Prove $x_2$ is a sol of $x'=Ax$ given $x_1$ is a sol.

Question

Suppose $2 \times 1$ vector $x_1(t) = [u(t), v(t)]^T$ is a solution of $x' = Ax$ where $A$ is a $2 \times 2$ matrix of real numbers.

Prove that $2 \times 1$ vector $x_2(t) = [ku(k^2t), kv(k^2t)]^T$ is also a solution of $x' = Ax$.

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From #5a here.

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Apparently, solutions of $x' = Ax$ look like $x = e^{At}x(0)$ $(*)$

So given

$$\begin{bmatrix} u(t)\\ v(t) \end{bmatrix} = e^{At}\begin{bmatrix} u(0)\\ v(0) \end{bmatrix} \tag{**}$$

we must prove that

$$\begin{bmatrix} ku(k^2t)\\ kv(k^2t) \end{bmatrix} = e^{At}\begin{bmatrix} ku(0)\\ kv(0) \end{bmatrix}$$

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What I tried:

$$RHS = e^{At}\begin{bmatrix} ku(0)\\ kv(0) \end{bmatrix}$$

$$ = ke^{At}\begin{bmatrix} u(0)\\ v(0) \end{bmatrix}$$

$$= ke^{At}e^{-At}\begin{bmatrix} u(t)\\ v(t) \end{bmatrix}$$

$$= k\begin{bmatrix} u(t)\\ v(t) \end{bmatrix}$$

$$LHS = \begin{bmatrix} ku(k^2t)\\ kv(k^2t) \end{bmatrix}$$

$$= k \begin{bmatrix} u(k^2t)\\ v(k^2t) \end{bmatrix}$$

$$= k e^{A(k^2t)} \begin{bmatrix} u(0)\\ v(0) \end{bmatrix} $$

$$= k (e^{At})^{k^2} \begin{bmatrix} u(0)\\ v(0) \end{bmatrix} $$

$$= k (e^{At})^{k^2-1} e^{At}\begin{bmatrix} u(0)\\ v(0) \end{bmatrix} $$

$$= k (e^{At})^{k^2-1} \begin{bmatrix} u(t)\\ v(t) \end{bmatrix}$$

$$= (e^{At})^{k^2-1} RHS$$

I'm stuck. Where did I go wrong? How can I approach this?

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** Is there a way to prove this without using $(*)$? **

It seems we are given that

$$\begin{bmatrix} u'(t)\\ v'(t) \end{bmatrix} = A\begin{bmatrix} u(t)\\ v(t) \end{bmatrix}$$

and must prove that

$$\begin{bmatrix} k^3u'(k^2t)\\ k^3v'(k^2t) \end{bmatrix} = A\begin{bmatrix} ku(k^2t)\\ kv(k^2t) \end{bmatrix}$$

If I write $A=\begin{bmatrix} a_{11} \ a_{12}\\ a_{21} \ a_{22} \end{bmatrix}$ and multiply out the matrices, should I be able to prove the proposition? I tried that, but it doesn't look like I can prove it:

We are given that

$$\begin{bmatrix} u'(t)\\ v'(t) \end{bmatrix} = \begin{bmatrix} a_{11}u(t) + a_{12}v(t)\\ a_{21}u(t) + a_{22}v(t) \end{bmatrix}$$

and must prove that

$$\begin{bmatrix} k^3u'(k^2t)\\ k^3v'(k^2t) \end{bmatrix} = \begin{bmatrix} a_{11}ku(k^2t) + a_{12}kv(k^2t)\\ a_{21}ku(k^2t) + a_{22}kv(k^2t) \end{bmatrix}$$

Answer 1

The question is "for some" not "for all" k.

For the first proof, we can pick $k=0,1,-1$ since they satisfy $k(e^{At})^{k^2 - 1} = kI$.

For the second proof, we can pick $k=0,1,-1$ since they satisfy $k^3 = k$.