Solution.

The matrix

The coefficient of \(x^2\) goes in position \((1,1)\), that of \(y^2\) in \((2,2)\), and the cross coefficient is split equally between the two off-diagonal positions so that the matrix is symmetric: \[A=\begin {pmatrix}5&2\\2&2\end {pmatrix},\qquad Q(x,y)=\begin {pmatrix}x&y\end {pmatrix} \begin {pmatrix}5&2\\2&2\end {pmatrix} \begin {pmatrix}x\\y\end {pmatrix}.\]

Diagonalising

\[X_A(\lambda )=\begin {vmatrix}5-\lambda& 2\\2&2-\lambda \end {vmatrix} =(5-\lambda )(2-\lambda )-4=\lambda ^2-7\lambda +6=(\lambda -6)(\lambda -1),\] so the eigenvalues are \(6\) and \(1\). Solving \((A-6I)v=0\) gives the eigenvector \((2,1)\); solving \((A-I)v=0\) gives \((-1,2)\). As the theory predicts for a symmetric matrix, these are orthogonal: \((2)(-1)+(1)(2)=0\).

Normalising each to unit length and using them as columns, \[P=\frac {1}{\sqrt 5}\begin {pmatrix}2&-1\\1&2\end {pmatrix},\] which is orthogonal, since its columns are orthonormal. Then \[P^tAP=\begin {pmatrix}6&0\\0&1\end {pmatrix},\] and the substitution \(\begin {pmatrix}x\\y\end {pmatrix}=P\begin {pmatrix}u\\v\end {pmatrix}\) turns the form into \[Q=6u^2+v^2,\] with no cross term. Since both coefficients are positive, \(Q\) is positive definite and the curve \(Q=1\) is an ellipse.

Solution. The matrix of the form is \(A=\begin {pmatrix}3&1\\1&3\end {pmatrix}\), with \[X_A(\lambda )=(3-\lambda )^2-1=\lambda ^2-6\lambda +8=(\lambda -4)(\lambda -2),\] so the eigenvalues are \(4\) and \(2\), with orthogonal eigenvectors \((1,1)\) and \((-1,1)\) respectively. Normalising, \[P=\frac {1}{\sqrt 2}\begin {pmatrix}1&-1\\1&1\end {pmatrix},\] which is a rotation through \(45^{\circ }\). Under \(x=Pu\) the equation becomes \[4u^2+2v^2=8,\qquad \text {that is}\qquad \frac {u^2}{2}+\frac {v^2}{4}=1.\] This is an ellipse, with semi-axes \(\sqrt 2\) along the \(u\)-axis and \(2\) along the \(v\)-axis — the original axes rotated by \(45^{\circ }\).

Because \(P\) is orthogonal the substitution is a genuine rotation, so these lengths are the true semi-axes of the original curve and not an artefact of the change of variable.

Solution. By the spectral theorem there is an orthogonal \(P\) with \(P^tAP=D=\text {diag}(\lambda _1,\dots ,\lambda _n)\). Substituting \(x=Py\), \[Q(x)=y^tDy=\lambda _1y_1^2+\cdots +\lambda _ny_n^2,\] and because \(P\) is invertible, \(x\) ranges over all non-zero vectors exactly as \(y\) does.

If every \(\lambda _i>0\)

Then for \(y\neq 0\) at least one \(y_i\neq 0\), so the sum \(\sum _i\lambda _iy_i^2\) has a strictly positive term and no negative ones, hence \(Q>0\).

If some \(\lambda _k\leq 0\)

Take \(y\) to be the \(k\)-th standard basis vector, so \(y\neq 0\) and \(Q=\lambda _k\leq 0\). Then \(x=Py\neq 0\) gives \(Q(x)\leq 0\), so \(Q\) is not positive definite.

This is what was being checked by hand in the Section 7 problem on the value of \(c\): completing the square there was a two-variable version of the same diagonalisation.

Solution. The symmetric matrix of the form, splitting the \(-2xy\) term equally, is \[A=\begin {pmatrix}2&-1&0\\-1&2&0\\0&0&3\end {pmatrix}.\] Because the matrix is block diagonal, the characteristic polynomial factors: \[X_A(\lambda )=(3-\lambda )\left [(2-\lambda )^2-1\right ] =(3-\lambda )(\lambda -1)(\lambda -3),\] so the eigenvalues are \(3\), \(3\) and \(1\) — all positive. By the previous problem \(Q\) is positive definite.

The conclusion can be confirmed without eigenvalues by completing the square: \[2x^2-2xy+2y^2+3z^2=2\left (x-\tfrac {y}{2}\right )^2+\tfrac {3}{2}y^2+3z^2,\] a sum of terms with positive coefficients which vanish together only when \(z=0\), \(y=0\) and then \(x=0\).