Reducing the quadratic form to canonical form.

Method of eigenvectors:

Consider the quadratic form $A(x,x)=\sum _{i,j=1}^n{a}_{i,j}{x}_i{x}_j$ in the Euclidean space $R^n$. Since its matrix $A=(a_{i}j)$ is symmetric, it can be represented as $A=UD{U}^T$, 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 $B^{\prime }=(e_1,...,e_n)$, 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$$\left(\begin{array}{c}{x}_1\\ ⋮\\ x_n\end{array}\right)=U\left(\begin{array}{c}{x}_1^{\prime }\\ ⋮\\ x_n^{\prime }\end{array}\right).$$

Example.

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

1. $11x_1^2+5x_2^2+2x_3^2+16x_1{x}_2+4x_1{x}_3-20x_2{x}_3.$

Solution.

The matrix of the quadratic form is given by:$$\left(\begin{array}{ccc}11& 8& 2\\ 8& 5& -10\\ 2& -10& 2\end{array}\right).$$

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

$$A-\lambda E=\left(\begin{array}{ccc}11& 8& 2\\ 8& 5& -10\\ 2& -10& 2\end{array}\right)-\lambda \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=$$ $$=\left(\begin{array}{ccc}11-\lambda & 8& 2\\ 8& 5-\lambda & -10\\ 2& -10& 2-\lambda \end{array}\right).$$

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

From here, we find the eigenvalues:

$${\lambda }_1=9,{\lambda }_2=-9,{\lambda }_3=18.$$

Next, we find the eigenvectors:

The eigenvector for eigenvalue ${\lambda }_1=9$ is found from the system:$$(A-\lambda E)X=0,X\ne 0,\Rightarrow (A-9E)X=0,X\ne 0$$

$$(A-9E)X=\left(\begin{array}{ccc}11-9& 8& 2\\ 8& 5-9& -10\\ 2& -10& 2-9\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=$$ $$=\left(\begin{array}{c}2x_1+8x_2+2x_3\\ 8x_1-4x_2-10x_3\\ 2x_1-10x_2-7x_3\end{array}\right)=0.$$

Let's solve the homogeneous system of equations:

$$\{\begin{array}{l}2x_1+8x_2+2x_3=0\\ 8x_1-4x_2-10x_3=0\\ 2x_1-10x_2-7x_3=0\end{array}$$

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

We fix a non-zero minor of the second order $M_2=\left|\begin{array}{cc}2& 8\\ 8& -4\end{array}\right|=-8-64=-72\ne 0.$.

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

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

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

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

$\mathrm{\Delta }=\left|\begin{array}{cc}2& 8\\ 8& -4\end{array}\right|=-8-64=-72;$

${\mathrm{\Delta }}_1=\left|\begin{array}{cc}-2c& 8\\ 10c& -4\end{array}\right|=8c-80c=-72c;$

${\mathrm{\Delta }}_2=\left|\begin{array}{cc}2& -2c\\ 8& 10c\end{array}\right|=20c+16c=36c;$

$x_1=\frac{{\mathrm{\Delta }}_1}{\mathrm{\Delta }}=\frac{-72c}{-72}=c;$ $x_2=\frac{{\mathrm{\Delta }}_2}{\mathrm{\Delta }}=\frac{36c}{-72}=-c/2.$

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

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

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

To find the eigenvector for the eigenvalue ${\lambda }_2=-9$, we solve the system:$$(A-\lambda E)X=0,X\ne 0,\Rightarrow (A+9E)X=0,X\ne 0$$

$$(A+9E)X=\left(\begin{array}{ccc}11+9& 8& 2\\ 8& 5+9& -10\\ 2& -10& 2+9\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=$$ $$=\left(\begin{array}{c}20x_1+8x_2+2x_3\\ 8x_1+14x_2-10x_3\\ 2x_1-10x_2+11x_3\end{array}\right)=0.$$

Let's solve the homogeneous system of equations:

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

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

We fix a non-zero minor of the second order $M_2=\left|\begin{array}{cc}20& 8\\ 8& 14\end{array}\right|=280-64=216\ne 0$.

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

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

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

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

$\mathrm{\Delta }=\left|\begin{array}{cc}20& 8\\ 8& 14\end{array}\right|=280-64=216;$

${\mathrm{\Delta }}_1=\left|\begin{array}{cc}-2c& 8\\ 10c& 14\end{array}\right|=-28c-80c=-108c;$

${\mathrm{\Delta }}_2=\left|\begin{array}{cc}20& -2c\\ 8& 10c\end{array}\right|=200c+16c=216c;$

$x_1=\frac{{\mathrm{\Delta }}_1}{\mathrm{\Delta }}=\frac{-108c}{216}=-c/2;$ $x_2=\frac{{\mathrm{\Delta }}_2}{\mathrm{\Delta }}=\frac{216c}{216}=c.$

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

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

The corresponding orthonormalized eigenvector is:$$e_2^{\prime }=(\frac{-1}{\sqrt{4+1+4}},\frac{2}{\sqrt{4+1+4}},\frac{2}{\sqrt{4+1+4}})=(\frac{-1}{3},\frac{2}{3},\frac{2}{3}).$$

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

$$(A-18E)X=\left(\begin{array}{ccc}11-18& 8& 2\\ 8& 5-18& -10\\ 2& -10& 2-18\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=$$ $$=\left(\begin{array}{c}-7x_1+8x_2+2x_3\\ 8x_1-13x_2-10x_3\\ 2x_1-10x_2-16x_3\end{array}\right)=0.$$

Let's solve the homogeneous system of equations:

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

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

We fix a non-zero minor of the second order $M_2=\left|\begin{array}{cc}-7& 8\\ 8& -13\end{array}\right|=91-64=27\ne 0$.

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

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

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

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

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

$\mathrm{\Delta }=\left|\begin{array}{cc}-7& 8\\ 8& -13\end{array}\right|=91-64=27;$

${\mathrm{\Delta }}_1=\left|\begin{array}{cc}-2c& 8\\ 10c& -13\end{array}\right|=26c-80c=-54c;$

${\mathrm{\Delta }}_2=\left|\begin{array}{cc}-7& -2c\\ 8& 10c\end{array}\right|=-70c+16c=-54c;$

$x_1=\frac{{\mathrm{\Delta }}_1}{\mathrm{\Delta }}=\frac{-54c}{27}=-2c;$ $x_2=\frac{{\mathrm{\Delta }}_2}{\mathrm{\Delta }}=\frac{-54c}{27}=-2c.$

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

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

The corresponding orthonormalized eigenvector is: $$e_3^{\prime }=(\frac{-2}{\sqrt{4+4+1}},\frac{-2}{\sqrt{4+4+1}},\frac{1}{\sqrt{4+4+1}})=(\frac{-2}{3},\frac{-2}{3},\frac{1}{3}).$$

Thus, we have found the vectors

$$e_1^{\prime }=(\frac{2}{3},\frac{-1}{3},\frac{2}{3});$$

$$e_2^{\prime }=(\frac{-1}{3},\frac{2}{3},\frac{2}{3});$$

$$e_3^{\prime }=(\frac{-2}{3},\frac{-2}{3},\frac{1}{3}).$$ И следовательно,

$$U=\frac{1}{3}\left(\begin{array}{ccc}2& -1& -2\\ -1& 2& -2\\ 2& 2& 1\end{array}\right),U^T=\frac{1}{3}\left(\begin{array}{ccc}2& -1& 2\\ -1& 2& 2\\ -2& -2& 1\end{array}\right).$$

In the basis $B^{\prime }=(e_1^{\prime },e_2^{\prime },e_3^{\prime })$, 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:

$$x_1=2x_1^{\prime }-x_2^{\prime }-2x_3^{\prime }$$

$$x_2=-x_1^{\prime }+2x_2^{\prime }-2x_3^{\prime }$$

$$x_3=2x_1^{\prime }+2x_2^{\prime }+x_3^{\prime }$$

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

$$x_1=2x_1^{\prime }-x_2^{\prime }-2x_3^{\prime };$$

$$x_2=-x_1^{\prime }+2x_2^{\prime }-2x_3^{\prime };$$

$$x_3=2x_1^{\prime }+2x_2^{\prime }+x_3^{\prime }.$$

Tags: Method of eigenvectors, canonical form, geometry, quadratic form