Reducing the quadratic form to canonical form.

Literature: Collection of problems in mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.

Method of eigenvectors:

Consider the quadratic form $A(x,x) =\sum\limits_{i,j=1}^na_{i,j}x_ix_j$ in the Euclidean space $R^n$. Since its matrix $A=(a_ij)$ is symmetric, it can be represented as $A=UDU^{T}$, where $D$ is a diagonal matrix whose diagonal elements are the eigenvalues of the matrix, and $U$ is an orthogonal matrix. The columns of matrix $U$ represent the coordinates of some orthonormal basis $B'=(e_1, ..., e_n)$, in which the matrix $A$ has a diagonal form $D$, and therefore, the quadratic form has the desired canonical form. The corresponding coordinate transformations are determined by the equation$$\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}=U\begin{pmatrix}x_1'\\\vdots\\x_n'\end{pmatrix}.$$

Example.

Find the orthogonal transformation that converts the following forms to canonical form, and write down this canonical form:

4.213. $11x_1^2+5x_2^2+2x_3^2+16x_1x_2+4x_1x_3-20x_2x_3.$

Solution.

The matrix of the quadratic form is given by:$$\begin{pmatrix}11&8&2\\8&5&-10\\2&-10&2\end{pmatrix}.$$

Let's find the eigenvalues of this matrix. To do this, we'll write down the characteristic equation:

$$A-\lambda E=\begin{pmatrix}11&8&2\\8&5&-10\\2&-10&2\end{pmatrix}-\lambda\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=$$ $$=\begin{pmatrix}11-\lambda&8&2\\8&5-\lambda&-10\\2&-10&2-\lambda\end{pmatrix}.$$

$$det(A-\lambda E)=\begin{vmatrix}11-\lambda&8&2\\8&5-\lambda&-10\\2&-10&2-\lambda\end{vmatrix}=$$ $$=(11-\lambda)(5-\lambda)(2-\lambda)+2\cdot 8\cdot (-10)+2\cdot 8\cdot (-10)-$$ $$-2\cdot(5-\lambda)\cdot 2-(11-\lambda)\cdot(-10)\cdot(-10)-8\cdot 8\cdot(2-\lambda)=$$ $$=-\lambda^3+\lambda^2(2+5+11)-\lambda(10+22+55)+110-160-160-20+$$ $$+4\lambda-1100+100\lambda-128+64\lambda=$$ $$=-\lambda^3+18\lambda^2+81\lambda-1458=-\lambda(\lambda^2-81)+18(\lambda^2-81)=$$ $$=(\lambda-9)(\lambda+9)(-\lambda+18)=0.$$

From here, we find the eigenvalues:

$$\lambda_1=9,\quad \lambda_2=-9, \quad\lambda_3=18.$$

Next, we find the eigenvectors:

The eigenvector for eigenvalue $\lambda_1=9$ is found from the system:$$(A-\lambda E)X=0, X\neq 0, \Rightarrow (A-9E)X=0, X\neq 0$$

$$(A-9E)X=\begin{pmatrix}11-9&8&2\\8&5-9&-10\\2&-10&2-9\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=$$ $$=\begin{pmatrix}2x_1+8x_2+2x_3\\8x_1-4x_2-10x_3\\2x_1-10x_2-7x_3\end{pmatrix}=0.$$

Let's solve the homogeneous system of equations:

$$\left\{\begin{array}{lcl}2x_1+8x_2+2x_3=0\\ 8x_1-4x_2-10x_3=0\\2x_1-10x_2-7x_3=0\end{array}\right.$$

Let's calculate the rank of the coefficient matrix $A=\begin{pmatrix}2&8&2\\8&-4&-10\\2&-10&-7\end{pmatrix}$ using the method of leading minors:

We fix a non-zero minor of the second order $M_2=\begin{vmatrix}2&8\\8&-4\end{vmatrix}=-8-64=-72\neq 0.$.

Now, let's consider the leading minor of the third order: $\begin{vmatrix}2&8&2\\8&-4&-10\\2&-10&-7\end{vmatrix}=56-160-160+16-200+448=0$;

Thus, the rank of matrix $A$ is two.

Let's choose the minor $M=\begin{vmatrix}2&8\\8&-4\end{vmatrix}=-72\neq 0$ as the basis minor. Then, assuming $x_3=c$, we obtain:$$\left\{\begin{array}{lcl}2x_1+8x_2+2c=0\\ 8x_1-4x_2-10c=0\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}2x_1+8x_2=-2c\\8x_1-4x_2=10c\end{array}\right.$$

By Cramer's Rule, we find $x_1$ and $x_2$:

$\Delta=\begin{vmatrix}2&8\\8&-4\end{vmatrix}=-8-64=-72;$

$\Delta_1=\begin{vmatrix}-2c&8\\10c&-4\end{vmatrix}=8c-80c=-72c;$

$\Delta_2=\begin{vmatrix}2&-2c\\8&10c\end{vmatrix}=20c+16c=36c;$

$x_1=\frac{\Delta_1}{\Delta}=\frac{-72c}{-72}=c;$ $x_2=\frac{\Delta_2}{\Delta}=\frac{36c}{-72}=-c/2.$

Thus, the general solution of the system is $X(c)=\begin{pmatrix}c\\-c/2\\c\end{pmatrix}$.

From the general solution, we find the fundamental solution set: $E=X(1)=\begin{pmatrix}1\\-1/2\\1\end{pmatrix}$.

The corresponding orthonormalized eigenvector is:$$e_1'=\left(\frac{2}{\sqrt{4+1+4}},\frac{-1}{\sqrt{4+1+4}},\frac{2}{\sqrt{4+1+4}}\right)=\left(\frac{2}{3},\frac{-1}{3},\frac{2}{3}\right).$$

To find the eigenvector for the eigenvalue $\lambda_2=-9$, we solve the system:$$(A-\lambda E)X=0, X\neq 0, \Rightarrow (A+9E)X=0, X\neq 0$$

$$(A+9E)X=\begin{pmatrix}11+9&8&2\\8&5+9&-10\\2&-10&2+9\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=$$ $$=\begin{pmatrix}20x_1+8x_2+2x_3\\8x_1+14x_2-10x_3\\2x_1-10x_2+11x_3\end{pmatrix}=0.$$

Let's solve the homogeneous system of equations:

$$\left\{\begin{array}{lcl}20x_1+8x_2+2x_3=0\\ 8x_1+14x_2-10x_3=0\\2x_1-10x_2+11x_3=0\end{array}\right.$$

Let's compute the rank of the coefficient matrix $A=\begin{pmatrix}20&8&2\\8&14&-10\\2&-10&11\end{pmatrix}$ using the method of leading minors:

We fix a non-zero minor of the second order $M_2=\begin{vmatrix}20&8\\8&14\end{vmatrix}=280-64=216\neq 0$.

Now, let's consider the leading minor of the third order: $\begin{vmatrix}20&8&2\\8&14&-10\\2&-10&11\end{vmatrix}=3080-160-160-56-2000-704=0$;

Thus, the rank of matrix $A$ is two.

Let's choose the minor $M=\begin{vmatrix}20&8\\8&14\end{vmatrix}=216\neq 0$ as the basis minor. Then, assuming $x_3=c$, we obtain: $$\left\{\begin{array}{lcl}20x_1+8x_2+2c=0\\ 8x_1+14x_2-10c=0\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}20x_1+8x_2=-2c\\8x_1+14x_2=10c\end{array}\right.$$

By Cramer's Rule, we find $x_1$ and $x_2$:

$\Delta=\begin{vmatrix}20&8\\8&14\end{vmatrix}=280-64=216;$

$\Delta_1=\begin{vmatrix}-2c&8\\10c&14\end{vmatrix}=-28c-80c=-108c;$

$\Delta_2=\begin{vmatrix}20&-2c\\8&10c\end{vmatrix}=200c+16c=216c;$

$x_1=\frac{\Delta_1}{\Delta}=\frac{-108c}{216}=-c/2;$ $x_2=\frac{\Delta_2}{\Delta}=\frac{216c}{216}=c.$

Thus, the general solution of the system is $X(c)=\begin{pmatrix}-c/2\\c\\c\end{pmatrix}$.

From the general solution, we find the fundamental solution set: $E=X(1)=\begin{pmatrix}-1/2\\1\\1\end{pmatrix}$.

The corresponding orthonormalized eigenvector is:$$e'_2=\left(\frac{-1}{\sqrt{4+1+4}},\frac{2}{\sqrt{4+1+4}},\frac{2}{\sqrt{4+1+4}}\right)=\left(\frac{-1}{3},\frac{2}{3},\frac{2}{3}\right).$$

To find the eigenvector for the eigenvalue $\lambda=18$, we solve the system: $$(A-\lambda E)X=0, X\neq 0, \Rightarrow (A-18E)X=0, X\neq 0$$

$$(A-18E)X=\begin{pmatrix}11-18&8&2\\8&5-18&-10\\2&-10&2-18\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=$$ $$=\begin{pmatrix}-7x_1+8x_2+2x_3\\8x_1-13x_2-10x_3\\2x_1-10x_2-16x_3\end{pmatrix}=0.$$

Let's solve the homogeneous system of equations:

$$\left\{\begin{array}{lcl}-7x_1+8x_2+2x_3=0\\ 8x_1-13x_2-10x_3=0\\2x_1-10x_2-16x_3=0\end{array}\right.$$

Let's compute the rank of the coefficient matrix $A=\begin{pmatrix}-7&8&2\\8&-13&-10\\2&-10&-16\end{pmatrix}$ using the method of leading minors:

We fix a non-zero minor of the second order $M_2=\begin{vmatrix}-7&8\\8&-13\end{vmatrix}=91-64=27\neq 0$.

Now, let's consider the leading minor of the third order: $\begin{vmatrix}-7&8&2\\8&-13&-10\\2&-10&-16\end{vmatrix}=-1456-160-160+52+700+1024=0$;

Thus, the rank of matrix $A$ is two.

Let's choose the minor $M=\begin{vmatrix}-7&8\\8&-13\end{vmatrix}=27\neq 0$ as the basis minor. Then, assuming $x_3=c$, we obtain:

$$\left\{\begin{array}{lcl}-7x_1+8x_2+2c=0\\ 8x_1-13x_2-10c=0\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}-7x_1+8x_2=-2c\\8x_1-13x_2=10c\end{array}\right.$$

By Cramer's Rule, we find $x_1$ and $x_2$:

$\Delta=\begin{vmatrix}-7&8\\8&-13\end{vmatrix}=91-64=27;$

$\Delta_1=\begin{vmatrix}-2c&8\\10c&-13\end{vmatrix}=26c-80c=-54c;$

$\Delta_2=\begin{vmatrix}-7&-2c\\8&10c\end{vmatrix}=-70c+16c=-54c;$

$x_1=\frac{\Delta_1}{\Delta}=\frac{-54c}{27}=-2c;$ $x_2=\frac{\Delta_2}{\Delta}=\frac{-54c}{27}=-2c.$

Thus, the general solution of the system is $X(c)=\begin{pmatrix}-2c\\-2c\\c\end{pmatrix}$.

From the general solution, we find the fundamental solution set: $E=X(1)=\begin{pmatrix}-2\\-2\\1\end{pmatrix}$.

The corresponding orthonormalized eigenvector is: $$e'_3=\left(\frac{-2}{\sqrt{4+4+1}},\frac{-2}{\sqrt{4+4+1}},\frac{1}{\sqrt{4+4+1}}\right)=\left(\frac{-2}{3},\frac{-2}{3},\frac{1}{3}\right).$$

Thus, we have found the vectors

$$e_1'=\left(\frac{2}{3},\frac{-1}{3},\frac{2}{3}\right);$$

$$e_2'=\left(\frac{-1}{3},\frac{2}{3},\frac{2}{3}\right);$$

$$e_3'=\left(\frac{-2}{3},\frac{-2}{3},\frac{1}{3}\right).$$ И следовательно,

$$U=\frac{1}{3}\begin{pmatrix}2&-1&-2\\-1&2&-2\\2&2&1\end{pmatrix}, \quad U^T=\frac{1}{3}\begin{pmatrix}2&-1&2\\-1&2&2\\-2&-2&1\end{pmatrix}.$$

In the basis $B'=(e_1', e_2', e_3')$, the given quadratic form takes the form $$A(x, x)=9x_1^2-9x_2^2+18x_3^2,$$ and the corresponding change of coordinates is:

$$x_1=2x_1 '-x_2'-2x_3'$$

$$x_2=-x_1 '+2x_2'-2x_3'$$

$$x_3=2x_1 '+2x_2'+x_3'$$

Ответ: $A(x, x)=9x_1^2-9x_2^2+18x_3^2;$

$$x_1=2x_1 '-x_2'-2x_3';$$

$$x_2=-x_1 '+2x_2'-2x_3';$$

$$x_3=2x_1 '+2x_2'+x_3'.$$

Tags: