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

True — the proof forms the real scalar ${\mathbf{x}}^{\ast }A\mathbf{x}$; symmetry forces it to equal its own conjugate, so $\lambda =\bar{\lambda }$.

False — $\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)$ has real eigenvalues $1,2$ but is not symmetric; real spectrum is necessary, not sufficient, for symmetry.

False — the orthogonality proof only works when $\lambda \ne \mu$; inside one eigenspace you must choose an orthogonal basis via Gram-Schmidt Process.

False — symmetric matrices are never defective; every eigenvalue's geometric multiplicity equals its algebraic multiplicity, so a full eigenbasis always exists.

True — orthonormal columns give $Q^{\top }Q=I$, which is exactly the definition of an orthogonal matrix whose transpose is its inverse.

False — both diagonalize $A$, but $Q^{\top }=Q^{-1}$ holds only for orthonormal columns; a generic $P$ from unnormalized/non-orthogonal eigenvectors has $P^{\top }\ne P^{-1}$.

True — these are normal matrices; over $\mathbb{C}$ they are unitarily diagonalizable, though a real normal matrix may need complex eigenvectors (e.g. a rotation).

True — $I$ has one eigenvalue $1$ of full multiplicity, so every orthonormal basis is an eigenbasis; $Q$ is completely free.

False — nothing forbids repeats (e.g. $2I$ repeats $2$); it guarantees a full orthonormal eigenset, not distinctness.

The error: symmetric matrices cannot have non-real eigenvalues, so $2i$ is impossible — the matrix in question is not truly symmetric.

The error: you forgot to normalize the eigenvectors; unit-length orthogonal columns give $Q^{\top }=Q^{-1}$, and the theorem holds once you fix the lengths.

The error: that matrix isn't symmetric ($a_{12}=1\ne 0=a_{21}$), so the orthogonality guarantee never applied in the first place.

The error: non-perpendicular basis vectors are normal inside a repeated eigenspace — apply Gram-Schmidt Process to orthogonalize them; the matrix is still fully diagonalizable.

The error: ${\mathbf{q}}_i{\mathbf{q}}_i^{\top }$ is an outer product (column times row) giving an $n\times n$ rank-1 matrix; ${\mathbf{q}}_i^{\top }{\mathbf{q}}_i$ would be the scalar $1$.

The error: for any square matrix trace $=\sum {\lambda }_i$; for a symmetric $A$ this lets you sanity-check computed eigenvalues instantly.

The error: if $A=A^{\top }$ and $A$ is invertible, then $A^{-1}=(A^{-1}{)}^{\top }$ too — inverting preserves symmetry (its eigenvalues just become $1/{\lambda }_i$).

Because $\lambda$ might a priori be complex, so $\mathbf{x}$ is complex; ${\mathbf{x}}^{\ast }\mathbf{x}=\sum |x_i{|}^2$ stays real and positive only with conjugation, which is what forces $\lambda =\bar{\lambda }$.

The proof pivots on moving $A$ across the inner product via $A^{\top }=A$; that swap is what yields $(\lambda -\mu )({\mathbf{u}}^{\top }\mathbf{v})=0$, and it fails without symmetry.

Its columns are orthonormal, so $Q^{\top }Q=I$; geometrically it rotates/reflects without stretching, meaning changing to eigen-coordinates preserves lengths and angles.

Reading right to left: $Q^{\top }$ rotates into eigen-axes, $\Lambda$ only stretches along those axes, $Q$ rotates back — a pure stretch along perpendicular directions, no shear.

${\mathbf{q}}_i{\mathbf{q}}_i^{\top }$ is the orthogonal projector onto that axis only when ${\mathbf{q}}_i^{\top }{\mathbf{q}}_i=1$; otherwise the projector is scaled wrong and the pieces won't sum to $A$.

Symmetric (hence normal) matrices are never defective, so geometric multiplicity equals algebraic; a general matrix like a Jordan block can have algebraic multiplicity $2$ but only a $1$-dimensional eigenspace.

A quadratic form ${\mathbf{x}}^{\top }A\mathbf{x}$ with symmetric $A$ becomes $\sum {\lambda }_i{y}_i^2$ in eigen-coordinates $\mathbf{y}=Q^{\top }\mathbf{x}$, so the eigenvalues are the pure stretch coefficients of the form.

Every eigenvalue is $0$ ($\Lambda =0$), and every orthonormal basis works as $Q$; the "stretch" collapses all directions to a single point.

Yes trivially — it is symmetric, its single eigenvalue is $a\in \mathbb{R}$, and $Q=[1]$; it's the degenerate base case of the general statement.

Nothing changes — $\lambda =0$ is a perfectly valid distinct eigenvalue, so its eigenvectors are still orthogonal to those of other eigenvalues; the matrix is just singular.

Yes — a negative ${\lambda }_i$ means $A$ flips the sign along that eigen-axis (stretch by $|{\lambda }_i|$ plus a reflection); this ties to Positive Definite Matrices failing when any ${\lambda }_i<0$.

It must be $cI$ — the full-multiplicity eigenspace is all of ${\mathbb{R}}^n$, so $A$ acts as uniform scaling and $Q$ can be anything orthogonal.

All eigenvalues are strictly positive, so $A=Q\Lambda Q^{\top }$ has an invertible square root $A^{1/2}=Q{\Lambda }^{1/2}{Q}^{\top }$, connecting to Positive Definite Matrices and the Singular Value Decomposition.

The Singular Value Decomposition $A=U\Sigma V^{\top }$ uses two different orthonormal bases and non-negative singular values, recovering the "orthogonal stretch" picture without needing $A=A^{\top }$.

They can be simultaneously orthogonally diagonalized — a shared orthonormal eigenbasis exists, so one $Q$ diagonalizes both.