Matrix Diagonalization - Examples

Example 0.

Show that the matrix \(\ \boldsymbol{A}\ =\ \left[\begin{array}{cc} 1 & a \\ 0 & 1 \end{array}\right] \in M_2(R),\quad a\neq 0,\ \) is not diagonalizable by a similarity transformation, \(\ \) that is, there is no non-degenerate matrix \(\ \boldsymbol{P}\ \) such that

(1)\[\boldsymbol{P}^{-1}\boldsymbol{A}\,\boldsymbol{P}\ =\ \boldsymbol{D},\]

where \(\ \boldsymbol{D}\ \) is diagonal.

Proof.

I. \(\,\) The matrix \(\,\boldsymbol{A}\in M_n(K)\,\) is diagonalizable by a similarity transformation (1) if and only if its eigenvectors span the space \(\,K^n,\ \) that is, eigenvectors of the matrix \(\,\boldsymbol{A}\) comprise a basis of the space \(\,K^n.\)

The characteristic equation of the matrix \(\,\boldsymbol{A}:\)

\[\begin{split}\det{(\boldsymbol{A}-\lambda\,\boldsymbol{I}_n)}\ =\ \left|\begin{array}{cc} 1-\lambda &a \\ 0 & 1-\lambda \end{array}\right|\ =\ (1-\lambda)^2 =\,0\end{split}\]

provides one eigenvalue \(\,\lambda=1\,\) whose algebraic multiplicity is equal to 2.

Eigenvectors \(\,\boldsymbol{x}\,\) associated with this eigenvalue may be determined from the equation \(\ (\boldsymbol{A}-\lambda\,\boldsymbol{I}_n)\,\boldsymbol{x}\,=\, \boldsymbol{0},\ \) that is

\[\begin{split}\left[\begin{array}{cc} 0 & a \\ 0 & 0 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right]\ =\ \left[\begin{array}{c} 0 \\ 0 \end{array}\right] \quad : \quad \begin{cases} \ \ x_1 \text{ arbitrary,} \\ \ \ a\,x_2=0\,; \end{cases} \ : \quad \begin{cases} \ \ x_1 \text{ arbitrary,} \\ \ \ x_2=0\,. \end{cases} \,,\end{split}\]

Hence, the eigenvectors are of the form

\[\begin{split}\boldsymbol{x}\,=\, \left[\begin{array}{c} \beta \\ 0 \end{array}\right]\ =\ \beta\ \left[\begin{array}{c} 1 \\ 0 \end{array}\right],\quad \beta\in R\setminus\{0\}\end{split}\]

and comprise (together with a zero vector) 1-dimensional subspace: geometric multiplicity of the eigenvalue \(\,\lambda=1\,\) is 1 \(\,\) and thus \(\,\) it is different from the algebraic multiplicity.

Hence, since any two eigenvectors of the matrix \(\,\boldsymbol{A}\ \) are linearly dependent, they do not form a basis of the space \(\,R^2.\ \) This means that it is not possible to diagonalize the matrix \(\,\boldsymbol{A}\ \) by a non-degenerate matrix \(\,\boldsymbol{P}.\)

II. \(\,\) Similar matrices have the same chracteristic polynomial, and thus also the same set of eigenvalues with the same corresponding algebraic multiplicities.

It is easy to see that the matrix \(\,\boldsymbol{A}\,\) has an eigenvalue \(\,\lambda=1\,\) with algebraic multiplicity 2.

If there exists a matrix \(\,\boldsymbol{P}\,\) satisfying the condition (1), \(\,\) then the matrix \(\,\boldsymbol{D}\,\) in that equation is a diagonal matrix of size 2 having an eigenvalue \(\,\lambda=1\ \) with algebraic multiplicity 2. These conditions uniquely determine an identity matrix: \(\ \boldsymbol{D}\,=\,\boldsymbol{I}_2.\)

Therefore we would have \(\ \boldsymbol{P}^{-1}\boldsymbol{A}\,\boldsymbol{P}\ =\ \boldsymbol{I}_2,\ \) and thus \(\ \boldsymbol{A}\,=\, \boldsymbol{P}\,\boldsymbol{I}_2\boldsymbol{P}^{-1}\ =\ \boldsymbol{P}\boldsymbol{P}^{-1}\ =\ \boldsymbol{I}_2.\)

This gives a contradiction as \(\ \boldsymbol{A}\neq\boldsymbol{I}_2.\ \) Hence, there is no matrix \(\,\boldsymbol{P}\,\) for which the equation (1) would hold.

Example 1.

The matrix \(\ \boldsymbol{\sigma}_x\,=\, \left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]\in M_2(C)\ \) is real and symmetric, and thus Hermitian.

We will verify that:

• its eigenvalues are real,

• eigenvectors associated with distinct eigenvalues are orthogonal,

• algebraic multiplicity of every eigenvalue is equal to its geometric multiplicity,

• eigenvectors of the matrix \(\ \boldsymbol{\sigma}_x\ \) comprise an orthonormal basis of the space \(\,C^2,\)

• the matrix \(\ \boldsymbol{\sigma}_x\ \) is diagonalizable by a real orthogonal similarity transformation.

The characteristic equation of the matrix \(\,\boldsymbol{\sigma}_x\,:\)

\[\begin{split}w(\lambda)\,=\, \left|\begin{array}{cc} -\lambda & 1 \\ 1 & -\lambda \end{array}\ \right|\ =\ (-\lambda)^2-1\ =\ \lambda^2-1\ =\ (\lambda-1)(\lambda+1)\ =\ 0\end{split}\]

provides two eigenvalues with their algebraic multiplicities:

\[\lambda_1\,=\ 1\quad(1)\,,\qquad\lambda_2\,=\ -1\quad(1)\,.\]

We substitute them to the equation

\[(\boldsymbol{\sigma}_x -\lambda\,\boldsymbol{I}_2)\,\boldsymbol{x}\ = \ \boldsymbol{0}\]

to determine the associated eigenvectors \(\ \boldsymbol{x}.\)

Eigenvectors associated with the eigenvalue \(\,\lambda_1=\,1:\)

\[\begin{split}\begin{array}{l} \left[\begin{array}{rr} -1 & 1 \\ 1 & -1 \end{array}\ \right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \ =\ \left[\begin{array}{c} 0 \\ 0 \end{array}\right] \quad : \quad \begin{cases}\ \begin{array}{r} -x_1+x_2=0 \\ x_1-x_2=0 \end{array} \end{cases} : \quad x_1=x_2=\alpha,\ \ \alpha\in C. \\ \\ \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \ =\ \left[\begin{array}{c} \alpha \\ \alpha \end{array}\right]\ =\ \alpha\ \left[\begin{array}{c} 1 \\ 1 \end{array}\right]\,, \quad\alpha\in C\setminus\{0\}. \end{array}\end{split}\]

Eigenvectors associated with the eigenvalue \(\,\lambda_2=\,-1:\)

\[\begin{split}\begin{array}{l} \left[\begin{array}{rr} 1 & 1 \\ 1 & 1 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \ =\ \left[\begin{array}{c} 0 \\ 0 \end{array}\right] \quad : \quad \begin{cases}\ \begin{array}{r} x_1+x_2=0 \\ x_1+x_2=0 \end{array} \end{cases} : \quad x_1=-x_2=\beta,\ \ \beta\in C. \\ \\ \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \ =\ \left[\begin{array}{c} \beta \\ -\beta \end{array}\right]\ =\ \beta\ \left[\begin{array}{r} 1 \\ -1 \end{array}\right]\,, \quad\beta\in C\setminus\{0\}. \end{array}\end{split}\]

Geometric multiplicity of the two eigenvalues is equal to 1 and agrees with the algebraic multiplicity. Moreover, every two eigenvectors associated with distinct eigenvalues are orthogonal:

\[\begin{split}\left\langle\ \left[\begin{array}{c} \alpha \\ \alpha \end{array}\right],\ \left[\begin{array}{c} \beta \\ -\beta \end{array}\right]\ \right\rangle\ =\ \alpha^*\beta\,+\,\alpha^*(-\beta)\ =\ \alpha^*\beta\,-\,\alpha^*\beta\ =\ 0.\end{split}\]

Hence, normalised eigenvectors of the matrix \(\ \boldsymbol{\sigma}_x\,\) comprise an orthonormal basis \(\ \mathcal{F}\ \) of the space \(\,C^2\,:\)

\[\begin{split}\mathcal{F}\ =\ \,(\,\boldsymbol{f}_1,\,\boldsymbol{f}_2\,)\,, \quad\text{where}\quad \boldsymbol{f}_1\ =\ \textstyle{\frac{1}{\sqrt{2}}}\, \left[\begin{array}{c} 1 \\ 1 \end{array}\right]\,,\quad \boldsymbol{f}_2\ =\ \textstyle{\frac{1}{\sqrt{2}}}\, \left[\begin{array}{c} 1 \\ -1 \end{array}\right]\,.\end{split}\]

The basis vectors form a matrix \(\,\boldsymbol{P}\ \) which diagonalizes the matrix \(\,\boldsymbol{\sigma}_x:\)

\[\begin{split}\boldsymbol{P}\,=\ [\ \boldsymbol{f}_1\ |\ \boldsymbol{f}_2\ ]\ =\ \textstyle{\frac{1}{\sqrt{2}}}\, \left[\begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array}\right]\,.\end{split}\]

Note that \(\,\boldsymbol{P}\,\) is a real matrix which is both \(\ \) Hermitian (symmetric) \(\ \) and \(\ \) unitary (orthogonal):

\[ \begin{align}\begin{aligned}\boldsymbol{P}^+\ =\ \,\boldsymbol{P}^T\ =\ \,\boldsymbol{P} \,,\qquad \boldsymbol{P}^+\boldsymbol{P}\ =\ \boldsymbol{P}^T\boldsymbol{P}\ =\ \boldsymbol{I}_2\,;\\\text{moreover:}\qquad \boldsymbol{P}^2\ =\ \boldsymbol{I}_2,\quad \boldsymbol{P}^{-1}\ =\ \boldsymbol{P}^+\ =\ \boldsymbol{P}\,.\end{aligned}\end{align} \]

Numerical diagonalization of the matrix \(\,\boldsymbol{\sigma}_x:\)

sage: P = (1/sqrt(2))*matrix(RR,[[1, 1],
                           [1,-1]])
sage: P*P

[ 1.00000000000000  0.000000000000000]
[ 0.000000000000000 1.00000000000000 ]

sage: S_x = matrix(RR,[[0, 1],
                       [1, 0]])

sage: P = (1/sqrt(2))*matrix(RR,[[1, 1],
                           [1,-1]])
sage: P*S_x*P

[ 1.00000000000000   0.000000000000000]
[ 0.000000000000000 -1.00000000000000 ]

Example 2.

The matrix \(\ \boldsymbol{R}_\phi\ =\ \left[\begin{array}{cc} \cos{\phi} & -\sin{\phi} \\ \sin{\phi} & \cos{\phi} \end{array}\right] \in M_2(C)\ \) is real orthogonal, and thus unitary:

\[\boldsymbol{R}_\phi^+\,\boldsymbol{R}_\phi\ =\ \boldsymbol{R}_\phi^T\,\boldsymbol{R}_\phi\ =\ \boldsymbol{I}_2.\]

We will verify that:

• its eigenvalues are complex numbers of modulus 1,

• eigenvectors associated with distinct eigenvalues are orthogonal,

• algebraic multiplicity of every eigenvalue is equal to its geometric multiplicity,

• eigenvectors of the matrix \(\ \boldsymbol{R}_\phi\ \) comprise an orthonormal basis of the space \(\,C^2,\)

• the matrix \(\ \boldsymbol{R}_\phi\ \) is diagonalizable by a unitary similarity transformation.

If \(\,\phi=0\ \) or \(\,\phi=\pi,\ \) then the matrix \(\ \boldsymbol{R}_\phi\ \) is diagonal and equals \(\,\boldsymbol{I}_2\ \) or \(\ -\,\boldsymbol{I}_2,\ \) correspondingly. We will assume then that \(\ \phi\neq 0,\,\pi.\)

A unitary matrix diagonalizing the matrix \(\ \boldsymbol{R}_\phi\ \) is a transition matrix from the canonical basis \(\,\mathcal{E}\,=\, (\boldsymbol{e}_1,\, \boldsymbol{e}_2)\ \) of the space \(\,C^2\ \) to an orthonormal basis \(\,\mathcal{F}\,=\, (\boldsymbol{f}_1,\boldsymbol{f}_2)\ \) consisting of normalised eigenvectors of the matrix \(\ \boldsymbol{R}_\phi.\)

We solve an eigenproblem of the matrix \(\ \boldsymbol{R}_\phi\ \) and \(\,\) construct the basis \(\,\mathcal{F}.\)

The characteristic polynomial \(\,w(\lambda)\,=\, \det(\boldsymbol{R}_\phi -\,\lambda\,\boldsymbol{I}_2)\ \) is given by

\[\begin{split}\begin{array}{rl} w(\lambda) \!\! & =\ \ \left|\begin{array}{cc} \cos{\phi}-\lambda & -\sin{\phi} \\ \sin{\phi} & \cos{\phi}-\lambda \end{array}\right| \ =\ (\cos{\phi}-\lambda)^2\,+\ {\sin}^2\phi\ = \\[12pt] & =\ \, {\cos}^2\phi - 2\,\lambda\,\cos{\phi}\ + \lambda^2 +\ {\sin}^2\phi\ =\ 1\, -\, 2\,\lambda\,\cos{\phi}\ +\,\lambda^2. \end{array}\end{split}\]

Eigenvalues of the matrix \(\ \boldsymbol{R}_\phi\ \) are roots of the characteristic equation \(\,w(\lambda)=0:\)

\[\begin{split}\begin{array}{rl} \vartriangleright & \lambda^2\,-\ 2\,\cos{\phi}\cdot\lambda\ +\ 1\ =\ 0 \\[8pt] & \Delta\ =\ 4\,{\cos}^2\phi\ -\ 4\ =\ -\ 4\,(1-{\cos}^2\phi)\ =\ -\ 4\,{\sin}^2\phi\,<\,0\,; \\[8pt] & \sqrt{\Delta}\ =\ \pm\,2\,i\,\sin{\phi}; \\[8pt] & \lambda_{1,2}\ =\ \frac{1}{2}\ (2\,\cos{\phi}\,\pm\,2i\sin{\phi})\ =\ \cos{\phi}\,\pm\,i\,\sin{\phi}\ =\ e^{\ \pm\,i\,\phi}\,. \end{array}\end{split}\]

We obtained two eigenvalues with algebraic multiplicities equal to 1:

\[\blacktriangleright\quad \lambda_1\,=\ e^{\ +\,i\,\phi}\,,\qquad \lambda_2\,=\ e^{\ -\,i\,\phi}\,.\]

We substitute these values to the equation

\[(\boldsymbol{R}_\phi-\lambda\,\boldsymbol{I}_2)\ \boldsymbol{x}\ = \ \boldsymbol{0}\]

in order to determine the associated eigenvectors \(\ \boldsymbol{x}.\)

For an eigenvalue \(\ \lambda_1 =\, e^{\ +\,i\,\phi}\,=\,\cos{\phi}\,+\,i\,\sin{\phi}\ :\)

\[ \begin{align}\begin{aligned}\begin{split}\left[\begin{array}{cc} -i\,\sin{\phi} & -\sin{\phi} \\ \sin{\phi} & -i\,\sin{\phi} \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \ =\ \left[\begin{array}{c} 0 \\ 0 \end{array}\right]\,;\end{split}\\\begin{split}\sin{\phi}\,\cdot\, \left[\begin{array}{cc} -i & -1 \\ 1 & -i \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \ =\ \left[\begin{array}{c} 0 \\ 0 \end{array}\right],\quad \text{where }\sin{\phi}\neq 0\,;\end{split}\\\begin{split}\left[\begin{array}{cc} -i & -1 \\ 1 & -i \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \ =\ \left[\begin{array}{c} 0 \\ 0 \end{array}\right] \quad : \quad \begin{cases}\ \begin{array}{r} -i\,x_1 - x_2 = 0 \\ x_1 - i\,x_2 = 0 \end{array} \end{cases} : \quad x_1=i\,x_2\end{split}\\\begin{split}\blacktriangleright\quad \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right]\ =\ \left[\begin{array}{r} i\,\alpha \\ \alpha \end{array}\right]\ =\ \alpha\,\left[\begin{array}{r} i \\ 1 \end{array}\right], \quad\alpha\in C\setminus\{0\}.\end{split}\end{aligned}\end{align} \]

For an eigenvalue \(\ \lambda_2 =\, e^{\ -\,i\,\phi}\,=\,\cos{\phi}\,-\,i\,\sin{\phi}\ :\)

\[ \begin{align}\begin{aligned}\begin{split}\left[\begin{array}{cc} i\,\sin{\phi} & -\sin{\phi} \\ \sin{\phi} & i\,\sin{\phi} \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \ =\ \left[\begin{array}{c} 0 \\ 0 \end{array}\right]\,;\end{split}\\\begin{split}\sin{\phi}\,\cdot\, \left[\begin{array}{cc} i & -1 \\ 1 & i \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \ =\ \left[\begin{array}{c} 0 \\ 0 \end{array}\right],\quad \text{where }\sin{\phi}\neq 0\,;\end{split}\\\begin{split}\left[\begin{array}{cc} i & -1 \\ 1 & i \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] \ =\ \left[\begin{array}{c} 0 \\ 0 \end{array}\right] \quad : \quad \begin{cases}\ \begin{array}{r} i\,x_1 - x_2 = 0 \\ x_1 + i\,x_2 = 0 \end{array} \end{cases} : \quad x_2=i\,x_1\end{split}\\\begin{split}\blacktriangleright\quad \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right]\ =\ \left[\begin{array}{r} \beta \\ i\,\beta \end{array}\right]\ =\ \beta\,\left[\begin{array}{r} 1 \\ i \end{array}\right], \quad\beta\in C\setminus\{0\}.\end{split}\end{aligned}\end{align} \]

Note that the eigenvectors do not depend on the angle \(\,\phi.\ \) Geometric multiplicities of the eigenvalues \(\ \lambda_1,\,\lambda_2\ \) are equal to 1 and agree with the algebraic multiplicities.

Furthermore, any two vectors associated with distinct eigenvalues are orthogonal:

\[\begin{split}\left\langle\ \left[\begin{array}{c} i\,\alpha \\ \alpha \end{array}\right],\ \left[\begin{array}{c} \beta \\ i\,\beta \end{array}\right]\ \right\rangle\ =\ -i\,\alpha^*\beta\,+\,\alpha^*\,i\,\beta\ =\ 0.\end{split}\]

Hence, we may now construct an orthonormal basis \(\ \mathcal{F}\ \) of the space \(\,C^2\ \) consisting of normalized eigenvectors of the matrix \(\ \boldsymbol{R}_\phi\ \) associated with distinct eigenvalues:

\[\begin{split}\mathcal{F}\ =\ (\,\boldsymbol{f}_1,\,\boldsymbol{f}_2\,), \quad\text{where}\quad \boldsymbol{f}_1\ =\ \textstyle{\frac{1}{\sqrt{2}}}\, \left[\begin{array}{c} i \\ 1 \end{array}\right],\quad \boldsymbol{f}_2\ =\ \textstyle{\frac{1}{\sqrt{2}}}\, \left[\begin{array}{c} 1 \\ i \end{array}\right]\,.\end{split}\]

The transition matrix \(\ \boldsymbol{P}\ \) from the canonical basis \(\ \mathcal{E}\ \) to the basis \(\ \mathcal{F}\ \), which we form from the columns \(\ \boldsymbol{f}_1,\,\boldsymbol{f}_2:\)

\[\begin{split}\boldsymbol{P}\ =\ [\ \boldsymbol{f}_1\ |\ \boldsymbol{f}_2\ ]\ = \textstyle{\frac{1}{\sqrt{2}}}\,\left[\begin{array}{cc} i & 1 \\ 1 & i \end{array}\right]\end{split}\]

is unitary and diagonalizes the matrix \(\ \boldsymbol{R}_\phi:\)

\[\boldsymbol{P}^+\boldsymbol{P}\ =\ \boldsymbol{I}_2,\qquad \boldsymbol{P}^{-1}\boldsymbol{R}_\phi\,\boldsymbol{P}\ =\ \text{diag}(e^{\ +\,i\,\phi},\,e^{\ -\,i\,\phi})\,.\]

The matrix \(\ \boldsymbol{R}_\phi\ \) represents an operation of rotation of a vector on a plane by the angle \(\,\phi.\ \) This interpretation explains why, apart from the cases when \(\,\phi=0,\,\pi\,,\) the eigenvalues are imaginary and not real: a vector rotated by an angle \(\,\phi\ \) is not parallel to the initial vector. In this situation diagonalization of a real matrix \(\ \boldsymbol{R}_\phi\ \) requires a unitary non-real matrix \(\ \boldsymbol{P}.\)

Example 3.

The matrix \(\ \boldsymbol{A}\ =\ \left[\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right] \in M_3(R)\ \) is real and symmetric, and thus Hermitian.

We will verify that:

• its eigenvalues are real,

• eigenvectors associated with distinct eigenvalues are orthogonal,

• algebraic multiplicity of every eigenvalue is equal to its geometric multiplicity,

• eigenvectors of the matrix \(\ \boldsymbol{A}\ \) comprise an orthonormal basis of the space \(\,R^3,\)

• the matrix \(\ \boldsymbol{A}\ \) is diagonalizable by a real orthogonal similarity transformation.

An orthogonal matrix which diagonalizes the matrix \(\ \boldsymbol{A}\ \) is a transition matrix from the canonical basis \(\,\mathcal{E}\,\) of the space \(\,R^3\,\) to an orthonormal basis \(\,\mathcal{F}^0\,\) consisting of normalized eigenvectors of this matrix. Hence, we have to solve an eigenproblem of the matrix \(\ \boldsymbol{A}\ \) and then construct an orthonormal basis consisting of its eigenvectors.

The characteristic equation:

\[\begin{split}\det{(\boldsymbol{A}-\lambda\,\boldsymbol{I}_n)}\ = \ \left|\begin{array}{ccc} -\lambda & 1 & 1 \\ 1 & -\lambda & 1 \\ 1 & 1 & -\lambda \end{array}\ \,\right|\ = \ -\lambda^3+3\,\lambda + 2\ = -(\lambda-2)(\lambda+1)^2\ =\ 0\end{split}\]

provides the eigenvalues with their algebraic multiplicities:

\[\lambda_1=\,2 \quad (1)\,,\qquad \lambda_2=\,-1 \quad (2)\,.\]

In general, vectors \(\,\boldsymbol{x}\,\) associated with an eigenvalue \(\,\lambda\,\) may be determined from the equation

(2)\[(\boldsymbol{A}-\lambda\,\boldsymbol{I}_n)\ \boldsymbol{x}\ = \ \boldsymbol{0}.\]

For \(\ \lambda\,=\,\lambda_1=\,2\ \) the equation (2) is of the form

\[\begin{split}\left[\begin{array}{ccc} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right] \ =\ \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\,.\end{split}\]

The method rref() of Sage transforms matrix of this homogeneous linear problem to reduced row echelon form:

sage: A = matrix(QQ,[[-2, 1, 1],
                     [ 1,-2, 1],
                     [ 1, 1,-2]])
sage: A.rref()

[ 1  0 -1]
[ 0  1 -1]
[ 0  0  0]

We obtain an equivalent linear problem

\[\begin{split}\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right] \ =\ \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\end{split}\]

corresponding to a system of equations

\begin{alignat*}{7} x_1 & \ \ & & \ - \ \ & x_3 & \ = \ \ & 0 \\ & \ \ & x_2 & \ - \ \ & x_3 & \ = \ \ & 0 \end{alignat*}

whose general solution is \(\ x_1=x_2=x_3=\alpha, \ \ \alpha\in R.\ \) \(\\\) Hence, the eigenvectors associated with the eigenvalue \(\,\lambda_1=2\,\) are of the form

\[\begin{split}\boldsymbol{x}\ =\ \left[\begin{array}{c} \alpha \\ \alpha \\ \alpha \end{array}\right] \ =\ \alpha\ \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right], \quad\alpha\in R\setminus\{0\}.\end{split}\]

Geometric multiplicity of this eigenvalue is 1 and agrees with the algebraic multiplicity.

Substitution of \(\ \lambda\,=\,\lambda_2=\,-1\ \) to the equation (2) gives

\[\begin{split}\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right] \ =\ \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\,.\end{split}\]

The function right_kernel_matrix() of Sage returns a matrix whose rows comprise a basis for the solution space of a homogeneous linear problem. In our case we obtain

sage: A = matrix(QQ,[[1, 1, 1],
                     [1, 1, 1],
                     [1, 1, 1]])

sage: A.right_kernel_matrix()

[ 1  0 -1]
[ 0  1 -1]

Hence, the eigenvectors associated with the eigenvalue \(\ \lambda_2\,=-1\ \) are of the form

\[\begin{split}\boldsymbol{x}\ =\ \alpha\ \,\left[\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right]\ +\ \beta \ \,\left[\begin{array}{r} 0 \\ 1 \\ -1 \end{array}\right]\ =\ \left[\begin{array}{c} \alpha \\ \beta \\ -\alpha-\beta \end{array}\right]\,, \quad\alpha,\beta\in R,\ \ \alpha^2+\,\beta^2>0.\end{split}\]

Geometric multiplicity of this eigenvalue is 2 and agrees with the algebraic multiplicity.

Note that eigenvectors associated with distinct eigenvalues are orthogonal:

\[\begin{split}\left\langle\ \ \left[\begin{array}{c} \alpha \\ \alpha \\ \alpha \end{array}\right],\ \left[\begin{array}{c} \beta \\ \gamma \\ -\,\beta-\gamma \end{array}\right]\ \ \right\rangle\ \ =\ \ \alpha\beta\,+\,\alpha\gamma\,+\,\alpha\,(-\,\beta-\gamma)\ =\ 0\,.\end{split}\]

Denote three linearly independent eigenvectors of the matrix \(\ \boldsymbol{A}\ \) as:

\[\begin{split}\boldsymbol{g}_1\,=\ \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right],\quad \boldsymbol{g}_2\,=\ \left[\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right],\quad \boldsymbol{g}_3\,=\ \left[\begin{array}{r} 0 \\ 1 \\ -1 \end{array}\right],\end{split}\]

so that \(\,\mathcal{G}= (\boldsymbol{g}_1,\boldsymbol{g}_2,\boldsymbol{g}_3)\ \) is a basis of the space \(\,R^3.\ \)

Matrix \(\ \boldsymbol{P}\,=\, [\ \boldsymbol{g}_1\,|\,\boldsymbol{g}_2\,|\,\boldsymbol{g}_3\ ]\ =\ \left[\begin{array}{rrr} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & -1 & -1 \end{array}\right]\ \) diagonalizes the matrix \(\ \boldsymbol{A}:\)

sage: A = matrix(QQ,[[0, 1, 1],
                     [1, 0, 1],
                     [1, 1, 0]])

sage: P = matrix(QQ,[[1, 1, 0],
                     [1, 0, 1],
                     [1,-1,-1]])
sage: P.I*A*P

[ 2  0  0]
[ 0 -1  0]
[ 0  0 -1]

The matrix \(\ \boldsymbol{P}\ \) is not orthogonal because the inner product \(\ \langle\boldsymbol{g}_2,\boldsymbol{g}_3\rangle = 1 \neq 0.\) To construct an orthogonal basis \(\ \mathcal{F}=(\boldsymbol{f}_1,\boldsymbol{f}_2,\boldsymbol{f}_3)\), and later an orthonormal basis \(\ \mathcal{F}^0= (\boldsymbol{f}_1^0,\boldsymbol{f}_2^0,\boldsymbol{f}_3^0)\), we have to apply the Gram-Schmidt ortogonalization process. Put

\[\begin{split}\boldsymbol{f}_1=\,\boldsymbol{g}_1=\, \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right],\quad \boldsymbol{f}_2=\,\boldsymbol{g}_2=\, \left[\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right],\quad \boldsymbol{f}_3=\,\boldsymbol{g}_3+\eta\,\boldsymbol{f}_2=\, \left[\begin{array}{c} \eta \\ 1 \\ -1-\eta \end{array}\right],\end{split}\]

where \(\,\eta\,\) may be determined from the orthogonality condition \(\ \langle\boldsymbol{f}_2,\boldsymbol{f}_3\rangle\,=\,2\eta+1\,=\,0,\ \) so that

\[\begin{split}\eta = - \textstyle\frac{1}{2}\,,\qquad \boldsymbol{f}_3\ =\ \left[\begin{array}{r} -\frac{1}{2} \\ 1 \\ -\frac{1}{2} \end{array}\right] \ =\ -\frac{1}{2}\ \left[\begin{array}{r} 1 \\ -2 \\ 1 \end{array}\right] \ \ \sim\ \ \left[\begin{array}{r} 1 \\ -2 \\ 1 \end{array}\right]\,.\end{split}\]

Hence, the orthogonal basis \(\ \mathcal{F}\ \) consists of eigenvectors

\[\begin{split}\boldsymbol{f}_1\,=\ \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right],\quad \boldsymbol{f}_2\,=\ \left[\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right],\quad \boldsymbol{f}_3\,=\ \left[\begin{array}{r} 1 \\ -2 \\ 1 \end{array}\right]\,,\end{split}\]

and the orthonormal basis \(\ \mathcal{F}^0\ \) may be obtained by dividing each vector by its norm:

\[\begin{split}\boldsymbol{f}_1^0\,=\ \textstyle{\frac{1}{\sqrt{3}}} \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right],\quad \boldsymbol{f}_2\,=\ \textstyle{\frac{1}{\sqrt{2}}} \left[\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right],\quad \boldsymbol{f}_3\,=\ \textstyle{\frac{1}{\sqrt{6}}} \left[\begin{array}{r} 1 \\ -2 \\ 1 \end{array}\right]\,.\end{split}\]

The column vectors \(\ \boldsymbol{f}_1^0,\,\boldsymbol{f}_2^0,\boldsymbol{f}_3^0\ \) comprise an orthogonal transition matrix \(\ \boldsymbol{P}^0\ \) from the canonical basis \(\,\mathcal{E}\,\) to the basis \(\ \mathcal{F}^0\ \) of the space \(\,R^3.\ \) This is the matrix which diagonalizes the matrix \(\,\boldsymbol{A}\,\) by an orthogonal similarity transformation:

\[\begin{split}\boldsymbol{P}^0\ =\ \,\textstyle \left[\begin{array}{rrr} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \\ \frac{1}{\sqrt{3}} & 0 & -\,\frac{2}{\sqrt{6}} \\ \\ \frac{1}{\sqrt{3}} & -\,\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \end{array}\right]\,.\end{split}\]

Numerical verification:

sage: P0 = matrix(RR,[[1/sqrt(3), 1/sqrt(2), 1/sqrt(6)],
                      [1/sqrt(3), 0,        -2/sqrt(6)],
                      [1/sqrt(3),-1/sqrt(2), 1/sqrt(6)]])

sage: P0.T*P0.n(digits=4)

[1.000  0.0000 0.0000]
[0.0000 1.000  0.0000]
[0.0000 0.0000 1.000 ]

sage: A = matrix(RR,[[0, 1, 1],
                     [1, 0, 1],
                     [1, 1, 0]])

sage: P0 = matrix(RR,[[1/sqrt(3), 1/sqrt(2), 1/sqrt(6)],
                      [1/sqrt(3), 0,        -2/sqrt(6)],
                      [1/sqrt(3),-1/sqrt(2), 1/sqrt(6)]])

sage: P0.I*A*P0.n(digits=4)

[2.000   0.0000  0.0000]
[0.0000 -1.000   0.0000]
[0.0000  0.0000 -1.000 ]