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Let $A= \begin{bmatrix} 1&3\\ 2&2 \end{bmatrix}$, compute $\exp(At)$.

I need help help on computing a matrix exponential. solution according to wolfamalpha:

The process that I settled on was.

1.get the eigenvalues 2. get the eigenvector 3. Diagonalization of A 4.get exponential matrix with $D = C^{-1}AC$

Eigenvalues

$\det{(\lambda I-A)} = \begin{bmatrix} \lambda-1&-3\\ -2&\lambda-2 \end{bmatrix} \lambda_1 = 4$,$\lambda_2 = -1$

Eigenvectors I reworked it and $v_2 = (-3,2)$

$Av_1 = \lambda_1v_1 \Rightarrow (A-\lambda_1)v_1 = 0$

$\Rightarrow\begin{bmatrix} \lambda_1-1&-3\\ -2&\lambda_1-2 \end{bmatrix}\cdot v_1 = 0 \Rightarrow \begin{bmatrix} 3&-3\\ -2&2 \end{bmatrix}\begin{bmatrix} v_{1,1}\\ v_{1,2} \end{bmatrix}=0$

$3v_{1,1} - 3v_{1,2} = 0 \Rightarrow v_{1,1} = v_{1,1}$

$v_1 = \begin{bmatrix} 1\\ 1 \end{bmatrix}$

Diagonalized Edited after the comment

$C$ is having eigenvectors as its columns.

$C =\begin{bmatrix} 1&-3\\ 1&2 \end{bmatrix}$ To diagonalize $D = C^{-1}AC = C^{-1}(AC)$

$D= \begin{bmatrix} 4&0\\ 0&-1 \end{bmatrix}$

Compute Exp(At)

$\frac{1}{5}\begin{bmatrix} 2&3\\ -1&1 \end{bmatrix} \bigg(\begin{bmatrix} e^{4t}&0\\ 0&e^{-t} \end{bmatrix}\begin{bmatrix} 1&-3\\ 1&2 \end{bmatrix}\bigg)=$

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    $D$ should be a diagonal matrix. – 2017-02-10
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    Also, your $v_2$ seems incorrect. You should check your calculation again. – 2017-02-10
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    In the title, you ask about computing $\exp(At)$; at the beginning of the question, you’re asking about $\exp(A^T)$ instead; and in your work you do neither, computing $\exp(A)$. Which is it? – 2017-02-10
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    the first $exp(At)$ – 2017-02-10
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    Well, that’s not at all what you ended up with. What happened to $t$? – 2017-02-10

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