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For $n=2$, takeThey are commuting operators because every matrix commutes with the identity matrix. However $A$ has only the eigenvalue zero, and $\ker A=\mathrm{span}\{(1,0{)}^T\}$ has dimension one. It has no eigenbasis, so it cannot appear in any simultaneous diagonalization. Equivalently, a diagonal matrix similar to this nonzero nilpotent matrix would have to be zero, an impossibility. For any prescribed $n\ge 2$, use $A=E_{12}$ and $B=I_n$; its zero eigenspace has dimension $n-1$. For $n=1$ no such example exists.

The zero matrix has the required pattern, and matrix addition and scalar multiplication preserve every prescribed zero entry. This proves the vector subspace condition. A basis consists of the matrix unitsThey independently select every permitted entry of a tridiagonal matrix, so the dimension is $n+2(n-1)=3n-2$ for $n\ge 1$.

They are similar matrices, so their eigenvalues, with algebraic multiplicities, are identical. Explicitly,Moreover their eigenvectors correspond by $Av=\lambda v⟺(D^{-1}AD)(D^{-1}v)=\lambda D^{-1}v$. Since the transformed matrix is a real symmetric matrix, the spectral theorem for real symmetric matrices also shows that $A$ has real eigenvalues and is a diagonalizable matrix.

The update must place the sum of the two current entries in the first coordinate and retain the old first coordinate in the second. Thus choose the matrixBy matrix multiplication, $A(L_n,L_{n-1}{)}^T=(L_n+L_{n-1},L_n{)}^T=(L_{n+1},L_n{)}^T$ for every $n\ge 1$, using the given recurrence relation. This is the companion-matrix representation of the Lucas numbers.

Use mathematical induction on $n$. For $n=0$, the identity matrix gives $(1,2{)}^T=(L_1,L_0{)}^T$. If $A^n(1,2{)}^T=(L_{n+1},L_n{)}^T$, applying $A$ and using part (a) gives $A^{n+1}(1,2{)}^T=(L_{n+2},L_{n+1}{)}^T$. ThereforeThe base case is essential: the starting vector represents consecutive Lucas numbers, not two arbitrary initial values.

Choose $A=I_2$ and $B=3I_2$. Then $AB=3I_2=A^2+A+I_2$. Both are real $2\times 2$ matrices, as required.

Rearrange the identity intoTaking determinants gives $\det A\det (B-A-I)=1$. Hence $\det A\ne 0$, so $A$ is an invertible matrix, with $A^{-1}=B-A-I$. This uses that the matrices are finite and square: a right inverse then implies a two-sided inverse. Applying the identity to a vector in the kernel without first establishing commutation would not justify cancelling an $A$ on the wrong side. The argument is a case of invertibility forced by a polynomial product identity.

Let $Av=\lambda v$ for a real eigenvalue $\lambda$ and a nonzero real eigenvector $v$. Part (b) gives $\lambda \ne 0$ and $A^{-1}v={\lambda }^{-1}v$. The expression for $B$ in part (c) therefore givesConsequently $\lambda +1+{\lambda }^{-1}$ is a real eigenvalue of $B$, with the same eigenvector $v$. For a real matrix and real eigenvalue, a nonzero real vector can be chosen in the real kernel of $A-\lambda I$; no complex-vector assumption is needed.

For scalar $a=A\ne 0$, part (c) givesThe numerator is strictly positive for every real $a$, and the denominator is nonzero. Therefore $B\ne 0$ and hence $B$ is invertible. This proof uses the order structure of the real scalars; it does not say that the matrix $A^2+A+I$ must be nonsingular in higher dimension.

An integer-entry example isDirect matrix multiplication gives $A^2=\left(\begin{array}{cc}-1& 1\\ -1& 0\end{array}\right)$, so adding $A$ and $I_2$ gives the zero matrix. Equivalently, this is the companion matrix of $t^2+t+1$ and has the two nonreal cube roots of unity as eigenvalues. The direct multiplication is already a complete proof. It also shows why part (e) does not extend to all real square matrices: with this $A$, choosing $B=0$ satisfies the original product identity.

Assume $n\ge 1$. With $u=(1,\dots ,1{)}^T$, the all-ones matrix satisfies $Av=({\sum }_j{v}_j)u$. Hence $n$ is an eigenvalue, and for $n\ge 2$ the other eigenvalue is $0$. At $n=1$ the matrix is $[1]$, so 1 is its only eigenvalue; zero is not an eigenvalue with a nonzero eigenvector in that case.

The all-ones matrix is diagonalizable for every $n\ge 1$. The vector $u$ together with the $n-1$ vectors from part (ii) is an eigenbasis. Independence follows because a multiple of $u$ has coordinate sum $n$ times that multiple, whereas every other basis vector has coordinate sum zero. In this basis the diagonal matrix is $\mathrm{diag}(n,0,\dots ,0)$. At $n=1$ the one-vector basis suffices. Alternatively, the spectral theorem for real symmetric matrices applies. If dimension zero is allowed by the convention for $\mathbb{N}$, the empty $0\times 0$ matrix has no eigenvalues, an empty basis and a vacuous diagonalization.

Write $B=QD{Q}^{-1}$ with $D$ a diagonal matrix. Then $0=B^k=Q{D}^k{Q}^{-1}$, so each diagonal entry ${\lambda }_i$ satisfies ${\lambda }_i^k=0$ and therefore ${\lambda }_i=0$. Hence $D=0$ and $B=0$. Conversely the zero matrix is diagonalizable and nilpotent. This proves the existing result that a diagonalizable nilpotent matrix is zero; nilpotence alone does not imply a zero matrix, because nonzero nilpotent Jordan blocks exist.

The matrix $C=A-b{b}^T$ is an invertible matrix. Indeed, on the line spanned by $b$ it has eigenvalue $-1/2$, and on $b^{\perp }$ it agrees with the positive-definite restriction of $A$. Therefore the full Newton step is well defined and givesThis is the unique stationary point because $C$ is an invertible matrix. The algebra applies to every starting point, but it does not make the answer a minimum: for $d\ge 2$ it is the saddle point just classified, and for $d=1$ it is a maximum. A safeguarded minimization algorithm that replaces an indefinite matrix used as its Hessian matrix or damps the step need not take this exact step; the assertion concerns the ordinary full Newton step formula.

The $m$ columns of $L$ form a basis of $\ker A^{\top }$, so that null space has dimension $m$. The rank-nullity theorem applied to $A^{\top }:{\mathbb{R}}^n\to {\mathbb{R}}^n$ gives $\mathrm{rank}{A}^{\top }=n-m$. A matrix and its transpose have equal matrix rank, hence

Both $L$ and $R$ have $m$ linearly independent columns. The Gram matrix $R^{\top }R$ is invertible becauseConsequently $R^{\top }$ is a surjective linear map: for any $y\in {\mathbb{R}}^m$, the choice $x=R(R^{\top }R{)}^{-1}y$ satisfies $R^{\top }x=y$. HenceThe latter image of a linear map has dimension $m$, giving $\mathrm{rank}B=m$. If $m=0$, $B=0$ and the conclusion is immediate. This argument explains why the product retains its full possible matrix rank; a general product of two matrices need not do so.

Split $B$ into its symmetric part and skew-symmetric part:Taking the transpose gives $S^{\top }=S$ and $K^{\top }=-K$. Thus $S$ is a symmetric matrix and $K$ a skew-symmetric matrix. These formulas also show uniqueness, since a real matrix that is both symmetric and skew-symmetric is zero.

For a real eigenvalue $\lambda$, choose a nonzero real eigenvector $v$ with $Av=\lambda v$. By (e),Since $v^{\top }v=\Vert v{\Vert }^2>0$, $\lambda =0$. If eigenvectors were initially allowed to be complex, the real matrix $A-\lambda I$ has a nonzero real null space whenever it has a nonzero complex null space, so this choice of $v$ remains justified.

If $(I+A)x=0$, take its dot product with $x$ and apply (e):Hence $x=0$. A square matrix with zero null space is an invertible matrix by the rank-nullity theorem. Therefore $I+A$ is nonsingular. This proves invertibility of identity plus a real skew-symmetric matrix.

Interpret the norms as Euclidean norms and set $B=A^{\mathsf{T}}A$. This is a real symmetric matrix with ${\mathbf{x}}^{\mathsf{T}}B\mathbf{x}=\Vert A\mathbf{x}{\Vert }^2\ge 0$, hence a positive semidefinite matrix. Squaring the nonnegative objective does not change its argmin. Let ${\lambda }_{\min }$ be the smallest eigenvalue of $B$ and $E_{\min }$ its eigenspace.

First minimize on the unit sphere. The Lagrange multiplier function $L(\mathbf{x},\lambda )={\mathbf{x}}^{\mathsf{T}}B\mathbf{x}-\lambda ({\mathbf{x}}^{\mathsf{T}}\mathbf{x}-1)$ has stationarity condition $B\mathbf{x}=\lambda \mathbf{x}$. Multiplication by ${\mathbf{x}}^{\mathsf{T}}$ shows that the squared objective at such a point is $\lambda$. The spectral theorem for real symmetric matrices also gives, for every unit vector, ${\mathbf{x}}^{\mathsf{T}}B\mathbf{x}\ge {\lambda }_{\min }$, with equality exactly in $E_{\min }$. This proves global minimality, rather than merely finding stationary candidates.

If ${\lambda }_{\min }>0$, every point with $\Vert \mathbf{x}\Vert >1$ has a strictly smaller objective after rescaling to unit length. ThereforeIf ${\lambda }_{\min }=0$, the kernel of a linear map satisfies $\ker B=\ker A$, since $B\mathbf{x}=0$ implies $\Vert A\mathbf{x}{\Vert }^2=0$, and the converse is immediate. The minimum is then zero, attained precisely atIn this case minimizers need not have unit length. This distinction is essential for a rank-deficient $A$, including any real $m\times n$ matrix with $m<n$. The result is the minimum image norm outside a Euclidean unit ball, governed by the least singular value.