Review : Eigenvalues & Eigenvectors — Question 10

Question 10 – Solution

Strategy. Separate the constant direction, which the matrix cannot detect, from the two directions of variation.

Step 1: Verify a complete orthogonal basis. Direct multiplication gives $$L\mathbf 1=0,\qquad Lv=v,\quad v=(1,0,-1)^T, \qquad Lw=3w,\quad w=(1,-2,1)^T.$$ The three vectors are nonzero and mutually perpendicular, so they form a basis of $\mathbb R^3$. Thus the complete eigenvalue list is $\boxed{0,1,3}$, with the corresponding one-dimensional eigenspaces.

Step 2: Interpret the nonnegative quadratic form. Expanding gives $$\boxed{x^TLx=(x_1-x_2)^2+(x_2-x_3)^2\ge 0.}$$ It vanishes exactly when $x_1=x_2=x_3$, the constant direction. The zero eigenvalue reflects the fact that adding the same constant to all three entries changes no adjacent difference.

Step 3: Solve subject to the compatibility condition. Since $\mathbf 1^TL=0$, solvability requires $\mathbf 1^Tb=0$. Conversely, every such $b$ can be written $b=\alpha v+\beta w$, and $x=\alpha v+(\beta/3)w+c\mathbf 1$ solves the system for every real $c$. For the stated $b$, dot products give $\alpha=3/2$ and $\beta=1/2$. Therefore all solutions are $$\boxed{x=(5/3,-1/3,-4/3)^T+c\mathbf 1,\qquad c\in\mathbb R.}$$ The displayed particular vector has zero mean; the total sum is $3c$, so zero mean uniquely selects $c=0$.

Step 4: Prove the sharp restricted bounds. A zero-mean vector has form $x=a v+b w$. Orthogonality and $\|v\|^2=2$, $\|w\|^2=6$ give $$\|x\|^2=2a^2+6b^2,\qquad x^TLx=2a^2+18b^2.$$ The lower equality holds exactly on the line through $v$; the upper exactly on the line through $w$, with zero included in both. This proves both sharp bounds. A nonzero constant vector has positive length but zero quadratic form, so the lower bound cannot hold on all of $\mathbb R^3$.