q D Q^T$ with $Q$ orthogonal and $D$ real.

Consequences that get examined directly:

• Hermitian $\Rightarrow$ all eigenvalues real.
• Unitary $\Rightarrow$ all eigenvalues have modulus 1.
• Skew-Hermitian $\Rightarrow$ all eigenvalues purely imaginary.
• For a normal matrix, eigenvectors belonging to distinct eigenvalues are orthogonal — not merely independent, which is all you get in general.

Each converse holds too, given normality: a normal matrix with all-real eigenvalues is Hermitian, and so on. Without normality the converses fail — $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has a real eigenvalue and is not Hermitian.

Real orthogonal matrices

A rotation $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ is orthogonal, hence normal, hence unitarily diagonalisable over $\mathbb{C}$ with eigenvalues $\pm i$ — both of modulus $1$, as required. Over $\mathbb{R}$ it is not diagonalisable at all, because $x^2+1$ has no real root.

So "normal" does not mean "diagonalisable over the reals". The real spectral theorem needs symmetric, and orthogonal is a different condition entirely. Note also $\det = \pm 1$ for orthogonal matrices, not $+1$: reflections have determinant $-1$.

Where intuition breaks

"Symmetric, so diagonalisable." Over $\mathbb{R}$ yes. Over $\mathbb{C}$, $\begin{pmatrix}1&i\\i&-1\end{pmatrix}$ is symmetric, nonzero and squares to zero. The complex analogue of real symmetry is Hermitian, and the option will say "symmetric" and let you supply "real" yourself.

"Normal, so diagonalisable over $\mathbb{R}$." The rotation matrix. Normal gives unitary diagonalisability over $\mathbb{C}$, which says nothing about $\mathbb{R}$.

"Orthogonal, so $\det = 1$." $\det = \pm1$. The $+1$ ones form the rotations; the $-1$ ones include every reflection.

"Hermitian, so positive definite." $-I$ is Hermitian with eigenvalues $-1$. Hermitian fixes the eigenvalues to be real, not positive.

"Eigenvectors for distinct eigenvalues are orthogonal." True for normal matrices; false in general, where they are only independent. Take $\begin{pmatrix}1&1\\0&2\end{pmatrix}$: eigenvectors $(1,0)$ and $(1,1)$ are not orthogonal.

"$A$ and $A^*$ have the same eigenvalues." They are conjugates of each other. For Hermitian matrices this is the same thing, which is why the error survives.

"Gram–Schmidt works on any spanning set." It needs independence. On a dependent list it produces a zero vector, which cannot be normalised — the fix is to discard it, but the naive algorithm divides by zero.

"Unitarily diagonalisable is the same as diagonalisable." $\begin{pmatrix}1&1\\0&2\end{pmatrix}$ is diagonalisable — distinct eigenvalues — and not normal, so not unitarily diagonalisable. The eigenbasis exists; an orthonormal one does not.

The exam's angle

1. Compute $AA^*$ and $A^*A$ first. Normality is a two-line check and it decides whether the spectral theorem is available at all.
2. Match eigenvalue location to family. Real $\to$ Hermitian, modulus one $\to$ unitary, purely imaginary $\to$ skew-Hermitian — but only once you know the matrix is normal.
3. Read $\mathbb{R}$ or $\mathbb{C}$. Symmetric matters over $\mathbb{R}$; over $\mathbb{C}$ it is nearly worthless and Hermitian is the condition.
4. For "orthonormal basis of eigenvectors", the question is normality, not diagonalisability.
5. Remember orthogonal matrices are normal but rarely real-diagonalisable. Rotations are the standard counterexample and appear constantly.

The night before

• Normal $\iff$ unitarily diagonalisable, over $\mathbb{C}$. Real symmetric $\iff$ orthogonally diagonalisable, over $\mathbb{R}$.
• Hermitian: real eigenvalues. Unitary: modulus $1$. Skew-Hermitian: purely imaginary. Converses hold given normality.
• Normal eigenvectors for distinct eigenvalues are orthogonal; in general only independent.
• Complex symmetric is worthless: $\begin{pmatrix}1&i\\i&-1\end{pmatrix}$ is symmetric and nilpotent. Use Hermitian.
• Orthogonal $\Rightarrow \det = \pm1$. Rotation: normal, not diagonalisable over $\mathbb{R}$, eigenvalues $\pm i$.
• Cauchy–Schwarz with equality $\iff$ dependence.
• Gram–Schmidt needs independence; gives $QR$.
• Orthogonal projection is the best approximation; $V = W\oplus W^\perp$.
• Diagonalisable $\ne$ unitarily diagonalisable.