The Spectral Theorem and SVD: Examples and Counterexamples

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

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

One checks that $H(i,1{)}^{\mathsf{T}}=3(i,1{)}^{\mathsf{T}}$ and $H(-i,1{)}^{\mathsf{T}}=(-i,1{)}^{\mathsf{T}}$. After dividing by $\sqrt{2}$, these eigenvectors are orthonormal, so the columns of $U$ form the orthonormal eigenbasis guaranteed by [L1].

Because the columns of $U$ are orthonormal eigenvectors with eigenvalues $3$ and $1$, the conjugated matrix $U^{\ast }HU$ is $\mathrm{diag}(3,1)$.

Direct multiplication gives $A^{\mathsf{T}}A=A{A}^{\mathsf{T}}=I_2$, so $A$ is normal by [L1]. Its characteristic polynomial is $x^2+1$, so over $\mathbb{C}$ its eigenvalues are $i$ and $-i$.

The matrix $A$ itself has the form $\left(\begin{array}{cc}a& b\\ -b& a\end{array}\right)$ with $a=0$ and $b=-1\mathrm{\ne }0$, so it is exactly one of the $2\times 2$ blocks allowed by [L2].

The matrix is symmetric because $A^{\mathsf{T}}=A$, and direct multiplication gives $A^2=0$ while $A\mathrm{\ne }0$, so $A$ is a nonzero nilpotent matrix of index $2$.

Its conjugate transpose is $A^{\ast }=\left(\begin{array}{cc}1& -i\\ -i& -1\end{array}\right)$, and direct multiplication gives $A^{\ast }A=\left(\begin{array}{cc}2& 2i\\ -2i& 2\end{array}\right)$ but $A{A}^{\ast }=\left(\begin{array}{cc}2& -2i\\ 2i& 2\end{array}\right)$. Hence $A^{\ast }A\mathrm{\ne }A{A}^{\ast }$, so $A$ is not normal by [L1].

One has $A(1,1{)}^{\mathsf{T}}=4(1,1{)}^{\mathsf{T}}$ and $A(1,-1{)}^{\mathsf{T}}=(1,-1{)}^{\mathsf{T}}$, so after normalising these vectors, [L1] gives an orthonormal eigenbasis with eigenvalues $4$ and $1$. Therefore $A$ is non-negative.

In the same eigenbasis, the non-negative square root has eigenvalues $2$ and $1$, which corresponds in the standard basis to $R=\left(\begin{array}{cc}3\mathrm{/}2& 1\mathrm{/}2\\ 1\mathrm{/}2& 3\mathrm{/}2\end{array}\right)$. Direct multiplication gives $R^2=A$, and direct algebra gives $R=\frac{1}{3}A+\frac{2}{3}I$, exactly as [L2] predicts.

Direct multiplication gives $T^{\ast }T=\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right)$, so its non-negative square root is $U=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)$.

The matrix $S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ satisfies $S^{\mathsf{T}}S=I$, so it is orthogonal, and $SU=\left(\begin{array}{cc}0& 2\\ 1& 0\end{array}\right)=T$. Hence this is the polar decomposition predicted by [L1].

Because $T^{\ast }T=T$, the non-negative square root is $U=T$. Also $S_1$ and $S_2$ are both orthogonal because $S_1^{\mathsf{T}}{S}_1=S_2^{\mathsf{T}}{S}_2=I$.

Direct multiplication gives $S_{1}U=U=S_{2}U=T$, while $S_1\mathrm{\ne }{S}_2$. Hence this singular operator has more than one polar isometric factor, exactly as [L1] allows.

Direct multiplication gives $A^{\mathsf{T}}A=\mathrm{diag}(4,1,0)$, so the singular values are $2,1,0$, with standard basis vectors as right-singular vectors.

The images of the first two standard basis vectors are $2e_1$ and $e_2$, so the standard basis of ${\mathbb{R}}^2$ gives the left-singular vectors. Therefore $U=I_2$, $V=I_3$, and the displayed diagonal matrix $\mathrm{\Sigma }$ give an SVD of $A$, exactly as [L1] guarantees.

By [L1], deleting the second singular direction replaces $A$ by $A_1=\left(\begin{array}{ccc}2& 0& 0\\ 0& 0& 0\end{array}\right)$, which has rank $1$, and $A-A_1=\left(\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\end{array}\right)$ has operator norm $1$.

The next singular value after the retained one is $1$, so [L2] says exactly that no rank-at-most-one matrix lies closer to $A$ in operator norm than $A_1$. Therefore $A_1$ is the Eckart-Young minimiser.

For every nonzero $(x,y,z)$, the displayed formula shows $1\le R_D(x,y,z)\le 4$, with the value $4$ at $e_1$, the value $2$ at $e_2$, and the value $1$ at $e_3$. On $\mathrm{span}(e_2,e_3)$ the quotient is at most $2$, and on $\mathrm{span}(e_1,e_2)$ it is at least $2$.

Therefore ${\max }_{v\mathrm{\ne }0}{R}_D(v)=4$, ${\min }_{v\mathrm{\ne }0}{R}_D(v)=1$, and the $k=2$ Courant-Fischer min-max and max-min values are both $2$. This matches the ordered eigenvalues $4,2,1$, exactly as [L1] predicts.

The characteristic polynomial of $A$ is $(x-3{)}^2(x-1)$, so its ordered eigenvalues are $3,3,1$. The characteristic polynomial of $B$ is $(x-3)(x-1)$, so its ordered eigenvalues are $3,1$.

These satisfy $3={\lambda }_1(A)\ge {\mu }_1(B)=3\ge {\lambda }_2(A)=3$ and $3={\lambda }_2(A)\ge {\mu }_2(B)=1\ge {\lambda }_3(A)=1$, exactly the pattern asserted by [L1].

The row sums off the diagonal are $1,1,0$, so the displayed three Gershgorin disks are exactly the ones attached to $A$.

Because $A$ is upper triangular, its eigenvalues are the diagonal entries $2,3,4$, and each of these lies in the displayed union of disks. This is the concrete containment asserted abstractly by [L1].

By [L1], the quarter-turn $A$ is a normal real operator.

The same example [L1] shows that $A$ has no real eigenvalue, so it cannot have a real eigenbasis and is not diagonalisable over $\mathbb{R}$. This refutes the field-free claim.

By [L1], the matrix $A$ is indeed complex symmetric.

If $A$ were unitarily diagonalisable, [L2] would make it normal. That contradicts [L1]. Therefore the claim is false.

For every $v=(x,y{)}^{\mathsf{T}}$, one has $Av=(-y,x{)}^{\mathsf{T}}$ and therefore $⟨Av,v⟩=-xy+xy=0\ge 0$. Thus the hypothesis of the claim holds for $A$.

By [L1], the same operator $A$ is not self-adjoint. Hence the displayed inequality alone does not force self-adjointness, and the claim is false.

The operator $I$ is non-negative, and [L1] says its non-negative square root is uniquely $I$ itself.

Nevertheless $(-I{)}^2=I^2=I$ and $-I\mathrm{\ne }I$. So $I$ has at least two square roots, and uniqueness fails once the non-negative condition on the square root is dropped.

By [L1], the singular matrix $T$ admits both decompositions $T=S_{1}U$ and $T=S_{2}U$, with $S_1\mathrm{\ne }{S}_2$.

Therefore the isometry in a polar decomposition need not be unique for a singular operator. This refutes the claim.

The characteristic polynomial of $N$ is $x^2$, so both eigenvalues are $0$ and their absolute values are $0,0$.

Direct multiplication gives $N^{\mathsf{T}}N=\mathrm{diag}(0,4)$, so the singular values are $2$ and $0$. Because $2\mathrm{\ne }0$, the singular values are not the absolute values of the eigenvalues.

The characteristic polynomial of $N$ is $x^2$, so every eigenvalue of $N$ has modulus $0$.

For a unit vector $(x,y{)}^{\mathsf{T}}$, one has $\parallel N(x,y{)}^{\mathsf{T}}\parallel =2\mathrm{\mid }y\mathrm{\mid }\le 2$, with equality at $(0,1{)}^{\mathsf{T}}$. Thus $\parallel N\parallel =2$, which by [L1] is also its largest singular value. Because $2\mathrm{\ne }0$, the operator norm is not the largest eigenvalue modulus.