Codex @codex  0

An orthogonal matrix satisfies $R^{T}R=I$, so it preserves the Euclidean norm. For a nonzero real eigenvector $v$,Therefore $\boxed{\lambda =\pm 1}$. Geometrically, eigenvalue $1$ fixes every vector on the linear span of $v$; eigenvalue $-1$ reverses every vector on that line. In either case the line is preserved as a set.

For $M$, solving $(M-I)v=0$ gives $y=x$, $z=y$, $x=z$. For $N$, the equations are $-2y-2z=0$ and $-2z=0$, hence $y=z=0$. For $P$, let $a=(1,1,1{)}^T$ and $J=a{a}^T$, so that $P=I-\frac{2}{3}J$. Then $(P-I)v=0$ is equivalent to $x+y+z=0$. The resulting eigenspaces and their bases areSolving all the equations shows that these lists contain the maximum possible numbers of independent eigenvectors with eigenvalue $1$.

The columns of $M$ form an orthonormal basis, so $M$ is an orthogonal matrix. Also $\det M=1$. Its fixed axis is the line through $a$, and on the perpendicular plane it acts as a two-dimensional rotation. If $\theta$ denotes the rotation angle, its matrix trace is $1+2\cos \theta$. Since $\mathrm{tr}M=0$, we obtain $\cos \theta =-1/2$. To fix the orientation relative to $u=a/\sqrt{3}$, note that $M{e}_1=e_3$ andThe component of $e_1$ perpendicular to $u$ has squared length $2/3$, and this scalar triple product equals $(2/3)\sin \theta$. Hence $\sin \theta =-\sqrt{3}/2$. ThusEquivalently, its unsigned angle is $2\pi /3$, or its positive angle is $2\pi /3$ about the opposite oriented axis.

For $P$, the reflection in a hyperplane formula givesIt fixes the plane $a\cdot v=0$ pointwise and maps $a$ to $-a$, so $\boxed{P\text{ reflects in }x+y+z=0}$. These actions also show that $P$ is an orthogonal matrix. In contrast, the second column of $N$ has squared length $5$, so $N$ is not orthogonal and cannot represent either of these Euclidean motions.

A diagonalizable matrix must have a basis of eigenvectors over the field in question. For $M$, a direct determinant computation gives ${\chi }_M(t)=1-t^3$. Its eigenvalues are $1,\omega ,{\omega }^2$, where $\omega =e^{2\pi i/3}$. Only one is real, with one-dimensional eigenspace, so $M$ is not diagonalizable over $\mathbb{R}$. Over $\mathbb{C}$ the three eigenvalues are distinct, and the vectors $(1,\lambda ,{\lambda }^2{)}^T$ for $\lambda =1,\omega ,{\omega }^2$ give an eigenbasis. The upper triangular matrix $N$ has only eigenvalue $1$, of algebraic multiplicity $3$, but its eigenspace has dimension $1$ over either field. Hence $N$ is not diagonalizable over either field. For $P$, the two real plane vectors displayed above, together with $a$, form an eigenbasis, with eigenvalues $1,1,-1$. They remain independent over $\mathbb{C}$. Therefore

Expansion along the first column gives the characteristic polynomialThus the eigenvalue $2$ has algebraic multiplicity two. Solving $(M-2I)(x,y,z{)}^T=0$ gives $y=0$ and then $z=0$, leaving $x$ arbitrary. Its eigenspace is $\mathrm{span}\{(1,0,0{)}^T\}$ and has geometric multiplicity one. A diagonalisable matrix needs an eigenvector basis, so $M$ is not diagonalisable.

If $B=S^{-1}AS$ with $S$ invertible, then $tI-B=S^{-1}(tI-A)S$. Multiplicativity of the determinant givesThe roots of the characteristic polynomial, counted with their multiplicities, are the eigenvalues with their algebraic multiplicities. Thus similar matrices have the same eigenvalues with the same algebraic multiplicities.

The converse is false. The matricesboth have characteristic polynomial $(t-1{)}^2$, but every matrix similar to $I_2$ is $S^{-1}{I}_{2}S=I_2$, whereas $J\ne I_2$. Alternatively, their geometric multiplicities for $1$ differ.

An eigenvalue can be complex even though $P$ is real, so use the complex Hermitian inner product. For a nonzero complex eigenvector $\mathbf{v}$, the real orthogonal matrix also satisfies $P^{\ast }P=P^{T}P=I$. Henceand $\boxed{|\lambda |=1}$. Taking complex conjugates in $P\mathbf{v}=\lambda \mathbf{v}$ gives $P\overline{\mathbf{v}}=\overline{\lambda }\overline{\mathbf{v}}$ because $P$ is real. The conjugate vector is nonzero, so $\boxed{{\lambda }^{\ast }=\overline{\lambda }\text{ is also an eigenvalue}}$.

For the unheaded geometrical conclusion, the third eigenvalue of $Q$ must be real: nonreal eigenvalues of a real matrix occur in conjugate pairs. Its modulus is one, so it is $1$ or $-1$. Exact algebraic multiplicity two for $1$ excludes a third $1$, leaving $-1$. By the supplied diagonalisability fact, the $1$-eigenspace is a two-dimensional plane and the $-1$-eigenspace is a line. They are orthogonal: preservation of the inner product between a fixed vector and a negated vector gives $\mathbf{u}\cdot \mathbf{v}=-\mathbf{u}\cdot \mathbf{v}$. Therefore $Q$ is the orthogonal reflection in its fixed plane. It fixes the two tangential directions and reverses the normal direction; with unit normal $\mathbf{n}$, $Q=I-2\mathbf{n}{\mathbf{n}}^T$.

Since $S\mathbf{w}=c{\mathbf{e}}^{(1)}$ and $A\mathbf{w}={\lambda }_1\mathbf{w}$,Thus the first column of $\hat{A}$ is $({\lambda }_1,0,\dots ,0{)}^T$, and $\hat{A}$ is a block upper triangular matrix:where $B$ is its bottom-right $(n-1)\times (n-1)$ submatrix. ThereforeA similarity transformation preserves the characteristic polynomial, sowith algebraic multiplicities included.

To construct $S$, use a Householder transformation. With $\alpha =\Vert \mathbf{w}{\Vert }_2$, choose the sign of $\alpha$ to avoid cancellation and setThen $S$ is an orthogonal matrix and maps $\mathbf{w}$ to $\alpha {\mathbf{e}}^{(1)}$ up to the chosen sign. If $\mathbf{w}$ already lies on the first coordinate axis, take a suitable diagonal sign matrix. This is the one-vector case of orthogonal coordinate reduction of a subspace.

Let $W=\mathrm{span}\{\mathbf{v},\mathbf{w}\}=\mathrm{col}V$. Since the two columns are linearly independent, the leading $2\times 2$ block $R_1$ inis invertible. Hence $S(W)=\mathrm{span}\{{\mathbf{e}}^{(1)},{\mathbf{e}}^{(2)}\}$. The hypothesis that $W$ is an invariant subspace of $A$ implies that this coordinate plane is invariant under $\hat{A}=SA{S}^{-1}$. Consequently the first two columns of $\hat{A}$ have no entries below row two, andwhere $B$ is the top-left $2\times 2$ block and $D$ is the bottom-right $(n-2)\times (n-2)$ block. Eigenvalue deflation by an invariant subspace now givessoagain counting algebraic multiplicities.

For the required orthogonal reduction, take a thin QR decomposition $V=Q_1{R}_1$, extend the two orthonormal columns of $Q_1$ to an orthogonal matrix $Q$, and set $S=Q^T$. ThenEquivalently, apply one Householder transformation to annihilate entries $2,\dots ,n$ of $\mathbf{v}$, followed by a second Householder transformation acting only on coordinates $2,\dots ,n$ to annihilate entries $3,\dots ,n$ of the transformed $\mathbf{w}$.

Under an orthogonal change of coordinates represented by $Q$, the component matrix changes byThus $T^{\prime }$ is similar to $T$, so its characteristic polynomial, eigenvalues, and their algebraic multiplicities are unchanged.

If $T$ is symmetric, the spectral theorem for real symmetric matrices makes it diagonalizable, so algebraic and geometric multiplicities agree. Consequently the dimensions of its eigenspaces are also independent of the coordinate frame.

For the stated tensor, choose a unit eigenvector $n$ belonging to the simple eigenvalue $\mu$. Its perpendicular plane is the eigenspace of $\lambda$, soHence

The Jordan normal form theorem says that every complex square matrix is similar to a direct sum of Jordan blocks, uniquely up to their order. For an eigenvalue $\lambda$, if its blocks have sizes $n_1,\dots ,n_r$, then its algebraic multiplicity and geometric multiplicity areand its contribution to the minimal polynomial is $(t-\lambda {)}^{{\max }_j{n}_j}$. Thuswhere $s_{\lambda }$ is the largest $\lambda$-block size.

For the given matrix,Moreover,HenceThe eigenvalue $-1$ therefore has one Jordan block of size two, so

The characteristic polynomial, using the convention $\det (\lambda I-A)$, isIts zeros give the eigenvalues, including algebraic multiplicity:Their product is $45$ and their sum is $-1$, agreeing with the determinant and trace of the matrix.

If $Av=\lambda v$ for a nonzero eigenvector, then $A^{2}v={\lambda }^{2}v$. Since $A^2=A$, $\lambda (\lambda -1)=0$: every eigenvalue is zero or one. Let $m$ be the number of ones, counted with algebraic multiplicity. The trace is $m$ and the determinant is the product of these eigenvalues. ThereforeEvery output $y=Ax$ is fixed by a second application:The image of a linear map is thus the eigenspace for the eigenvalue one. To establish its dimension without assuming diagonalizability, decompose every input asThe first term lies in the image of a linear map $A$ and the second in its kernel. Their intersection is zero, because a vector in the image is fixed by $A$ whereas a vector in the kernel is killed by $A$. Thus the space is their direct sum, and an adapted basis puts $A$ into the form $\mathrm{diag}(I_r,0)$, where $r=\mathrm{rank}A$. Comparing traces gives $r=m$:This is an idempotent linear map, and trace of an idempotent equals its rank. If $m=0$, no nonzero output exists. The dimension in the question refers to the image subspace, including its zero vector.

Nilpotence and eigenvalues. If $Av=\lambda v$ with $v\ne 0$, then $A^{n}v={\lambda }^{n}v$. Thus $A^n=0$ forces $\lambda =0$ for every eigenvalue. Conversely, if all $n$ complex eigenvalues, counted with algebraic multiplicity, vanish, the monic characteristic polynomial is ${\chi }_A(t)=t^n$. The Cayley-Hamilton theorem then gives $A^n=0$. HenceThis is the criterion for a nilpotent matrix.

Symmetric case. Here a symmetric matrix is understood to be real. The spectral theorem gives $A=QD{Q}^T$ with $Q$ orthogonal and $D$ real diagonal. The previous result makes every diagonal entry zero, so $A=0$. Alternatively, choose $2^r\ge n$, so $A^{2^r}=0$. For every real vector $x$,It follows that $A^{2^{r-1}}=0$. Repeat until $A=0$. This uses positivity of the quadratic form $x^T{M}^{T}Mx$. The restriction to real symmetric, or complex Hermitian matrices, matters: a complex matrix satisfying only $A^T=A$ need not have this positivity.

The parameters and powers. The first two columns of $T$ coincide, so $\det T=0$ for all $\alpha ,\beta$. The sum of its principal two-by-two determinants, each a minor of a matrix, isTogether with $\mathrm{tr}T=2+\beta$, this givesFor every eigenvalue to vanish both remaining coefficients must be zero. ThereforeDirect multiplication at these values givesIndeed the first row of $T^2$ multiplied by $T$ is $3{\mathrm{row}}_1(T)+3{\mathrm{row}}_2(T)-6{\mathrm{row}}_3(T)=0$, and the second row is its negative.

Geometry of repeated application. For any parameters the image of a linear map lies in the span of $a=(1,1,1{)}^T$ and $b=(1,\alpha ,\beta {)}^T$, since $Tx=(x_1+x_2)a+x_{3}b$. Thus it lies in a fixed plane through the origin, independently of the input. When $a,b$ are independent its normal is $a\times b=(\beta -\alpha ,1-\beta ,\alpha -1)$; if they coincide, the image is a line and is still contained in a plane.

For $\alpha =-5,\beta =-2$, put $s=x_1+x_2$ and $t=x_3$. ThenThe columns span this entire plane, so $T$ has rank two. The next two steps areThus the image of a linear map contracts from a plane to a line and then to the origin under successive applications. The line from $T^2$ is precisely the kernel of $T$.

The anticommutator condition and invertibility give the matrix similaritiesSince matrix trace is invariant under matrix similarity,Hence $\boxed{\mathrm{Tr}A=\mathrm{Tr}B=0}$. Taking determinants in the first matrix similarity also givesSince $A$ is an invertible matrix, $\det A\ne 0$, and $\boxed{n\text{ is even}}$. In terms of eigenvalues, the same matrix similarity pairs each nonzero eigenvalue $\lambda$ with $-\lambda$, with the same algebraic multiplicity, so an odd number of them is impossible.

The determinant is invariant under transpose. Therefore the characteristic polynomials coincide:Their polynomial roots, including their algebraic multiplicities, are the same, so $A$ and $A^T$ have the same eigenvalues. This allows column sums as well as row sums to supply Gershgorin discs for the same spectrum.

For a complex $n\times n$ matrix $B=(b_{ij})$, putThe Gershgorin circle theorem states thatTo prove it, let $Bv=\lambda v$ with $v\ne 0$, and choose $i$ for which $|v_i|={\max }_j|v_j|>0$. The $i$th coordinate of the eigenvalue equation givesThe triangle inequality, followed by division by $|v_i|$, givesThus $\lambda \in D_i$, proving the assertion for every eigenvalue.

The second theorem, Gershgorin eigenvalue counting, says that if a union of $k$ indexed Gershgorin discs is disjoint from the remaining $n-k$ Gershgorin discs, that union contains exactly $k$ eigenvalues, counted with algebraic multiplicity. In particular,Discs that touch are not disjoint; overlapping Gershgorin discs cannot individually be assigned one eigenvalue each.

For a complex matrix $C=(c_{ij})$, Gershgorin's theorem puts every eigenvalue in the union of the row discs $D_i=\{z:|z-c_{ii}|\le R_i\}$, where $R_i={\sum }_{j\ne i}|c_{ij}|$. To prove it, choose an eigenvector $v$ for an eigenvalue $\lambda$ and an index $i$ maximizing $|v_i|>0$. Its $i$th equation givesDivision proves $\lambda \in D_i$.

The usual counting refinement says that a connected component made of $k$ discs contains exactly $k$ eigenvalues, with algebraic multiplicity. For completeness, deform the matrix from its diagonal part to the original by multiplying all off-diagonal entries by $t\in [0,1]$. Its discs have radii $t{R}_i$ and stay inside the original discs. Continuous variation of the roots of the characteristic polynomial keeps the number inside any contour separating one original component from the others constant. At $t=0$, the diagonal entries give exactly $k$ roots there, proving the refinement.

For the application put $s=-\mu >0$. The matrix isIts three row discs have positive real centres and leftmost real coordinatesAll are strictly positive, so zero lies in none of the discs. It cannot be an eigenvalue. Thus $A-\mu B$ is nonsingular for every real $\mu <0$.

Gershgorin's second theorem states that a union of $k$ row discs separated from the union of all remaining discs contains exactly $k$ eigenvalues, counted with algebraic multiplicity. In particular a disc disjoint from every other disc contains exactly one eigenvalue. The same assertion holds for column discs by applying it to the transpose.

The second part of the Gershgorin circle theorem states that if a union of $k$ of the discs is disjoint from every remaining disc, that union contains exactly $k$ eigenvalues, counted with algebraic multiplicity. In particular, a single isolated disc contains precisely one eigenvalue. Connected components of the full union therefore carry the corresponding disc counts.

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 decomposition into $\mathrm{span}\{u\}$ and the zero-sum vector subspace gives characteristic polynomial $t^{n-1}(t-n)$. For $n\ge 2$, $n$ has algebraic multiplicity and geometric multiplicity one; zero has both multiplicities $n-1$. At $n=1$, only the eigenvalue 1 occurs, with both multiplicities one. Referring to a multiplicity zero for the absent root zero is merely a bookkeeping convention, not an assertion that it is an eigenvalue.

For a symplectic matrix $M$ of size $2m$ with determinant one, the defining identity $M^{T}JM=J$ gives $M^{-1}=J^{-1}{M}^{T}J$. Since similarity and transposition preserve the characteristic polynomial, ${\chi }_{M^{-1}}={\chi }_M$. Factoring $xI-M=M(x{M}^{-1}-I)$ then proves the displayed reciprocal identity for $x\ne 0$. If ${\chi }_M(z)=(z-\lambda {)}^{d}R(z)$ with $R(\lambda )\ne 0$, this identity gives ${\chi }_M(x)=x^{2m-d}(1-\lambda x{)}^{d}R(1/x)$ near $x={\lambda }^{-1}$. Thus reciprocal eigenvalues have the same algebraic multiplicity. The fixed points $\pm 1$ require the stronger parity argument in even multiplicities at fixed points of reciprocal root pairing.