We show how to define a function of a square matrix using a
diagonalization procedure. This method is applicable only for
diagonalizable square matrices, and
is not suitable for defective matrices. Recall that a matrix A is called diagonalizable if there exists a nonsingular matrix S such that
\( {\bf S}^{-1} {\bf A} {\bf S} = {\bf \Lambda} , \) a diagonal matrix. In other words, the matrix A is similar to a diagonal matrix.
An \( n \times n \) square matrix is diagonalizable if and only if there exist n linearly independent eigenvectors, so geometrical
multiplicity of each eigenvalue is the same as its algebraic multiplicity. Then the matrix S can be built from eigenvectors of A ,
column by column.
Let A be a square \( n \times n \) diagonalizable matrix, and let \( {\bf \Lambda} \) be the corresponding diagonal matrix of its eigenvalues:
\[
{\bf \Lambda} = \begin{bmatrix} \lambda_1 & 0 & 0 & \cdots & 0 \\ 0&\lambda_2 & 0& \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0&0&0& \cdots & \lambda_n \end{bmatrix} ,
\]
where \( \lambda_1 , \lambda_2 , \ldots , \lambda_n \) are eigenvalues (that may be equal) of the matrix A .
Let
\( {\bf x}_1 , {\bf x}_2 , \ldots , {\bf x}_n \) be linearly independent eigenvectors, corresponding to the eigenvalues
\( \lambda_1 , \lambda_2 , \ldots , \lambda_n .\) We build the nonsingular matrix
S from these eigenvectors (every column is an eigenvector):
\[
{\bf S} = \begin{bmatrix} {\bf x}_1 & {\bf x}_2 & {\bf x}_3 & \cdots & {\bf x}_n \end{bmatrix} .
\]
For any reasonable (we do not specify this word, it is sufficient to be smooth) function defined on the spectrum (set of all eigenvalues) of the diagonalizable matrix
A , we define the function of this matrix by the formula:
\[
f \left( {\bf A} \right) = {\bf S} f\left( {\bf \Lambda} \right) {\bf S}^{-1} , \qquad \mbox{where } \quad f\left( {\bf \Lambda} \right) =
\begin{bmatrix} f(\lambda_1 ) & 0 & 0 & \cdots & 0 \\ 0 & f(\lambda_2 ) & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0&0&0& \cdots & f(\lambda_n ) \end{bmatrix} .
\]
Example: All eigenvalues are distinct.
Example: Consider the \( 3 \times 3 \) matrix \( {\bf A} = \begin{bmatrix} 1&4&16 \\ 18&20&4 \\ -12&-14&-7 \end{bmatrix} \) that has three distinct eigenvalues
A = {{1,4,16},{18,20,4},{-12,-14,-7}}
Eigenvalues[A]
Out[2]= 9, 4, 1
Eigenvectors[A]
Out[3]= {{1, -2, 1}, {4, -5, 2}, {4, -4, 1}}
Using eigenvectors, we build the transition matrix
S of its
eigenvectors:
\[
{\bf S} = \begin{bmatrix} 1&4&4 \\ -2&-5&-4 \\ 1&2&1 \end{bmatrix} , \quad\mbox{with} \quad {\bf S}^{-1} = \begin{bmatrix} -3&-4&-4 \\ 2&3&4 \\ -1&-2&-3 \end{bmatrix} .
\]
Then we are ready to construct eight (it is 23 roots
because each square root of an eigenvalue has two values; for
instance, \( \sqrt{9} = \pm 3 \) ) square roots of this positive definite matrix:
\[
\sqrt{\bf A} = {\bf S} \sqrt{\Lambda} {\bf S}^{-1} = \begin{bmatrix} 1&4&4 \\ -2&-5&-4 \\ 1&2&1 \end{bmatrix} \begin{bmatrix} \pm 3&0&0 \\ 0&\pm 2&0 \\ 0&0&\pm 1 \end{bmatrix} \begin{bmatrix} -3&-4&-4 \\ 2&3&4 \\ -1&-2&-3 \end{bmatrix} ,
\]
with appropriate choice of roots on the diagonal. In particular,
\[
\sqrt{\bf A} = \begin{bmatrix} 3&4&8 \\ 2&2&-4 \\ -2&-2&1 \end{bmatrix} , \quad
\begin{bmatrix} 21&28&32 \\ -34&-46&-52 \\ 16&22&25 \end{bmatrix} , \quad
\begin{bmatrix} -11&-20&-32 \\ 6&14&28 \\ 0&-2&-7 \end{bmatrix} , \quad
\begin{bmatrix} 29&44&56 \\ -42&-62&-76 \\ 18&26&31 \end{bmatrix} .
\]
We check with
Mathematica for specific roots of eigenvalues:
3, 2, and 1. However, we can take any combination of these roots using
\( \pm 3, \pm 2, \pm 1 \) next time.
S = Transpose[Eigenvectors[A]]
Out[7]= {{3, 4, 8}, {2, 2, -4}, {-2, -2, 1}}
Example: Some eigenvalues are the same.
Example: \( 3 \times 3 \) matrix
\( {\bf A} = \begin{bmatrix} -20&-42&-21 \\ 6&13&6 \\ 12&24&13 \end{bmatrix} \) that has two distinct eigenvalues
A:=matrix([[-20,-42,-21],[6,13,6],[12,24,13]])
\( \left(\begin{array}{ccc} -20 & -42 & -21\\ 6 & 13 & 6\\ 12 & 24 & 13 \end{array}\right)\)
eigofA:=linalg::eigenvectors(A)
\(
\left[\left[4,1,\left[\left(\begin{array}{c} -\frac{7}{4}\\
\frac{1}{2}\\ 1
\end{array}\right)\right]\right],\left[1,2,\left[\left(\begin{array}{c}
-2\\ 1\\ 0 \end{array}\right),\left(\begin{array}{c} -1\\ 0\\ 1
\end{array}\right)\right]\right]\right] \)
Eigenvectors:= eigofA[1][3][1], eigofA[2][3][1], eigofA[2][3][2]
\(
\left(\begin{array}{c} -\frac{7}{4}\\ \frac{1}{2}\\ 1
\end{array}\right),\left(\begin{array}{c} -2\\ 1\\ 0
\end{array}\right),\left(\begin{array}{c} -1\\ 0\\ 1
\end{array}\right) \)
Since the double eigenvalue
\( \lambda =1
\) has two linearly independent eigenvectors, the given matrix is
diagonalizable, and we are able to build the transition matrix of its
eigenvectors:
\[
{\bf S} = \begin{bmatrix} -7&-1&-2 \\ 2&0&1 \\ 4&1&0 \end{bmatrix} , \quad\mbox{with} \quad {\bf S}^{-1} = \begin{bmatrix} 1&2&1 \\ -4&-8&-3 \\ -2&-3&-2 \end{bmatrix} .
\]
We construct the transition matrix of the eigenvectors corresponding to
eigenvalues:
S:=Eigenvectors[1].Eigenvectors[2].Eigenvectors[3]
\(
\left(\begin{array}{ccc} -\frac{7}{4} & -2 & -1\\ \frac{1}{2} & 1 &
0\\ 1 & 0 & 1 \end{array}\right) \)
We buld the diagonal matrix of eigenvalues:
eigenvalues:=matrix([[4,0,0],[0,1,0],[0,0,1]])
\(
\left(\begin{array}{ccc} 4 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1
\end{array}\right) \)
Check to verify that
A*S = S*eigenvalues :
A*S
\(
\left(\begin{array}{ccc} -7 & -2 & -1\\ 2 & 1 & 0\\ 4 & 0 & 1
\end{array}\right) \)
S*eigenvalues
\(
\left(\begin{array}{ccc} -7 & -2 & -1\\ 2 & 1 & 0\\ 4 & 0 & 1
\end{array}\right) \)
Therefore, the given matrix
A is diagonalizable.
For three functions,
\( f(\lambda ) = e^{\lambda \,t} , \quad \Phi (\lambda ) = \frac{\sin \left( \sqrt{\lambda} \,t \right)}{\sqrt{\lambda}} ,
\quad \Psi (\lambda ) = \cos \left( \sqrt{\lambda} \,t \right) \) we construct the corresponding matrix-functions:
\begin{align*}
f({\bf A}) &= {\bf S} e^{{\bf \Lambda}\,t} {\bf S}^{-1} = e^{2t} \begin{bmatrix} -7 & -14 & -7 \\ 2&4&2 \\ 4&8&4 \end{bmatrix} + e^t
\begin{bmatrix} 8&14&7 \\ -2&-3&-2 \\ -4&-8&-3 \end{bmatrix} ,
\\
{\bf \Phi} ({\bf A}) &= {\bf S} \frac{\sin \left( \sqrt{\bf \Lambda} \,t \right)}{\sqrt{\bf \Lambda}} {\bf S}^{-1} =
\sin 2t \begin{bmatrix} -7/2 & -7 & -7/2 \\ 1&2&1 \\ 2&4&2 \end{bmatrix} + \sin t \begin{bmatrix} 8&14&7 \\ -2&-3&-2 \\ -4&-8&-3 \end{bmatrix} ,
\\
{\bf \Psi} ({\bf A}) &= {\bf S} \cos \left( {\bf \Lambda}\,t \right) {\bf S}^{-1} = \cos 2t \begin{bmatrix} -7 & -14 & -7 \\ 2&4&2 \\ 4&8&4 \end{bmatrix} +
\cos t \begin{bmatrix} 8&14&7 \\ -2&-3&-2 \\ -4&-8&-3 \end{bmatrix} .
\end{align*}
These matrix functions are unique solutions of the following initial
value problems:
\[
\frac{\text d}{{\text d}t}\,e^{{\bf A}\,t} = {\bf A}\,e^{{\bf A}\,t} ,
\qquad \lim_{t\to 0} \,e^{{\bf A}\,t} = {\bf I} , \quad \mbox{where }
{\bf I} \mbox{ is the identity matrix};
\]
\[
\frac{{\text d}^2}{{\text d}t^2}\,{\bf \Phi} ({\bf A}) + {\bf A}\,{\bf
\Phi} ({\bf A}) = {\bf 0} ,
\qquad \lim_{t\to 0} \,{\bf \Phi} ({\bf A}) = {\bf 0} , \quad
\quad \lim_{t\to 0} \,\dot{\bf \Phi} ({\bf A}) = {\bf I} , \quad \mbox{where }
{\bf I} \mbox{ is the identity matrix};
\]
\[
\frac{{\text d}^2}{{\text d}t^2}\,{\bf \Psi} ({\bf A}) + {\bf A}\,{\bf
\Psi} ({\bf A}) = {\bf 0} ,
\qquad \lim_{t\to 0} \,{\bf \Psi} ({\bf A}) = {\bf I} , \quad
\quad \lim_{t\to 0} \,\dot{\bf \Psi} ({\bf A}) = {\bf 0} .
\]
Example: Matrix with complex eigenvalues.
Example: Consider the
\( 3 \times 3 \) matrix
\( {\bf A} = \begin{bmatrix} 1 &2&3 \\ 2 &3&4 \\ 2&-6&-4 \end{bmatrix} \)
that has two complex conjugate eigenvalues
\( \lambda = 1 \pm 2{\bf j} \) and one real eigenvalue
\( \lambda = -2 .\) Mathematica confirms:
A = {{1, 2, 3}, {2, 3, 4}, {2, -6, -4}}
Eigenvalues[A]
Out[2]= {1 + 2 I, 1 - 2 I, -2}
Eigenvectors[A]
Out[3]= {{-1 - I, -2 - I, 2}, {-1 + I, -2 + I, 2}, {-7, -6, 11}}
We build the transition matrix of its eigenvectors:
\[
{\bf S} = \begin{bmatrix} -1-{\bf j} & -1+{\bf j} &-7 \\ -2-{\bf j} & -2+{\bf j} &-6 \\ 2&2&1 \end{bmatrix} , \quad \mbox{with} \quad
{\bf S}^{-1} = \frac{1}{6} \begin{bmatrix} 1 - 10{\bf j} & -1 + 13{\bf j} & 1 + 8{\bf j} \\
1 +10 {\bf j} & -1 -13{\bf j} & 1 -8{\bf j} \\
-4 & 4 & 2 \end{bmatrix} .
\]
Now we are ready to define a function of the given square matrix. For example, if
\( f(\lambda ) = e^{\lambda \, t} , \) we obtain the corresponding exponential matrix:
\begin{align*}
e^{{\bf A}\,t} &= {\bf S} \begin{bmatrix} e^{(1+2{\bf j})\,t} & 0&0 \\
0& e^{(1-2{\bf j})\,t} & 0 \\ 0&0&e^{-2t} \end{bmatrix} {\bf S}^{-1}
\\
&= \begin{bmatrix} -1-{\bf j} & -1+{\bf j} &-7 \\ -2-{\bf j} & -2+{\bf j} &-6 \\ 2&2&1 \end{bmatrix} \,
\begin{bmatrix} e^{t} \left( \cos 2t + {\bf j}\,\sin 2t \right) & 0&0 \\
0& e^{t} \left( \cos 2t - {\bf j}\,\sin 2t \right) & 0 \\ 0&0&e^{-2t} \end{bmatrix} \,
\frac{1}{6} \begin{bmatrix} 1 - 10{\bf j} & -1 + 13{\bf j} & 1 + 8{\bf j} \\
1 +10 {\bf j} & -1 -13{\bf j} & 1 -8{\bf j} \\
-4 & 4 & 2 \end{bmatrix}
\\
&= \frac{1}{3} \, e^{-2t} \begin{bmatrix} 14 & -14& -7 \\ 12&-12& -6 \\ -2&2&1 \end{bmatrix} +
\frac{1}{3} \, e^{t} \,\cos 2t \begin{bmatrix} -11&14&7 \\ -12&15&6 \\ 2&-2&2 \end{bmatrix} +
\frac{1}{3} \, e^{t} \,\sin 2t \begin{bmatrix} -9&12&9 \\ -19&25&17 \\ 20&-26&-16 \end{bmatrix} .
\end{align*}
Here we use Euler's formula:
\( e^{a+b{\bf j}} =
e^a \left( \cos b + {\bf j} \sin b \right) . \)
Mathematica confirms
S = {{-1-I, -1+I, -7}, {-2-I, -2+I, -6}, {2, 2, 1}}
The matrix function
\( e^{{\bf A}\,t} \) is the unique solution of the following
matrix initial value problem:
\[
\frac{\text d}{{\text d}t}\,e^{{\bf A}\,t} = {\bf A}\,e^{{\bf A}\,t} , \qquad \lim_{t\to 0} \,e^{{\bf A}\,t} = {\bf I} ,
\]
where
I is the identity matrix.