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If $T^2=I$, every $v\in V$ has the decompositionwhose first term belongs to $E_1$ and second to $E_{-1}$. Their intersection is zero by ii, so $V=E_1\oplus E_{-1}$ is a direct sum. Choosing bases of the two eigenspaces gives a diagonal matrix with entries $1$ and $-1$.

Since $T^2\ne T$ while $T^2=I$, we have $T\ne I$. Thus $\dim E_{-1}$ is $1$, $2$, or $3$. Reordering the basis gives exactlyFor $-I_3$, $T$ is point inversion through the origin. For $\mathrm{diag}(1,1,-1)$, it fixes the plane $E_1$ pointwise and reverses the complementary line $E_{-1}$: if $v=p+\ell$ in this direct sum, then $Tv=p-\ell$. For $\mathrm{diag}(1,-1,-1)$, it fixes the line $E_1$ pointwise and reverses the complementary plane $E_{-1}$.

These are instances of oblique symmetry of complementary subspaces, allowing oblique eigenspaces. They are a Euclidean reflection in a hyperplane and a rotation in three dimensions through $\pi$ about a line, respectively, when the two eigenspaces are perpendicular. The hypotheses do not require an orthogonal transformation or an orthonormal basis, so that additional metric property cannot be inferred just from the displayed diagonal forms.

Write $r=(r_1,\dots ,r_n{)}^T$. The linear map represented by $A$ has the formThe image of a linear map here is contained in the one-dimensional linear span of $e_i$, so its matrix rank is at most one. The rank-nullity theorem yieldsThus the solution space of $Ax=0$, namely the zero eigenspace when zero is an eigenvalue, contains at least $n-1$ linearly independent vectors. If $r=0$ its dimension is $n$; otherwise it is $n-1$. When $n=1$ the asserted lower bound is zero, even if zero is not an eigenvalue.

Let $X_c$ be the column-centered data matrix. Its sample covariance matrix is $S=X_c^T{X}_c/(n-1)$. The loadings matrix $V=(v_1,\dots ,v_p)$ has orthonormal eigenvectors of $S$ as columns, ordered by nonincreasing eigenvalues:Each principal component loading specifies the coefficients of a principal direction in the original variables. Signs are arbitrary, and equal eigenvalues permit orthogonal changes of basis within their eigenspaces.

If $Av=\lambda v$, then $A^{2}v={\lambda }^{2}v$. Alternatively, direct matrix multiplication givesThus the normalized eigenvectors $v_{+},v_{-}$ from part (b) have respective eigenvaluesIf $\sin 2\theta =0$, then $B=I$, and the whole two-dimensional space is its eigenspace. This degeneracy includes cases where the two eigenvalues of $A$ were distinct but opposite.

The characteristic polynomial is $\det (\lambda I-M)=\lambda ({\lambda }^2-2)$. Solving the three eigenvector equations givesEach eigenspace consists of the scalar multiples of its displayed nonzero vector. Their pairwise dot products vanish: $v_{+}\cdot v_{-}=1-2+1=0$ and $v_{\pm }\cdot v_0=1-1=0$. This also agrees with the orthogonal diagonalization of a real symmetric matrix.

Compute the characteristic polynomial using $\det (\lambda I-M)$:Its three distinct real roots are $1$ and $3\pm \sqrt{15}$. For $\lambda =1$, the first row of $(M-I)v=0$ gives $z=3y$ and the second then gives $x=y$, so take $v_1=(1,1,3{)}^T$.

For either root of ${\lambda }^2-6\lambda -6=0$, take $z=1$ and solve the row equations to get $v=(\lambda -8,5-\lambda ,1{)}^T$. Substitution makes the first two row residuals opposite multiples of the quadratic equation and the third zero. Thus a complete choice isEvery nonzero scalar multiple is another valid eigenvector in the corresponding one-dimensional eigenspace.

The characteristic polynomial isFor $\lambda =2$, the eigenvector equation reduces to $x=y$; for $\lambda =8$, it reduces to $x=-y$. Choose the following unit vectors consistently for the remaining parts:Each eigenspace consists of all scalar multiples of its displayed vector, including zero; an eigenvector itself must be nonzero.

For $\lambda =5$, the eigenvalue equation reduces to a one-dimensional eigenspace spanned by $(1,2,-1)$. For $\lambda =-3$, the rows of $A+3I$ are multiples of $(1,2,-3)$, so its eigenspace is the plane $x+2y-3z=0$. One possible set of three independent eigenvectors isEvery eigenvector for $-3$ is a nonzero linear combination of the last two. Their two-dimensional eigenspace shows that the repeated root is not defective: the matrix is diagonalizable.

For the particular eigenvectors chosen in (iii), the dot products areThus only ${\mathbf{e}}_1$ and ${\mathbf{e}}_2$ are orthogonal for this choice. The matrix is not symmetric, so distinct eigenvalues alone do not imply orthogonality. Another basis within the repeated eigenspace may change which chosen vectors are perpendicular; the dot products above specify the convention completely.

A vector perpendicular to ${\mathbf{e}}_1$ has the form $\mathbf{v}=(s,t,-s)$. Direct matrix multiplication givesThus every nonzero such vector is an eigenvector with $\boxed{\lambda =-2}$. The zero vector must be excluded from the literal claim about “any” perpendicular vector, since an eigenvector is nonzero. The eigenspace is two-dimensional. Equivalently, $A=-2I+6{\mathbf{e}}_1{\mathbf{e}}_1^{\mathsf{T}}$ exhibits the spectrum of a scalar matrix plus a rank-one projector.

Choose two perpendicular unit directions in the $-2$ eigenspace:Their pairwise dot products vanish and their Euclidean norms are one. They form an orthonormal basis of eigenvectors, with eigenvalues $4,-2,-2$, as guaranteed for a real symmetric matrix by the spectral theorem for real symmetric matrices.

Using the ordered orthonormal basis chosen in (c), the column matrix isOther signs or another orthonormal choice within the repeated eigenspace are equally valid, provided the same choice is retained in the remaining parts.

Use the common eigenvectors of $A$ and $A^{\mathsf{T}}$. For each $k$,Thus, without constructing $S$ explicitly, its eigenvalues areThe eigenvalue $1$ has direction $(1,1,1{)}^{\mathsf{T}}$. The repeated eigenvalue $-1/2$ has a real eigenspace consisting of vectors whose coordinates sum to zero. This is the general rule for the symmetric part of an orthogonal matrix. Adding unrelated lists of eigenvalues would not suffice; here the shared eigenvectors justify the pairing.

By the spectral theorem for real symmetric matrices, choose the normalized eigenvectors to form an orthonormal basis. With $P_n={\mathbf{e}}_n{\mathbf{e}}_n^{\mathsf{T}}$, orthonormality gives $P_n{P}_m={\delta }_{nm}{P}_n$ and ${\sum }_n{P}_n=I$. For every vector $\mathbf{x}$, expand $\mathbf{x}={\sum }_n({\mathbf{e}}_n^{\mathsf{T}}\mathbf{x}){\mathbf{e}}_n$ and apply $A$:Since this equality holds for every $\mathbf{x}$, the spectral decomposition isEach $P_n$ is the rank-one orthogonal projection onto its eigenvector direction. In a degenerate eigenspace, normalized eigenvectors must be chosen mutually orthogonal; normalization alone does not guarantee this resolution of the identity.

A left eigenvector satisfies $(M-\lambda I{)}^T\mathbf{x}=0$. Here $M$ is a symmetric matrix, so its left and right eigenspaces have the same components. The characteristic polynomial factors asThe eigenvalues and all corresponding nonzero row eigenvectors areFor example, in the $xy$ block, equal components give the eigenvalue $3$ and opposite components give $-1$, while the $z$ coordinate is unaffected.

The two orthogonal directions $(1,1{)}^T$ and $(1,-1{)}^T$ are preserved by this symmetric matrix. Direct multiplication, using the hyperbolic function identities $\cosh u\pm \sinh u=e^{\pm u}$, gives the eigenvalue–eigenvector pairsEach eigenvector can be multiplied by any nonzero scalar. For $u\ne 0$ these give the two one-dimensional eigenspaces. When $u=0$, the matrix is the identity: the only eigenvalue is $1$, with multiplicity two, and every nonzero vector is an eigenvector.

Induction on $p$ gives $S^p{\mathbf{u}}_i={\lambda }_i^p{\mathbf{u}}_i$, including $p=0$. The absolutely convergent matrix exponential series may therefore be applied term by term to each eigenvector:ThusFor a repeated eigenvalue, every nonzero vector in its eigenspace remains an eigenvector. Because the real exponential function is injective, distinct real eigenvalues of $S$ do not merge under this transformation.

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.

Suppose $AB=BA$ and $A{v}_j=a_j{v}_j$. ThenBecause the eigenvalues of $A$ are distinct, the eigenspace for $a_j$ is one-dimensional. It follows that $B{v}_j=b_j{v}_j$, including the possibility $b_j=0$. Thus an orthonormal eigenbasis of $A$ is also an eigenbasis of $B$. The simplicity of both spectra means their eigenvectors coincide as directions, up to ordering, phases and nonzero scalar multiples.

Conversely, if they share a complete eigenbasis, choose the unitary matrix $U$ whose columns are these common normalized eigenvectors. Then $U^{†}AU=D_A$ and $U^{†}BU=D_B$ are diagonal, so $D_A{D}_B=D_B{D}_A$ and $AB=BA$. HenceThis is simultaneous diagonalization of the two Hermitian matrices.

The spectral theorem gives a real orthonormal eigenbasis $v_1,\dots ,v_n$ for a real symmetric matrix $A$. Distinct eigenspaces are orthogonal; within a repeated eigenspace, choose an orthonormal basis, for example by Gram-Schmidt orthogonalization. Put these eigenvectors in the columns of $T$. Then $T^{-1}=T^T$ andFor a change of variables $y=Sx$, choose $\boxed{S=T^T}$. Since $x=Ty$, the quadratic form becomes $x^{T}Ax={\sum }_i{\lambda }_i{y}_i^2$; the ${\lambda }_i$ are precisely the eigenvalues of $A$.

In the numerical case, the off-diagonal entries are half the cross-term coefficients, soA direct eigenvector calculation gives the unit eigenvectors $(1,1,0)/\sqrt{2}$, $(1,-1,1)/\sqrt{3}$, $(1,-1,-2)/\sqrt{6}$, with eigenvalues two, three and minus six. They are pairwise orthogonal. Therefore the required orthogonal diagonalization isand the transformed quadratic form is $\boxed{2y_1^2+3y_2^2-6y_3^2}$.