9.2 Diagonal Forms
Theorem 9.2.1. Any real quadratic form may be reduced to a diagonal form by an orthogonal substitution. That is, if \(A\) is the matrix of \(Q\) then there exists an orthogonal matrix \(P\) such that \(P^tAP=\text {diag}(\lambda _1,\lambda _2,\dots ,\lambda _n)\), where the \(\lambda _i\) are the eigenvalues of \(A\).
Proof. A quadratic form is written \(Q(x)=x^tAx\) where \(A\) may be taken symmetric: any matrix \(B\) with \(Q(x)=x^tBx\) can be replaced by \(\frac {1}{2}(B+B^t)\) without changing the form, since \(x^tBx\) is a scalar and therefore equals its own transpose \(x^tB^tx\).
Being real symmetric, \(A\) is orthogonally diagonalisable by the spectral theorem: there is an orthogonal \(P\) with \(P^tAP=D=\text {diag}(\lambda _1,\dots ,\lambda _n)\).
Now make the substitution \(x=Py\), legitimate because \(P\) is invertible, and called orthogonal because \(P\) is. Then \[Q(x)=x^tAx=(Py)^tA(Py)=y^t(P^tAP)y=y^tDy =\lambda _1y_1^2+\lambda _2y_2^2+\cdots +\lambda _ny_n^2,\] which is diagonal: it contains no cross terms \(y_iy_j\) with \(i\neq j\). □
Remark. Because \(P\) is orthogonal the substitution is a rotation, or a rotation composed with a reflection, so it preserves lengths and angles. That is why this is the right tool for identifying conic sections and quadric surfaces in the next subsection: the new axes are genuinely the old ones turned, and the shape of the curve is not distorted by the change of variable.
Example 9.2.2. Consider \(Q(x_1,x_2)=2x^2_1-4x_1x_2+5x_2^2\)
- i.
- Reduce \(Q\) to diagonal form.
- ii.
- What is the diagonal substitution.
Solution.
- i.
- \(Q(X)=(x_1,x_2) \begin {pmatrix} 2&-2\\-2&5\\ \end {pmatrix} \begin {pmatrix} x_1\\x_2 \end {pmatrix} =X^tAX\) \(\implies A= \begin {pmatrix} 2&-2\\-2&5\\ \end {pmatrix} \implies \lambda =1,6\) and \( V= \begin {pmatrix} 1\\-2\\ \end {pmatrix}, \begin {pmatrix} 2\\1\\ \end {pmatrix} \). So the diagonal for is \[D(X)=a^2+6b^2\]
- ii.
- \(P= \begin {pmatrix} \frac {1}{\sqrt {5}}&\frac {2}{\sqrt {5}}\\\\ \frac {-2}{\sqrt {5}}&\frac {1}{\sqrt {5}}\\ \end {pmatrix} \) So that the orthogonal substitution is
\[x_1=\frac {1}{\sqrt {5}}a-2/\sqrt {5}b\qquad \text {and}\qquad x_2=\frac {2}{\sqrt {5}}a+\frac {1}{2}b\]
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