Diagonalize the matrix \(\Mtrx{A}\) below if possible.
\[ \Mtrx{A} = \left[ \begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array} \right] \]We begin by looking for the eigenvalues of \(\Mtrx{A}\). First we find its characteristic polynomial:
\[ \begin{array}{rcl} p(\lambda) & = & \det\left( \left[ \begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array} \right]\ -\ \lambda\cdot \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] \right) \\ & = & \det \left[ \begin{array}{cc} 3 - \lambda & 1 \\ 1 & 3 - \lambda \end{array} \right] \\ & = & (3-\lambda)^2 - 1 \\ & = & \lambda^2 -6\lambda + 8 \\ & = & (\lambda - 4)(\lambda - 2) \end{array} \]So \(\Mtrx{A}\) has two distinct eigenvalues:
The eigenspace of \(\Mtrx{A}\) associated to \(\lambda_1\) consists of the solutions of the matrix equation \((\Mtrx{A} - \lambda_1 \IdMtrx{2})\Vect{x} = \Vect{0}\):
\[ \left[ \begin{array}{rr} 1 & 1 \\ 1 & 1 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \end{array} \right] \]The solutions of the corresponding system of linear equations are of the form
| \((x,y)\) | \(=\) | \(s(-1,1)\) |
where \(s\) in \(\RNr\) is arbitrary. Therefore the eigenspace of \(\lambda_1\) is
| \(E_1\) | \(\DefEq\) | \(\span(-1,1)\) |
and \(\EuScript{B}_1 \DefEq (-1,1)\) is a basis of \(E_1\).
The eigenspace of \(\Mtrx{A}\) associated to \(\lambda_2\) consists of the solutions of the matrix equation \((\Mtrx{A} - \lambda_2 \IdMtrx{2})\Vect{x} = \Vect{0}\):
\[ \left[ \begin{array}{rr} -1 & 1 \\ 1 & -1 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \end{array} \right] \]The solutions of the corresponding system of linear equations are of the form
| \((x,y)\) | \(=\) | \(t(1,1)\) |
where \(t\) in \(\RNr\) is arbitrary. Therefore the eigenspace of \(\lambda_2\) is
| \(E_2\) | \(\DefEq\) | \(\span(1,1)\) |
and \(\EuScript{B}_ {2} \DefEq (1,1)\) is a basis of \(E_2\).
Setting \(\Vect{b}_1\DefEq (-1,1)\) and \(\Vect{b}_2 \DefEq(1,1)\), we found that \(\RNrSpc{2}\) has the ordered basis \(\EuScript{B}\DefEq (\Vect{b}_1,\Vect{b}_2)\) of eigenvectors. Therefore \(\Mtrx{A}\) is diagonalizable. A diagonalizing matrix is
\[ \Mtrx{C} = \CoordTrafoMtrx{C}{\EuScript{S}}{\EuScript{B}} \DefEq \left[ \begin{array}{rr} -1 & 1 \\ 1 & 1 \end{array} \right] \]with
\[ \Mtrx{C}^{-1} = \CoordTrafoMtrx{C}{\EuScript{B}}{\EuScript{S}} = -\dfrac{1}{2}\, \left[ \begin{array}{rr} 1 & -1 \\ -1 & -1 \end{array} \right] \]Consequently,
\[ \left[ \begin{array}{cc} 2 & 0 \\ 0 & 4 \end{array} \right]\ =\ -\dfrac{1}{2}\, \left[ \begin{array}{rr} 1 & -1 \\ -1 & -1 \end{array} \right] \left[ \begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array} \right] \left[ \begin{array}{rr} -1 & 1 \\ 1 & 1 \end{array} \right] \]Diagonalize the matrix \(\Mtrx{A}\) below if possible.
\[ \Mtrx{A}\ =\ \dfrac{1}{6}\left[ \begin{array}{rrr} 5 & -1 & -2 \\ -3 & 3 & -6 \\ -1 & -1 & 4 \end{array}\right] \]We begin by looking for the eigenvalues of ; i.e. all those values of \(\lambda\) for which the characteristic polynomial below vanishes.
\[ \begin{array}{rcl} p(\lambda) & = & \text{det} \begin{bmatrix} \tfrac{5}{6} - \lambda & -\tfrac{1}{6} & -\tfrac{1}{3} \\ -\tfrac{1}{2} & \tfrac{1}{2} - \lambda & -1 \\ -\tfrac{1}{6} & -\tfrac{1}{6} & \tfrac{2}{3} - \lambda \end{bmatrix} \\ 6^3\cdot p(\lambda) & = & \text{det} \begin{bmatrix} 5 - 6\lambda & -1 & -2 \\ -3 & 3 - 6\lambda & -6 \\ -1 & -1 & 4 - 6\lambda \end{bmatrix} \\ & = & \lambda^3\ -\ 2 \lambda^2\ +\ \lambda \\ & = & (\lambda - 0)^1(\lambda-1)^2 \end{array} \]Thus \(\Mtrx{A}\) has two distinct eigenvalues:
The eigenspace of \(\Mtrx{A}\) associated to \(\lambda_1\) consists of the solutions of the matrix equation
\[ \dfrac{1}{6} \left[\begin{array}{rrr} 5 & -1 & -2 \\ -3 & 3 & -6 \\ -1 & -1 & 4 \end{array}\right]\, \begin{bmatrix} x \\ y \\ z \end{bmatrix}\ =\ \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]The solutions of the corresponding system of homogeneous linear equations are of the form
\[ \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = s \left[ \begin{array}{c} 1 \\ 3 \\ 1 \end{array} \right] \]with \(s\) an arbitrary number in \(\RNr\). Thus \(\EuScript{B}_1=(\Vect{b}_1)\) with \(\Vect{b}_1=(1,3,1)\) is a basis for \(E_1\), the eigenspace of \(\Mtrx{A}\) associated to \(\lambda_1\). Therefore the geometric multiplicity of \(\lambda_1\) is 1, and every nonzero vector in \(E_1\) is an eigenvector of \(\Mtrx{A}\) with eigenvalue \(0\). This means that \(\Mtrx{A}\) transforms all of \(E_1\) into the zerovector.
The eigenspace of \(\Mtrx{A}\) associated to \(\lambda_2\) consists of the solutions of the matrix equation
\[ \dfrac{1}{6} \left[\begin{array}{rrr} -1 & -1 & -2 \\ -3 & -3 & -6 \\ -1 & -1 & -2 \end{array}\right]\, \begin{bmatrix} x \\ y \\ z \end{bmatrix}\ =\ \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]The solutions of the corresponding system of homogeneous linear equations are of the form
\[ \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = s_1 \left[ \begin{array}{r} 1 \\ -1 \\ 0 \end{array} \right] + s_2 \left[ \begin{array}{r} 0 \\ -2 \\ 1 \end{array} \right] \]with \(s_1\) and \(s_2\) arbitrary numbers in \(\RNr\). Thus \(\EuScript{B}_2=(\Vect{b}_2,\Vect{b}_3)\) with
\[\Vect{b}_2 = (1,-1,0) \quad\text{and}\quad \Vect{b}_3=(0,-2,1)\]form a basis of \(E_2\), the eigenspace associated to \(\lambda_2\). Thus the geometric multiplicity of \(\lambda_2\) is 2, and every nonzero vector in \(E_2\) is an eigenvector of \(\lambda_2\) with eigenvalue \(1\). This means that \(\Mtrx{A}\) transforms each such vector into itself; i.e. \(\Mtrx{A}\) acts as the identity transformation on \(E_2\).
We conclude that \(\RNrSpc{3}\) has a basis \(\EuScript{B}=(\Vect{b}_1,\Vect{b}_2,\Vect{b}_3)\) consisting of eigenvectors of \(\Mtrx{A}\), and so \(\Mtrx{A}\) is diagonalizable. A diagonalizing matrix is
\[ \Mtrx{C} = \CoordTrafoMtrx{C}{\EuScript{S}}{\EuScript{B}} \DefEq \left[ \begin{array}{rrr} 1 & 1 & 0 \\ 3 & -1 & -2 \\ 1 & 0 & 1 \end{array} \right] \]The resulting diagonal matrix is
\[ \Mtrx{D} = \LinMapMtrx{A}{\EuScript{B}}{\EuScript{B}} = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \]This matrix describes the projection of \(\RNrSpc{3}\) onto the eigenspace \(E_2\). Accordingly, we have
| \(\Mtrx{D}\) | \(=\) | \(\CoordTrafoMtrx{C}{\EuScript{B}}{\EuScript{S}} \Mtrx{A} \CoordTrafoMtrx{C}{\EuScript{S}}{\EuScript{B}}\) |