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My professor wrote the solution to a system as

$$X = C_1 \begin{bmatrix}1 \\2 \end{bmatrix} e^{\lambda_1t} + C_2 \begin{bmatrix}3 \\4 \end{bmatrix} e^{\lambda_2t}$$

Where the column vectors are the basis to the eigenspace of the coefficient matrix and $\lambda_1$ and $\lambda_2$ are eigenvalues.

My question is why is it important to have the eigenvectors expressed in the solution? I thought all that we cared about was the eigenfunctions because those are what form our solution space?

I would've left the answer as

$$X=\mathbf{C_1}e^{\lambda_1t}+\mathbf{C_2}e^{\lambda_2t}$$

Since isn't all that we're doing anyways is finding the constant to multiply the basis vectors by? Is my way correct as well?

btw if it wasn't clear before

$$\mathbf{C_1} = \begin{bmatrix}C_{1 1} \\C_{1 2} \end{bmatrix}, \: \mathbf{C_2} = \begin{bmatrix}C_{2 1} \\C_{2 2} \end{bmatrix}$$

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  • $\begingroup$ Sanity check: do both expressions live in the same dimension? $\endgroup$ – GPerez May 6 '15 at 0:15
  • $\begingroup$ yea they do? $C_1$ and $C_2$ are bolded to indicate that they are vectors $\endgroup$ – bthmas May 6 '15 at 0:17
  • $\begingroup$ $\mathbf{C_1} = \begin{bmatrix}1C_1 \\2C_1 \end{bmatrix} = \begin{bmatrix}C_{1 1} \\C_{1 2} \end{bmatrix}$ $\endgroup$ – bthmas May 6 '15 at 0:22
  • $\begingroup$ the only straight line trajectories are the ones where eigenvectors are multiplied by the exponential of their eigenvalues. $\endgroup$ – abel May 6 '15 at 0:52
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    $\begingroup$ It depends, if $\mathbf{C_1},\mathbf{C_2}$ are allowed to be arbitrary vectors, then no, you are introducing more functions than those actually in the space. If they are defined as arbitrary multiples of the eigenvectors, then yes it's the same. $\endgroup$ – GPerez May 6 '15 at 20:14

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