Sine Vectors for a Path
Path eigenvalues describe the frequencies supported by a graph that has no branching: each interior vertex is adjacent to its predecessor and successor, while each endpoint has only one neighbor. The corresponding eigenvectors have a sine profile. The endpoints matter: the sine formula makes the entries just outside the path equal to zero, which is exactly the boundary condition needed for the adjacency-matrix calculation.
All graphs here are finite and simple. For \(n\ge 1\), let \(P_n\) be the path with vertex set \(V(P_n)=\{1,\dots,n\}\) and edge set \(E(P_n)=\{j(j+1):1\le j<n\}\). Its adjacency matrix \(\mathbf A(P_n)\) has entry \(1\) when two vertices are adjacent and \(0\) otherwise. For a vector \(x=(x_1,\dots,x_n)^{\mathsf T}\), its matrix product has entries \(x_2,x_1+x_3,\dots,x_{n-1}+x_{n}\) in the corresponding endpoint and interior positions; more precisely, the first entry is \(x_2\), the \(j\)th interior entry is \(x_{j-1}+x_{j+1}\), and the last entry is \(x_{n-1}\). For \(n=1\), the matrix is \((0)\).
Proof. Put \(\theta=k\pi/(n+1)\), and first omit the common normalization factor, writing \(y_j=\sin(j\theta)\). Set \(y_0=0\) and \(y_{n+1}=0\); the second equality holds because \(\sin((n+1)\theta)=\sin(k\pi)=0\). The sine addition identity gives \(y_{j-1}+y_{j+1}=2\cos(\theta)y_j\) for every \(1\le j\le n\). At \(j=1\), this equation reads \(y_2=2\cos(\theta)y_1\), which is the first row of the path matrix equation because \(y_0=0\). At \(j=n\), it reads \(y_{n-1}=2\cos(\theta)y_n\), because \(y_{n+1}=0\). For each interior \(j\), it is exactly the equation for the \(j\)th row. Multiplying \(y\) by \(\sqrt{2/(n+1)}\) therefore gives an eigenvector with eigenvalue \(2\cos\theta\).
To check its norm, the identity \(\sin^2 t=(1-\cos(2t))/2\) gives \(\sum_{j=1}^n\sin^2(j\theta)=\frac n2-\frac12\sum_{j=1}^n\cos(2j\theta)\). Since \(e^{2i\theta}\ne1\) and \(e^{2i(n+1)\theta}=1\), the finite geometric sum \(1+e^{2i\theta}+\cdots+e^{2in\theta}\) is \(0\). Taking real parts shows \(\sum_{j=1}^n\cos(2j\theta)=-1\). Thus \(\sum_{j=1}^n\sin^2(j\theta)=(n+1)/2\), and the squared norm after scaling is \((2/(n+1))((n+1)/2)=1\).
Finally, the angles \(k\pi/(n+1)\) are distinct and lie strictly between \(0\) and \(\pi\), where cosine is strictly decreasing. Hence the eigenvalues are distinct. Eigenvectors of a real symmetric matrix belonging to distinct eigenvalues are linearly independent: if \(Au=\lambda u\), \(Av=\mu v\), and \(A=A^{\mathsf T}\), then \(\lambda u^{\mathsf T}v=(Au)^{\mathsf T}v=u^{\mathsf T}Av=\mu u^{\mathsf T}v\), so \(u^{\mathsf T}v=0\) when \(\lambda\ne\mu\). Consequently these \(n\) vectors are mutually orthogonal and linearly independent. There are \(n\) of them in \(\mathbb R^n\), so they form a basis and account for the full spectrum. This includes \(n=1\): the formula gives eigenvalue \(0\) and the unit vector \((1)\). \(\square\)
Direct Calculation for \(P_3\)
Worked Example: The Complete Spectrum of \(P_3\)
Let \(V(P_3)=\{1,2,3\}\) and \(E(P_3)=\{12,23\}\). In this vertex order, the adjacency matrix is $$ \mathbf A(P_3)= \begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix}. $$ Here \(n+1=4\), so the normalization factor for each sine vector is \(\sqrt{2/4}=1/\sqrt2\). The three angles \(k\pi/4\) give the eigenvalues \(\sqrt2,0,-\sqrt2\), respectively.
For \(k=1\), the sine entries are \((\sin(\pi/4),\sin(\pi/2),\sin(3\pi/4))=(\sqrt2/2,1,\sqrt2/2)\). Thus \(x^{(1)}=(1/2,1/\sqrt2,1/2)^{\mathsf T}\), and multiplication gives \(\mathbf A(P_3)x^{(1)}=(1/\sqrt2,1,1/\sqrt2)^{\mathsf T}=\sqrt2\,x^{(1)}\). For \(k=2\), the entries are \((1,0,-1)\) before scaling, so \(x^{(2)}=(1/\sqrt2,0,-1/\sqrt2)^{\mathsf T}\). Multiplication gives \((0,0,0)^{\mathsf T}=0x^{(2)}\). For \(k=3\), the entries before scaling are \((\sqrt2/2,-1,\sqrt2/2)\), so \(x^{(3)}=(1/2,-1/\sqrt2,1/2)^{\mathsf T}\). Multiplication gives \((-1/\sqrt2,1,-1/\sqrt2)^{\mathsf T}=-\sqrt2\,x^{(3)}\).
Check: Each vector has squared norm \(1/4+1/2+1/4=1\) for \(k=1,3\), and \(1/2+0+1/2=1\) for \(k=2\). The eigenvalues are distinct, and three eigenvectors in \(\mathbb R^3\) give the complete spectrum \(\{\sqrt2,0,-\sqrt2\}\).
Direct Calculation for \(P_4\)
Worked Example: The Complete Spectrum of \(P_4\)
Let \(V(P_4)=\{1,2,3,4\}\) and \(E(P_4)=\{12,23,34\}\). Its adjacency matrix is $$ \mathbf A(P_4)= \begin{pmatrix} 0&1&0&0\\ 1&0&1&0\\ 0&1&0&1\\ 0&0&1&0 \end{pmatrix}. $$ Write \(a=\sin(2\pi/5)\), \(b=\sin(\pi/5)\), \(\lambda_1=2\cos(\pi/5)\), and \(\lambda_2=2\cos(2\pi/5)\). The normalization factor is \(\sqrt{2/5}\). Evaluating the sine entries and multiplying by the matrix gives the following four unnormalized vectors and products:
For \(k=1\), \(y^{(1)}=(b,a,a,b)^{\mathsf T}\), and \(\mathbf A(P_4)y^{(1)}=(a,a+b,a+b,a)^{\mathsf T}=\lambda_1(b,a,a,b)^{\mathsf T}\). The equalities follow from \(2\cos(\pi/5)\sin(\pi/5)=\sin(2\pi/5)=a\) and \(2\cos(\pi/5)\sin(2\pi/5)=\sin(3\pi/5)+\sin(\pi/5)=a+b\).
For \(k=2\), \(y^{(2)}=(a,b,-b,-a)^{\mathsf T}\), and \(\mathbf A(P_4)y^{(2)}=(b,a-b,b-a,-b)^{\mathsf T}=\lambda_2(a,b,-b,-a)^{\mathsf T}\). Indeed, \(2\cos(2\pi/5)a=\sin(4\pi/5)=b\) and \(2\cos(2\pi/5)b=\sin(3\pi/5)-\sin(\pi/5)=a-b\).
For \(k=3\), \(y^{(3)}=(a,-b,-b,a)^{\mathsf T}\), and \(\mathbf A(P_4)y^{(3)}=(-b,a-b,a-b,-b)^{\mathsf T}=-\lambda_2y^{(3)}\). For \(k=4\), \(y^{(4)}=(b,-a,a,-b)^{\mathsf T}\), and \(\mathbf A(P_4)y^{(4)}=(-a,a+b,-a-b,a)^{\mathsf T}=-\lambda_1y^{(4)}\). These last two equations also follow entry by entry from the \(k=2\) and \(k=1\) identities, respectively, with the indicated sign changes. Multiplication by \(\sqrt{2/5}\) gives the four normalized eigenvectors without changing their eigenvalues.
Check: For example, the squared norm of \(x^{(3)}\) is \((2/5)(a^2+b^2+b^2+a^2)\). The sine-sum identity proved above gives \(a^2+b^2=5/4\), so this equals \((2/5)(2(5/4))=1\); the same identity proves unit norm for each of the other three vectors. The four eigenvalues are \(2\cos(k\pi/5)\) for \(k=1,2,3,4\), hence distinct, so the calculations verify the full spectrum.
Direct Calculation for \(P_5\)
Worked Example: The Complete Spectrum of \(P_5\)
Let \(V(P_5)=\{1,2,3,4,5\}\) and \(E(P_5)=\{12,23,34,45\}\). Its adjacency matrix has \(1\) immediately above and below the diagonal and \(0\) elsewhere. Here the normalization factor is \(\sqrt{2/6}=1/\sqrt3\), and the eigenvalues \(2\cos(k\pi/6)\), in order, are \(\sqrt3,1,0,-1,-\sqrt3\).
Use the following unnormalized sine vectors, obtained by evaluating \(\sin(jk\pi/6)\) for \(j=1,\dots,5\): \(y^{(1)}=(1/2,\sqrt3/2,1,\sqrt3/2,1/2)^{\mathsf T}\); \(y^{(2)}=(\sqrt3/2,\sqrt3/2,0,-\sqrt3/2,-\sqrt3/2)^{\mathsf T}\); \(y^{(3)}=(1,0,-1,0,1)^{\mathsf T}\); \(y^{(4)}=(\sqrt3/2,-\sqrt3/2,0,\sqrt3/2,-\sqrt3/2)^{\mathsf T}\); and \(y^{(5)}=(1/2,-\sqrt3/2,1,-\sqrt3/2,1/2)^{\mathsf T}\).
For any vector \(y=(y_1,\dots,y_5)^{\mathsf T}\), the matrix product is \((y_2,y_1+y_3,y_2+y_4,y_3+y_5,y_4)^{\mathsf T}\). Substituting \(y^{(1)}\) gives \((\sqrt3/2,3/2,\sqrt3,3/2,\sqrt3/2)^{\mathsf T}=\sqrt3\,y^{(1)}\). Substituting \(y^{(2)}\) gives \((\sqrt3/2,\sqrt3/2,0,-\sqrt3/2,-\sqrt3/2)^{\mathsf T}=y^{(2)}\). Substituting \(y^{(3)}\) gives \((0,0,0,0,0)^{\mathsf T}=0y^{(3)}\). Substituting \(y^{(4)}\) gives \((-\sqrt3/2,\sqrt3/2,0,-\sqrt3/2,\sqrt3/2)^{\mathsf T}=-y^{(4)}\). Finally, substituting \(y^{(5)}\) gives \((-\sqrt3/2,3/2,-\sqrt3,3/2,-\sqrt3/2)^{\mathsf T}=-\sqrt3\,y^{(5)}\). Scaling each vector by \(1/\sqrt3\) gives the normalized eigenvectors.
Check: The squared norm of \(x^{(1)}\) is \((1/3)(1/4+3/4+1+3/4+1/4)=(1/3)3=1\). For \(x^{(2)}\), it is \((1/3)(3/4+3/4+0+3/4+3/4)=(1/3)3=1\). For \(x^{(3)}\), it is \((1/3)(1+0+1+0+1)=1\). The vectors for \(k=4,5\) have the same squared entries as those for \(k=2,1\), respectively, so their norms are also \(1\). The five listed eigenvalues are distinct; therefore these five directly checked eigenpairs give the complete spectrum.
A Sign Error to Watch For
The sine formula specifies every coordinate, including its sign. In \(P_4\), the \(k=3\) vector is proportional to \((\sin(3\pi/5),\sin(6\pi/5),\sin(9\pi/5),\sin(12\pi/5))=(a,-b,-b,a)\). It is not proportional to \((a,-b,b,-a)\). For the incorrect vector \(z=(a,-b,b,-a)^{\mathsf T}\), direct multiplication gives \(\mathbf A(P_4)z=(-b,a+b,-a-b,b)^{\mathsf T}\), which is not a scalar multiple of \(z\). For instance, the first coordinate would require the scalar to be \(-b/a\), while the second would then require \(a+b=(-b/a)(-b)=b^2/a\), an equality that does not hold. The corrected vector \((a,-b,-b,a)^{\mathsf T}\) gives \((-b,a-b,a-b,-b)^{\mathsf T}=-\lambda_2(a,-b,-b,a)^{\mathsf T}\), as required.
These calculations illustrate why both parts of an eigenpair need checking: a plausible pattern of alternating signs is not enough. The endpoint entries must satisfy the path boundary conditions, the interior entries must obey the two-neighbor recurrence, and the vector must be normalized if a unit eigenvector is required. The general sine formula handles all three tasks, while direct multiplication on small paths is an effective way to catch indexing and sign errors.
Check Your Understanding
Use the sine formula and, where requested, verify the matrix product directly.
- Write the normalized eigenvector for \(k=2\) on \(P_3\), and check its eigenvalue by multiplication.
- For \(P_4\), explain why the third and fourth coordinates of the \(k=3\) sine vector are both negative and positive, respectively.
- Use the theorem to give the five eigenvalues of \(P_5\) in increasing order.
- Verify directly that \((1,0,-1,0,1)^{\mathsf T}/\sqrt3\) is a unit eigenvector of \(P_5\), and state its eigenvalue.
- Why does having \(n\) distinct eigenvalues in these calculations show that the listed eigenpairs give the full spectrum of \(P_n\)?