Recall the concept of diagonalization of a square matrix. We have seen that an \(n\times n\) matrix \(A\) is diagonalizable if there is an eigenbasis of \(\R^n\text{.}\) In this section, we shall explore if we can find an eigenbasis which is also an orthonormal. First of all we shall define what is meaning of an orthogonal matrix.
Assume that \(P^{-1}=P^T\text{.}\) This implies \(PP^T=I\text{.}\) Let the columns of \(P\) are \(p_1, p_2,\ldots, p_n\text{.}\) Then \(\{p_1,\ldots, p_n\}\) is linearly independent. It is easy to see that the \(ij\)-th entry of \(PP^T\) is \(p_i\cdot p_j\text{.}\) Hence we have \(p_i\cdot p_j=\delta_{ij}=1\) if \(i=j\) and 0 otherwise. This proves rows of \(P\) and orthogonal and hence columns of \(P\) are orthogonal. The converse is easy.
An \(n\times n\) matrix is called orthogonally diagonalizable if there exists an orthogonal matrix \(P\) such that \(P^{-1}AP=P^TAP\) is a diagonal matrix.
Two \(n\times n\) matrices \(A\) and \(B\) are called orthogonally similar if there exists an orthogonal matrix \(P\) such that \(B =P^{-1}AP=P^TAP\) is a diagonal matrix.
Let \(A\) be a symmetric matrix and \(\lambda_1\) and \(\lambda_2\) are distinct eigenvalues of \(A\text{.}\) If \(v_1\) and \(v_2\) are eigenvectors corresponding to \(\lambda_1\) and \(\lambda_2\) respectively. Then \(v_1\) and \(v_2\) are orthogonal.
Let \(v_1,\ldots, v_n\) be orthogonormal eigenvectors of \(A\) such that \(Av_i=\lambda_i v_i\text{.}\) Then \(P=\begin{bmatrix} v_1\amp v_2\cdots v_n\end{bmatrix}\) is orthogonal. Hence
Suppose there exists an orthogonal matrix \(P\) susch that \(P^{-1}AP=D\text{.}\) Then \(AP=PD\text{.}\) Let \(D={\rm diag}\{\lambda_1,\ldots,\lambda_n\}\) and \(v_1,\ldots, v_n\) be columns of \(P\text{,}\) then \(\{v_1,\ldots, v_n\}\) is an orthonormal basis of \(\R^n\text{.}\) Also \(AP=PD\) implies \(Av_i=\lambda_i v_1\text{.}\) Hence \(\beta\) is an orthonormal eigenbasis of \(A\text{.}\)
We prove this result using induction on \(n\text{.}\) For \(n=1\text{.}\) Let \(A=[\alpha]\text{.}\) Then \(\{1\}\) is an orthonormal basis of \(\R\) and it is also an eigenvector.
Assume that the result is true for \(n-1\text{.}\) That is if \(A\) is an \((n-1)\times (n-1)\) real symmetric matrix then it is orthogonally diagonalizable.
Let us prove the result for \(n\text{.}\) Let \(A\) be an \(n\times n\) real symmetric matrix. By the fundamental theorem of algebra, we know that every real polynomial of has a root in \(\mathbb{C}\text{.}\) Hence the characteristic polynomila of \(A\) has a complex characteristics root. By TheoremΒ 5.3.4, all eigenvalues of \(A\) are real. Thus \(A\) has a real eigenvalue, say, \(\lambda\text{.}\) Let \(u\) be a unit eigenvector corresponding to the eigenvalue \(\lambda\) and \(W=\R u\text{.}\) Then \(W\) is a one dimensional subspace of \(\R^n\text{.}\) Hence \(W^\perp\) an \((n-1)\)-dimensional subspace of \(\R^n\text{.}\) Also \(W\) is \(A-\)invariannt. Hence by CheckpointΒ 6.3.12, \(W^\perp\) is \(A\)-invariant. Also \(\R^n=W\oplus W^\perp\text{.}\)
Let \(\beta = \{u,v_1,\ldots,v_{n-1}\}\) be an extended orthonormnal basis of \(A\text{.}\) Let \(P=[u~v_1~\cdots~v_{n-1}]\text{,}\) the orthogonal matrix whose columns are vectors \(u,v_1,\ldots,v_{n-1}\text{.}\) The the matrix \(M\) of \(A\) with respect to \(\beta\) is \(P^TAP\) which is of the form
where \(C\) is an \((n-1)\times (n-1)\) real symmetric matrix. (why)? Hence by induction, there exists an \((n-1)\times (n-1)\) orthogonal matrix \(Q\) such that \(Q^TCQ=D\text{,}\) a diagonal matrix. Hence
Consider a matrix \(A=\left(\begin{array}{rrr} 5 \amp -2 \amp -4 \\ -2 \amp 8 \amp -2 \\ -4 \amp -2 \amp 5 \end{array} \right)\text{.}\) Clearly \(A\) is symmetric and hence it is orthogonally diagonalizable. The characteristic polynomial of \(A\) is
Hence \(0, 9, 9\) are eigenvalues of \(A\text{.}\) Its is easy to find that \(v_1=(1, 1/2, 1)\) is an eigenvector corresponding to the eigenvalue 0. \(v_2=(1, 0, -1), v_2=(0, 1, -1/2)\) are eigenvectors corresponding to eigenvalue 9. Hence \(P:=\left(\begin{array}{rrr} 1 \amp 1 \amp 0 \\ \frac{1}{2} \amp 0 \amp 1 \\ 1 \amp -1 \amp -\frac{1}{2} \end{array} \right)\text{.}\) Then
Suppose \(f\colon \R^n \to \R^n\) is a map, that preserves the distance, that is \(\norm{f(x)-f(x)}=\norm{x-y}\) for all \(x,y\in \R^n\text{.}\) We would like to study such maps. Let us first look at a spacial case when \(f\) fixes the origin.
\begin{equation*}
f(x)\cdot f(y) = x\cdot y
\end{equation*}
for all \(x,y\text{.}\) That is, \(f\) preserves the dot product. This implies that \(f\) maps an orthonormal basis of \(\R^n\) to an orthonormal basis of \(\R^n\text{.}\) In particulatr, \(\{f(e_i)\}\) is an orthonormal basis of \(\R^n\text{,}\) where \(\{e_i\}\) is the standard basis. Hence
Let \(f\colon \R^n\to \R^n\) be such that \(\norm{f(x)-f(y)}=\norm{x-y}\) for all \(x,y\in \R^n\text{.}\) Then there exists a unique vector \(x_0\in \R^n\) and an orthogonal liear map \(A\) such that \(f(x)=Ax+x_0\text{.}\) In particular \(f\) is an affine linear transformation.
Let \(x_0:=f(0)\) and \(g(x)=f(x)-x_0\text{.}\) Then it is easy to check that \(g(0)=0\) and \(\norm{g(x)-g(y)}=\norm{x-y}\) for all \(x,y\text{.}\) Hence by LemmaΒ 6.4.12, \(g\) is linear. By TheoremΒ 6.4.11, \(g(x)=Ax\) for some orthogonal linear transformation \(A\text{.}\) Hence \(f(x)=Ax+x_0.\)
Since \(A\) is symmetric, it is orthogonally diagonalizable:
\begin{equation*}
A = Q DQ^T,
\end{equation*}
with \(D= \mathrm{diag}(\lambda_1,\dots,\lambda_n)\text{,}\) where all \(\lambda_i>0\text{.}\) Define \(D^{1/2} = \mathrm{diag}(\sqrt{\lambda_1},\dots,\sqrt{\lambda_n})\text{.}\) Then
Next we remove the middle third of the line, and replace it with two lines that each have the same length (1/3 or orgininal) as the remaining lines on each side. This new form is called the generator, because it specifies a rule that is used to generate a new form. Note that length of each seqment is 1/3. See FigureΒ 6.4.18.
Next we repeat the above steps to each of four segments in the generator. Then we get the cureve as in FigureΒ 6.4.19. Length of each segment in this case is \(16/9\text{.}\) If we apply the generator once again, we ge the curve as in FigureΒ 6.4.20. Length of each segment in this case is \(64/27\text{.}\)
If we keep apply this process, we get what is called the Koch Curve (named after the mathematician Helge von Koch in 1904). After applying the 7 iterators we get the curve as in FigureΒ 6.4.21
Now let us construct the Koch curve as an application of affine linear map. The construction begins with a single line segment from \((0,0)\) to \((1,0)\text{.}\) At each step, this segment is replaced by four smaller segments:
The Sierpinski triangle (named after the Polish mathematician Waclaw Sierpinski), also called the Sierpinski gasket, is a self-similar fractal subset \(S\) of the plane. It can be obtained by an iterative geometric construction starting from a filled equilateral triangle and applying iterated function system (IFS) consisting of affine maps.
Stage \(n+1\text{:}\) For each filled triangle from stage \(n\text{,}\) repeat the same process: divide into four, remove the central one. See FigureΒ 6.4.25 and FigureΒ 6.4.26 for two and three iterations.
The Sierpinski carpet is the planar fractal obtained by repeatedly removing the open central square from a subdivided unit square. Equivalently, it is the unique nonempty set \(C\) satisfying an iterated-function system (IFS) of eight contractive affine maps.
The Sierpinski pyramid (also called the Sierpinski tetra-pyramid when based on a triangle, or SierpiΕski square pyramid when based on a square) is a three-dimensional fractal obtained by repeatedly subdividing a pyramid into smaller self-similar pyramids. It provides a natural extension of the ideas behind the SierpiΕski triangle and SierpiΕski carpet to three dimensions.
The construction can be described as an application of affine linear maps. Starting from an initial pyramid \(P_0\text{,}\) we apply scaling by a factor of \(\tfrac{1}{2}\) followed by translations to position the smaller pyramids. In the square-based case, four pyramids are placed at the corners of the base, and one is placed on the top near the apex. This gives a total of five affine maps:
This equation can be written compactly in matrix notation as
\begin{equation*}
Q(x,y) =
\begin{bmatrix}x \amp y\end{bmatrix}
\begin{bmatrix} a \amp b \\ b \amp c \end{bmatrix}
\begin{bmatrix}x \\ y\end{bmatrix}+ \begin{bmatrix} d \amp e \end{bmatrix}
\begin{bmatrix} x \\ y \end{bmatrix} + f = 0.
\end{equation*}
Thus, the cross term \(2bxy\) is eliminated by using the orthogonal linear trasformation \(\begin{bmatrix} u \\ v \end{bmatrix}=P\begin{bmatrix} x \\ y \end{bmatrix} \) and the conic aligns with its principal axes, that is along the eigenvectors directions.
The transformation, \(P^{-1}\begin{pmatrix}x\\y
\end{pmatrix}+\begin{pmatrix} \frac{\alpha}{2\lambda_1}\\\frac{\beta}{2\lambda_2}\end{pmatrix} \) is called an affine linear transformation. Here \(P^{-1}\) is a orthogonl linear map. Thus an affine linear transformation on \(\R^n\) is a map of the form \(T(x)=Px+v\) ,where \(P\) is an orthogonal transformation and \(v\) is a called a translation vector. Such maps are also called isometries.
In case, \(\lambda_1\) and \(\lambda_2\) both are negative then, we can multiply the whole equation by \(-1\) and we get the a similar expression except, the right hand changes its sign.
Condiser the quadratic \((Q(x,y)=7 \, x^{2} - 6 \, x y + 7 \, y^{2} - 2 \, x + 3 \, y - 24\text{.}\) Let us convert this quadrtic into a conic section in canonial form.
It is easy to check the the eigenvalues of \(A\) are \(\lambda_1=10\) and \(\lambda_2=4\) with the corresponding eigenvectors \(v_1 = \left(\frac{1}{2} \, \sqrt{2},\,-\frac{1}{2} \, \sqrt{2}\right)\) and \(v_2=\left(\frac{1}{2} \, \sqrt{2},\,\frac{1}{2} \, \sqrt{2}\right)\text{.}\) Hence we have
\begin{equation*}
\begin{bmatrix} x \\ y \end{bmatrix} = P \begin{bmatrix} u \\ v \end{bmatrix}=\left(\begin{array}{rr}
\frac{1}{\sqrt{2}} u + \frac{1}{\sqrt{2}} v \\ -\frac{1}{\sqrt{2}} u + \frac{1}{\sqrt{2}v}
\end{array}\right).
\end{equation*}
Now substituting \(x=\frac{1}{\sqrt{2}} u + \frac{1}{\sqrt{2}} v\) and \(y=-\frac{1}{\sqrt{2}} u + \frac{1}{\sqrt{2}} v \) in the given quadratic, we get
\begin{equation*}
Q(u,v)=10 \, u^{2} + 4 \, v^{2} - 7 \, \sqrt{2} u + \sqrt{2} v - 56.
\end{equation*}
It is easy to check that he eigenvalues are \(\lambda_1=0.4384471871911698, \lambda_2=4.561552812808830\text{.}\) Since both the eigenvalues are positive, this quadratic is an ellipse. This is what the graph shows.
Consider the quadratic eqation given by \(-x^2+4xy-y^2-30x,+y+20=0\text{.}\) Use Sage to classify this and plot the curve along with the transformed coordinates system.
The matrix associated with the quadratic part of the above equation is \(A = \left(\begin{array}{rr}
3 \amp -3 \\
-3 \amp 3
\end{array}\right)\text{.}\) It is easy to check that the eigenvalues of \(A\) are \(\lambda_1=6, \lambda_2=0\text{.}\) Since one of the eigenvalues is 0, this curve is a parabola. Let us draw this curve along with the tranformed orgini and the two new coordinate directions in Sage.
For given quadratic equation \(Q(x,y)=ax^2+2bxy+cy^2+dx+ey+f=0\text{,}\) write down the corresponding canonical conics by describing the new orgin \((x_0,y_0)\text{,}\) and the new coordinate vectors by codisering different cases in a tabular form.
Subsection6.4.3Classification of Quadratic Surfaces in Three Variables
The classification of quadratic equation in three varibale can be done in a very similar manner as we have seen in case of two variable SubsectionΒ 6.4.2
\begin{equation*}
Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} + \mathbf{b}^T \mathbf{x} + j,
\quad
\mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad
A = \begin{bmatrix} a \amp d \amp f \\ d \amp b \amp e \\ f \amp e \amp c \end{bmatrix}, \quad
\mathbf{b} = \begin{bmatrix} g \\ h \\ i \end{bmatrix}.
\end{equation*}
Since \(A\) is symmetric, there exists an orthogonal matrix \(P\) such that
\begin{equation*}
P^T A P = D= \operatorname{diag}(\lambda_1,\lambda_2,\lambda_3).
\end{equation*}
After an orthogonal change of variables \(\mathbf{x} = P\mathbf{u}\) and a translation to eliminate linear terms, the quadratic form reduces to the canonical form.
