# Notation

• $$\lambda_{1,1}, \lambda_{2,2}, \ldots, \lambda_{n,n}$$ are all the eigenvalues of $$A_n$$.
• $$x_1, x_2, \ldots, x_n$$ are all the eigen vectors of $$A_n$$.
• $$\Lambda_n = \text{diag}\left(\lambda_{1,1}, \lambda_{2,2}, \ldots, \lambda_{n,n}\right)$$.
• $$S = \left[\begin{array}{cccc} | & | & & | \\ x_1 & x_2 & \cdots & x_n \\ | & | & & | \end{array}\right]$$.

# Question

q1: Is the solution to $$\frac{du}{dt} = Au$$, given $$u(0)$$, correctly described by...

$$u(t) = \sum_{k = 1}^{n} \left\{\mathrm{e}^{\lambda_{k,k}t}c_kx_k \right\} \style{font-family:inherit}{\text{ s.t. }} Sc = u(0)\tag{\style{font-family:inherit}{\text{e1}}}\label{eq1}$$

# Example

Given that...

• $$A_5 = \text{ones}\left(5\right)$$.
• $$u(0) = \left[\begin{array}{ccccc} 0 & 1 & 1 & 1 & 2 \end{array}\right]^T$$.

We determine that...

• $$\Lambda = \left[ \begin{array}{ccccc} 5 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$$.
• $$S = \left[ \begin{array}{ccccc} 1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \end{array}\right]$$.
• $$Sc = u(0) \Longrightarrow c = \left[\begin{array}{ccccc} 1 & 1 & 0 & 0 & 0 \end{array}\right]^T$$.

Using $$\ref{eq1}$$...

$$u(t) = \sum_{k = 1}^{5} \left\{\mathrm{e}^{\lambda_{k,k}t}c_kx_k \right\}\\ \Longrightarrow u(t) = \mathrm{e}^{(5)t}(1)x_1 + \mathrm{e}^{(0)t}(1)x_2 + \mathrm{e}^{(0)t}(0)x_3 + \mathrm{e}^{(0)t}(0)x_4 + \mathrm{e}^{(0)t}(0)x_5\\ \Longrightarrow u(t) = \mathrm{e}^{5t}\left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{array}\right] + \left[\begin{array}{c} -1 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}\right]$$

q2: If your answer to q1 was yes, is there a way to express $$\ref{eq1}$$ more compactly?

$$u(t) = \frac{S\text{e}^{\Lambda t}\text{adj}\left(S\right)u(0)}{|S|}\tag{\style{font-family:inherit}{\text{e2}}}\label{eq2}$$

I verified that e1 & e2 yield the same answer to my example problem (but they might not work for others?), using Mathematica...

adj[m_] :=
Map[Reverse, Minors[Transpose[m], Length[m] - 1], {0, 1}] *
Table[(-1)^(i + j), {i, Length[m]}, {j, Length[m]}];

A = ConstantArray[1,{5,5}];
\[CapitalLambda] = DiagonalMatrix[Eigenvalues[A]];
S = Transpose[Eigenvectors[A]];
u0 = {0,1,1,1,2};
c = LinearSolve[S,u0];

(* Answer w/ method 1 *)
a1 = Sum[Exp[Eigenvalues[A][[k]]]*c[[k]]*Eigenvectors[A][[k]],{k,1,Length[S]}];

(* Answer w/ method 2 *)

Print["A = ", A//MatrixForm];
Print["\[CapitalLambda] = ", \[CapitalLambda]//MatrixForm];
Print["S = ", S//MatrixForm];
Print["u(0) = ", u0//MatrixForm];
Print["c = ", c//MatrixForm];
Print["a1: u(t) = ", a1//MatrixForm];
Print["a2: u(t) = ", a2//MatrixForm];


Yes. The exponential of a square matrix $$A$$ (or, more generally, of a bounded linear operator $$A$$ on a normed vector space) is defined by the Taylor series: $$\exp(A) = \sum_{k \geq 0} {1 \over k!} A^{k}.$$ Notice that this is also a square matrix (of the same dimension as $$A$$). How would we express its eigenvalues in terms of those of $$A$$?

The solution to the IVP $$\dot{u} = Au, \quad u(0) = u_0$$ is: $$u(t) = \exp(t A) \, u_{0},$$ which can be verified directly. All that remains is to express $$u_0$$ as a linear combination of the eigenvectors of $$\exp(tA)$$, and to figure out how the eigenvalues of $$\exp(tA)$$ are related to those of $$A$$.

• Would you mind showing that your expression ($u(t) = \exp(t A) \, u_{0}$) resolves the same answer as mine, i.e., $u(t) = \mathrm{e}^{5t}\left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{array}\right] + \left[\begin{array}{c} -1 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}\right]$ Commented Apr 8, 2019 at 16:30
• To be honest, I am not following why, if $A_5$ is the 5-by-5 identity matrix (did I understand that correctly?), then its diagonalization should be what you have written as $\Lambda$, as opposed to $A_5$ itself (which is already diagonal).
– avs
Commented Apr 8, 2019 at 16:42
• $A_5$ is a 5x5 matrix containing all ones, but I am looking for an expression that would be valid for any square matrix A Commented Apr 8, 2019 at 17:25
• I see. Well, if $\lambda$ is an eigenvalue of $A$ with eigenvector $x$, then $e^{t \lambda}$ is an eigenvalue of $\exp(tA)$ with the same eigenvector. (This can be proved using the power series I used to define $\exp(A)$.) Thus, if $u_{0} = \sum_{k} c_{k} x_{k}$, where the $x_{k}$'s are the eigenvectors with the corresponding eigenvalues of $A$, then, indeed, $$\exp(t A) u_{0} = e^{t A} u_{0} = \sum_{k} c_{k} e^{t A} x_{k} = \sum_{k} c_{k} e^{t \lambda_{k}} x_{k}.$$
– avs
Commented Apr 8, 2019 at 17:37
• I wrote my summation to be 1 to n where n is the size of the matrix A; will there always be n (not necessarily distinct) eigenvalue/eigenvector pairs? Commented Apr 8, 2019 at 18:21