<\/p>\n
Problem 9.1<\/span> Solution.<\/span> <\/p>\n Problem 9.2<\/span> Solution.<\/span> <\/p>\n <\/p>\n Problem 9.3<\/span> Solution.<\/span> <\/p>\n Problem 9.4<\/span> Solution.<\/span> Denote two eigenvalues \\(\\lambda_1 = 1\\) and \\(\\lambda_2 = 4.\\) The eigenvectors corresponding to these eigenvalues are \\(v_1 = (1,\\,0)\\) and \\(v_2 = (2,\\,3),\\) respectively.\n<\/p>\n<\/div>\n <\/p>\n Problem 9.5<\/span> Solution.<\/span> <\/p>\n Problem 9.6<\/span> Solution.<\/span> <\/p>\n Problem 9.7<\/span> Solution.<\/span> <\/p>\n Problem 9.8<\/span> Solution.<\/span> <\/p>\n Problem 9.9<\/span> Solution.<\/span> <\/p>\n Problem 9.10<\/span> Solution.<\/span> First, this is YC Lee’s solution: Let \\(v\\) be a eigenvector corresponding to \\(\\lambda.\\) Then Next, this is I Seul Bee’s solution: Consider <\/p>\n Problem 9.11<\/span> Solution.<\/span> <\/p>\n Problem 9.12<\/span> Solution.<\/span> <\/p>\n Problem 9.13<\/span> Solution.<\/span> <\/p>\n Problem 9.14<\/span> Solution.<\/span> <\/p>\n Problem 9.15<\/span> Solution.<\/span> <\/p>\n Problem 9.16<\/span> Solution.<\/span> <\/p>\n Problem 9.17<\/span> Solution.<\/span> Problem 9.18<\/span> Solution.<\/span> Problem 9.19<\/span> Solution.<\/span> \n(2) The form \\(2x^2 + 2xy + 2y^2\\) is expressible as \n(3) The new coordinate is \n(4) The new coordinate is \n(5) The new coordinate is <\/p>\n","protected":false},"excerpt":{"rendered":" Problem 9.1 Describe geometrically the linear transformation \\(T_A : \\mathbb{R}^2 \\rightarrow \\mathbb{R}^2\\) given by \\(A = \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}\\) and then interpret the meanings of the eigenvalues and eigenvectors accordingly. Solution. \\(T_A\\) is a reflection about the line \\(y=x.\\) Hence \\(v\\) is an eigenvector of \\(T_A\\) if and only if \\(v\\) is parallel to or orthogonal to the…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_lmt_disableupdate":"","_lmt_disable":"","footnotes":""},"categories":[57],"tags":[],"class_list":["post-6537","post","type-post","status-publish","format-standard","hentry","category-linear-algebra"],"_links":{"self":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/6537","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/comments?post=6537"}],"version-history":[{"count":48,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/6537\/revisions"}],"predecessor-version":[{"id":6616,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/6537\/revisions\/6616"}],"wp:attachment":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/media?parent=6537"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/categories?post=6537"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/tags?post=6537"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nDescribe geometrically the linear transformation \\(T_A : \\mathbb{R}^2 \\rightarrow \\mathbb{R}^2\\) given by
\n\\[A = \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}\\]
\nand then interpret the meanings of the eigenvalues and eigenvectors accordingly.\n<\/p>\n
\n\\(T_A\\) is a reflection about the line \\(y=x.\\) Hence \\(v\\) is an eigenvector of \\(T_A\\) if and only if \\(v\\) is parallel to or orthogonal to the line \\(y=x.\\) \\(T_A\\) does not change the length of a vector, hence the eigenvalue is \\(1\\) or \\(-1.\\)\n<\/p>\n<\/div>\n
\nFind one or more eigenvalue\/eigenvector pairs other than exponentials of the form \\(Ce^{\\lambda x}\\) for the second derivative operator \\(D^2 : C^{\\infty} (\\mathbb{R}) \\rightarrow C^{\\infty}(\\mathbb{R}).\\)\n<\/p>\n
\nConsider the functions \\(y= \\sin kx,\\) \\(y=\\cos kx,\\) \\(y=\\sin kx + \\cos kx \\) for \\(k\\ne 0.\\) The eigenvalue corresponding to these functions is \\(-k^2.\\)\n<\/p>\n<\/div>\n
\nShow that if \\(T:V\\rightarrow V\\) is a linear transformation which is not injective, then \\(0\\) is an eigenvalue of \\(T.\\)\n<\/p>\n
\nSince \\(T\\) is not one-to-one, \\(\\operatorname{Ker}(T) \\ne \\left\\{ \\mathbf{0} \\right\\},\\) that is, there is a nonzero vector \\(v\\in\\operatorname{Ker}(T).\\) For this vector \\(v,\\) we have \\(T(v) = \\mathbf{0} = 0v.\\) Hence \\(0\\) is an eigenvalue.\n<\/p>\n<\/div>\n
\nFind the characteristic polynomial, eigenvalues and corresponding eigenvectors for the matrix
\n\\[A = \\begin{pmatrix} 1 & 2 \\\\ 0 & 4 \\end{pmatrix}.\\]\n<\/p>\n
\n\\[\\begin{align}
\ntI – A &= \\left[\\begin{array}{cc} t-1 & -2 \\\\ 0 & t-4 \\end{array}\\right],\\\\[5pt]
\np_A(t) &= \\det (tI-A) = (t-1)(t-4).\\end{align}\\]
\nThe equation \\(p_A (t)=0\\) has two solutions \\(t=1\\) and \\(t=4.\\)<\/p>\n
\nFor which real values of \\(a\\) does the matrix
\n\\[A = \\left( \\begin{array}{rc} 2 & a \\\\ -1 & 1 \\end{array} \\right)\\]
\nhave real eigenvalues? State your answer as an inequality.\n<\/p>\n
\nThe characteristic polynomial of \\(A\\) is
\n\\[p_A (t) = \\det\\left[\\begin{array}{cc} t-2 & -a \\\\ 1 & t-1 \\end{array}\\right] = t^2 – 3t + 2+a.\\]
\nSince \\(t^2 – 3t + 2+a =0\\) holds for some real number \\(t,\\) the discriminant has to be nonnegative, that is,
\n\\[D = 9 – 4(2+a) \\ge 0.\\]
\nTherefore \\[a \\le \\frac{1}{4}.\\]\n<\/p>\n<\/div>\n
\nCompute the characteristic polynomial and eigenvalues for the matrix
\n\\[A = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 4 & 2 \\\\ 0 & 2 & 1 \\end{pmatrix}.\\]\n<\/p>\n
\nIt is straightforward.
\n\\[p_A (t) = \\det\\left[\\begin{array}{ccc} t-1 & 0 & 0 \\\\ – & t-4 & -2 \\\\ 0 & -2 & t-1 \\end{array}\\right] = t(t-1)(t-5).\\]
\nHence the eigenvalues are
\n\\[\\lambda_1 = 0 ,\\,\\, \\lambda_2 = 1 ,\\,\\, \\lambda_3 = 5.\\]
\nThe eigenvectors corresponding to these eigenvalues are
\n\\[
\nv_1 = \\left[\\begin{array}{r} 0 \\\\ 1 \\\\ -2 \\end{array}\\right] ,\\,\\,
\nv_2 = \\left[\\begin{array}{r} 1 \\\\ 0 \\\\ 0 \\end{array}\\right] ,\\,\\,
\nv_3 = \\left[\\begin{array}{r} 0 \\\\ 2 \\\\ 1 \\end{array}\\right] .
\n\\]<\/p>\n<\/div>\n
\nShow that the following matrix has no real eigenvalues. Interpret this geometrically.
\n\\[A = \\left(\\begin{array}{rc} 0 & 1 \\\\ -1 & 0 \\end{array}\\right).\\]\n<\/p>\n
\n\\[A = \\left[\\begin{array}{cr} \\cos\\frac{3}{2}\\pi & -\\sin\\frac{3}{2}\\pi \\\\ \\sin\\frac{3}{2}\\pi & \\cos\\frac{3}{2}\\pi \\end{array}\\right].\\]
\nHence \\(A\\) represents a rotation by \\(-90 ^\\circ .\\) No vector \\(v\\) other than zero vector satisfies \\(T(v) = \\lambda v\\) for some scalar \\(\\lambda.\\)\n<\/p>\n<\/div>\n
\nLet \\(A\\in M_n(\\mathbb{R})\\) where \\(n\\) is odd. Show that \\(A\\) has at least one real eigenvalue.\n<\/p>\n
\nSince
\n\\[p_A (t) = \\det(tI-A),\\]
\n\\(p_A (t)\\) is a polynomial of degree \\(n,\\) and the coefficient of \\(t^n\\) is \\(1.\\) Hence
\n\\[\\lim_{t\\rightarrow\\infty} p_A(t) = \\infty \\,\\,\\text{and}\\,\\, \\lim_{t\\rightarrow -\\infty} p_A (t) = – \\infty.\\]
\nBy the definitions of limits,
\n\\[p_A (t_0 ) < 0 \\,\\,\\text{and}\\,\\, p_A (t_1) > 0\\]
\nfor some \\(t_0\\) and \\(t_1\\). Furthermore \\(t_0 \\ne t_1.\\) Since \\(p_A(t)\\) is a continuous function, by intermediate value theorem, \\(p_A(\\lambda)=0\\) for some \\(\\lambda\\) between \\(t_0\\) and \\(t_1.\\) \\(\\lambda\\) is a real eigenvalue of \\(A.\\)\n<\/p>\n<\/div>\n
\nFor any \\(A\\in M_n(\\mathbb{R}),\\) show that the number of imaginary roots of the characteristic polynomial is even. This gives an alternative approach to the previous problem.\n<\/p>\n
\nLet \\(p_A(t)\\) be the characteristic polynomial of \\(A.\\) If \\(\\lambda = a+bi,\\) \\(a\\in\\mathbb{R},\\) \\(b\\in\\mathbb{R}^\\times\\) and \\(p_A (\\lambda)=0,\\) then \\(\\overline{\\lambda} \\ne \\lambda\\) and \\(p_A (\\overline{\\lambda})=0.\\) Note that if \\(A\\) were a complex matrix, then we could not derive this result.\n<\/p>\n<\/div>\n
\nLet \\(\\lambda \\in K\\) be an eigenvalue of \\(A\\in M_n (K).\\) Show that \\(\\lambda ^r\\) is an eigenvalues of \\(A^r ,\\) the \\(r\\)th power of \\(A,\\) \\(r\\ge 0.\\)\n<\/p>\n
\nWe introduce two solutions.<\/p>\n
\n\\[A^r v = A^{r-1} (\\lambda v) = A^{r-2} (\\lambda^2 v) = \\cdots = \\lambda^r v.\\]
\nHence \\(\\lambda^r\\) is an eigenvalue of \\(A^r.\\)<\/p>\n
\n\\[\\lambda^r I – A^r = (\\lambda I-A) (\\lambda^{r-1} I + \\lambda^{r-2} A + \\lambda^{r-3} A^2 + \\cdots + A^{r-1} ).\\]
\nHence
\n\\[p_{A^r} (\\lambda^r ) = \\det(\\lambda I-A) \\det(\\ast) = 0 \\times \\det(\\ast) = 0.\\]
\nTherefore \\(\\lambda^r\\) is an eigenvalue of \\(A^r.\\)\n<\/p>\n<\/div>\n
\nLet \\(A\\in M_n (K)\\) be such that \\(A^r\\) is the zero matrix for some \\(r \\ge 1.\\) (In this case, \\(A^r\\) represents \\(A\\) to the power \\(r.\\)) Show that all of \\(A\\text{‘s}\\) eigenvalues are \\(0.\\)\n<\/p>\n
\nIf \\(\\lambda \\ne 0\\) and \\(\\lambda\\) is an eigenvalue of \\(A,\\) then \\(\\lambda^r\\) is an eigenvalue of \\(A^r.\\) But the eigenvalue of \\(O,\\) the zero matrix, is only \\(0.\\) Hence \\(\\lambda\\) must be zero.\n<\/p>\n<\/div>\n
\nShow that the eigenvalues of a triangular matrix are precisely the diagonal entries.\n<\/p>\n
\nLet \\(A = (a_{ij})_{n\\times n}\\) be an upper triangular matrix. Then \\(a_{ij} =0\\) for \\(i > j.\\)
\nSince \\(tI-A\\) is also an upper triangular matrix, the characteristic polynomial of \\(A\\) is
\n\\[p_A (t) = \\det(tI-A) = \\prod_{i=1}^n (t-a_{ii}).\\]
\nTherefore the eigenvalues of \\(A\\) are exactly \\(a_{11},\\) \\(a_{22},\\) \\(\\cdots,\\) \\(a_{nn},\\) the diagonal entries of \\(A.\\)\n<\/p>\n<\/div>\n
\nFind an example of a real \\(2\\times 2\\) matrix \\(A\\) which is not diagonalizable as an endomorphism of \\(\\mathbb{R}^2,\\) but is diagonalizable as an endomorphism of \\(\\mathbb{C}^2.\\)\n<\/p>\n
\nTake
\n\\[A = \\left[\\begin{array}{rc} 1 & 1 \\\\ -1 & 1 \\end{array}\\right].\\]
\nThen the characteristic polynomial of \\(A\\) is
\n\\[p_A (t) = (t-1)^2 +1.\\]
\n\\(p_A(t)\\) cannot assume zero unless \\(t\\) is an imaginary number.\n<\/p>\n<\/div>\n
\nFind all of the eigenvalues of the matrix
\n\\[A=\\left(\\begin{array}{rc} 0 & 2 \\\\ -2 & 4 \\end{array}\\right).\\]\n<\/p>\n
\nSince \\(p_A(t) = (t-2)^2,\\) the eigenvalue of \\(A\\) is only \\(2.\\)\n<\/p>\n<\/div>\n
\nShow that the matrix of the previous problem is not diagonalizable over either the real or complex fields.\n<\/p>\n
\nSince \\(v=(1,\\,1)\\) is the only eigenvector of \\(A,\\) the eigenvector of \\(A\\) does not constitute an eigenbasis. Hence \\(A\\) is not diagonalizable.\n<\/p>\n<\/div>\n
\nFind the eigenvalues of the matrix
\n\\[A = \\begin{pmatrix} 2 & 1 \\\\ 0 & 5 \\end{pmatrix}.\\]\n<\/p>\n
\nSince \\(p_A(t) = (t-2)(t-5),\\) the eigenvalues are \\(\\lambda_1 = 2\\) and \\(\\lambda_2 = 5 ,\\) and the eigenvectors corresponding to these eigenvalues are \\(v_1 = (1,\\,0)\\) and \\(v_2 = (1,\\,3).\\)\n<\/p>\n<\/div>\n
\nFor the matrix \\(A\\) of the previous problem, find an invertible matrix \\(P\\) such that \\(P^{-1} AP\\) is a diagonal matrix.\n<\/p>\n
\n\\[\\left[\\begin{array}{cc} 2 & 0 \\\\ 0 & 5 \\end{array}\\right]
\n=
\n\\left[\\begin{array}{cc} 1 & 1 \\\\ 0 & 3 \\end{array}\\right]^{-1}
\n\\left[\\begin{array}{cc} 2 & 1 \\\\ 0 & 5 \\end{array}\\right]
\n\\left[\\begin{array}{cc} 1 & 1 \\\\ 0 & 3 \\end{array}\\right].
\n\\]\n<\/p>\n<\/div>\n
\nLet a sequence \\(\\left\\{ a_n \\right\\}\\) be defined recursively by
\n\\[a_0 = 1 ,\\,\\, a_1 = 1 \\,\\,\\text{and}\\,\\, a_{n+2} = 5a_{n+1} – 6a_n \\,\\,\\text{for}\\,\\, n\\ge 0.\\]
\nFind the explicit formula for the general term \\(a_n.\\)\n<\/p>\n
\nThe recursive formula is expressible as
\n\\[\\left[\\begin{array}{c} a_n \\\\ a_{n-1} \\end{array}\\right]
\n=
\n\\left[\\begin{array}{cr} 5 & -6 \\\\ 1 & 0 \\end{array}\\right]
\n\\left[\\begin{array}{c} a_{n-1} \\\\ a_{n-2} \\end{array}\\right].\\]
\nHence
\n\\[\\left[\\begin{array}{c} a_n \\\\ a_{n-1} \\end{array}\\right]
\n=
\n\\left[\\begin{array}{cr} 5 & -6 \\\\ 1 & 0 \\end{array}\\right]^{n-1}
\n\\left[\\begin{array}{c} 1 \\\\ 1 \\end{array}\\right].\\]
\nTake
\n\\[A = \\left[\\begin{array}{cr} 5 & -6 \\\\ 1 & 0 \\end{array}\\right].\\]
\nThe eigenvalues of \\(A\\) are \\(\\lambda_1 = 2\\) and \\(\\lambda_2 = 3,\\) and the eigenvectors corresponding to these eigenvalues are
\n\\[v_1 = \\left[\\begin{array}{c} 2 \\\\ 1\\end{array}\\right] \\,\\,\\text{and}\\,\\,v_2 = \\left[\\begin{array}{c} 3 \\\\ 1\\end{array}\\right].\\]
\nThe diagonalization of \\(A\\) is
\n\\[
\n\\left[\\begin{array}{cr} 5 & -6 \\\\ 1 & 0 \\end{array}\\right]
\n=
\n\\left[\\begin{array}{cr} 2 & 3 \\\\ 1 & 1 \\end{array}\\right]
\n\\left[\\begin{array}{cr} 2 & 0 \\\\ 0 & 3 \\end{array}\\right]
\n\\left[\\begin{array}{cr} 2 & 3 \\\\ 1 & 1 \\end{array}\\right]^{-1}.
\n\\]
\nHence we have
\n\\[
\n\\left[\\begin{array}{c} a_n \\\\ a_{n-1} \\end{array}\\right]
\n=
\n\\left[\\begin{array}{cr} 2 & 3 \\\\ 1 & 1 \\end{array}\\right]
\n\\left[\\begin{array}{cr} 2^{n-1} & 0 \\\\ 0 & 3^{n-1} \\end{array}\\right]
\n\\left[\\begin{array}{cr} 2 & 3 \\\\ 1 & 1 \\end{array}\\right]^{-1}
\n\\left[\\begin{array}{c} 1 \\\\ 1 \\end{array}\\right].
\n\\]
\nAn evaluation gives us the formula
\n\\[a_n = 2^{n+1} – 3^{n}\\]
\nfor \\(n\\ge 0.\\)\n<\/p>\n<\/div>\n
\nFind new coordinates \\(x ‘ ,\\) \\(y ‘ \\) so that the following quadratic forms can be written as \\(\\lambda_1 (x ‘ )^2 + \\lambda_2 ( y ‘ )^2 .\\) Express the following forms in the form \\(\\lambda_1 (x ‘ )^2 + \\lambda_2 ( y ‘ )^2 .\\)
\n(1) \\(x^2 + 4xy + y^2 \\)
\n(2) \\(2x^2 + 2xy + 2y^2 \\)
\n(3) \\(x^2 -12xy -4y^2 \\)
\n(4) \\(3x^2 + 2xy + 3y^2 \\)
\n(5) \\(x^2 – 2xy + y^2 \\)\n<\/p>\n
\n(1) The form \\(x^2 + 4xy + y^2\\) is expressible as
\n\\[
\n\\left[\\begin{array}{cc} x & y \\end{array}\\right]
\n\\left[\\begin{array}{cc} 1 & 2 \\\\ 2 & 1 \\end{array}\\right]
\n\\left[\\begin{array}{c} x \\\\ y \\end{array}\\right]
\n=:
\n\\left[\\begin{array}{cc} x & y \\end{array}\\right]A\\left[\\begin{array}{c} x \\\\ y \\end{array}\\right] .\\]
\nThe eigenvalues of \\(A\\) are
\n\\[\\lambda_1 = 3 ,\\,\\, \\lambda_2 = -1\\]
\nand the corresponding eigenvectors are
\n\\[
\n\\left[\\begin{array}{c} 1 \\\\ 1\\end{array}\\right] ,\\,\\,
\n\\left[\\begin{array}{r} -1 \\\\ 1 \\end{array}\\right].
\n\\]
\nNormalizing these eigenvectors, we have an orthonormal eigenbasis:
\n\\[
\n\\left[\\begin{array}{c} \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} \\end{array}\\right] ,\\,\\,
\n\\left[\\begin{array}{r} -\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} \\end{array}\\right].
\n\\]
\nHence the new coordinate is
\n\\[\\left[\\begin{array}{c} x \\\\ y \\end{array}\\right]
\n=
\n\\left[\\begin{array}{cr} \\frac{1}{\\sqrt{2}} & – \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{array}\\right]
\n\\left[\\begin{array}{c} x’ \\\\ y’ \\end{array}\\right].\\]
\nAs a consequence, we have
\n\\[x^2 + 4xy + y^2 = 3 ( x ‘ )^2 – ( y ‘ )^2 .\\]\n<\/p>\n<\/div>\n
\n\\[
\n\\left[\\begin{array}{cc} x & y \\end{array}\\right]
\n\\left[\\begin{array}{cc} 2 & 1 \\\\ 1 & 2 \\end{array}\\right]
\n\\left[\\begin{array}{c} x \\\\ y \\end{array}\\right]
\n=:
\n\\left[\\begin{array}{cc} x & y \\end{array}\\right]A\\left[\\begin{array}{c} x \\\\ y \\end{array}\\right] .\\]
\nThe eigenvalues of \\(A\\) are
\n\\[\\lambda_1 = 3 ,\\,\\, \\lambda_2 = 1\\]
\nand the corresponding eigenvectors are
\n\\[
\n\\left[\\begin{array}{c} 1 \\\\ 1\\end{array}\\right] ,\\,\\,
\n\\left[\\begin{array}{r} -1 \\\\ 1 \\end{array}\\right].
\n\\]
\nNormalizing these eigenvectors, we have an orthonormal eigenbasis:
\n\\[
\n\\left[\\begin{array}{c} \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} \\end{array}\\right] ,\\,\\,
\n\\left[\\begin{array}{r} -\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} \\end{array}\\right].
\n\\]
\nHence the new coordinate is
\n\\[\\left[\\begin{array}{c} x \\\\ y \\end{array}\\right]
\n=
\n\\left[\\begin{array}{cr} \\frac{1}{\\sqrt{2}} & – \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{array}\\right]
\n\\left[\\begin{array}{c} x’ \\\\ y’ \\end{array}\\right].\\]
\nAs a consequence, we have
\n\\[2x^2 + 2xy + 2y^2 = 3 ( x ‘ )^2 + ( y ‘ )^2 .\\]\n<\/p>\n<\/div>\n
\n\\[\\left[\\begin{array}{c} x \\\\ y \\end{array}\\right]
\n=
\n\\left[\\begin{array}{rc} -\\frac{3}{\\sqrt{13}} & \\frac{2}{\\sqrt{13}} \\\\ \\frac{2}{\\sqrt{13}} & \\frac{3}{\\sqrt{13}} \\end{array}\\right]
\n\\left[\\begin{array}{c} x’ \\\\ y’ \\end{array}\\right].\\]
\nAs a consequence, we have
\n\\[x^2 – 12xy – 4y^2 = 5 ( x ‘ )^2 – 8 ( y ‘ )^2 .\\]\n<\/p>\n<\/div>\n
\n\\[\\left[\\begin{array}{c} x \\\\ y \\end{array}\\right]
\n=
\n\\left[\\begin{array}{cr} \\frac{1}{\\sqrt{2}} & – \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{array}\\right]
\n\\left[\\begin{array}{c} x’ \\\\ y’ \\end{array}\\right].\\]
\nAs a consequence, we have
\n\\[3x^2 + 2xy +3y^2 = 4 ( x ‘ )^2 +2 ( y ‘ )^2 .\\]\n<\/p>\n<\/div>\n
\n\\[\\left[\\begin{array}{c} x \\\\ y \\end{array}\\right]
\n=
\n\\left[\\begin{array}{rc} -\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{array}\\right]
\n\\left[\\begin{array}{c} x’ \\\\ y’ \\end{array}\\right].\\]
\nAs a consequence, we have
\n\\[x^2 – 2xy + y^2 = 2 ( x ‘ )^2 .\\]\n<\/p>\n<\/div>\n