Review : Eigenvalues & Eigenvectors — Question 4

Question 4 – Solution

Strategy. An orthonormal eigenvector basis turns both a quadratic form and an inverse into scalar calculations.

Step 1: Construct and check the basis. Take $$q_1=\frac 1{\sqrt 2}(1,1,0)^T,\quad q_2=\frac 1{\sqrt 2}(1,-1,0)^T,\quad q_3=(0,0,1)^T.$$ Direct multiplication gives eigenvalues $3,1,-1$, respectively. The columns are orthonormal, so $Q=(q_1\ q_2\ q_3)$ has $Q^TQ=I$. Thus $SQ=QD$ with $D=\operatorname{diag}(3,1,-1)$, and $\boxed{S=QDQ^T}$.

Step 2: Prove perpendicularity. For a real symmetric matrix, if $Sa=\lambda a$ and $Sb=\mu b$, then $$\lambda a^Tb=(Sa)^Tb=a^TS^Tb=a^TSb=\mu a^Tb.$$ When $\lambda\ne\mu$, this forces $a^Tb=0$. Normalizing the three perpendicular directions gives exactly the orthonormal basis above.

Step 3: Derive the sharp quadratic-form bounds. Write $x=c_1q_1+c_2q_2+c_3q_3$. For a unit vector, $c_1^2+c_2^2+c_3^2=1$, and $$x^TSx=3c_1^2+c_2^2-c_3^2.$$ This is a weighted average of $3,1,-1$, so $\boxed{-1\le x^TSx\le 3}$. The minimum occurs exactly at $x=\pm q_3$; the maximum exactly at $x=\pm q_1$. The coefficients of all other directions must vanish at an endpoint.

Step 4: Invert the nonzero eigenvalues. Since no eigenvalue is zero, $$\boxed{S^{-1}=Q\operatorname{diag}(1/3,1,-1)Q^T =\begin{pmatrix}2/3&-1/3&0\\-1/3&2/3&0\\0&0&-1\end{pmatrix}.}$$ Multiplying by $S$ gives $I_3$. Invertibility excludes zero eigenvalues, not negative ones. But $q_3^TSq_3=-1<0$, so $S$ is not positive definite; in fact its quadratic form takes both positive and negative values.