eigenvalues-eigenvectorslinear algebraordinary differential equations
So I came across this problem while studying and I'm a little confused.
$$
y'(t) = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}y(t)
$$
Usually I solve for the eigenvalues and the corresponding eigenvectors and then basically plug into:
$$
y = c_{1} e^{\lambda_{1} t} \begin{bmatrix} e1 \\ e2 \end{bmatrix}+ c_{2} e^{\lambda_{2} t} \begin{bmatrix} e3 \\ e4 \end{bmatrix}
$$
But in this case it isn't diagonalizable and I only get one eigenvector. How do you solve this?
I agree with all the comments. Namely,
In spite of points 2 and 3, I'd still call this a bug. Alpha is intended to guess the users intent. While clearly very hard, I don't think that interpreting numbers like $0.55$ as $55/100$ is too far out there. Even failing that, a small perturbation of the elements of the matrix don't change the fact that the matrix is diagonalizable.
Fortunately, there is an easy work around. Just enter:
diagonalize rationalize {{0.45,0.4},{0.55,0.6}}
It doesn't make much sense to compute the eigenvalues without their multiplicity and then say "since the eigenvalues are real, the matrix is diagonalizable" because it doesn't hold in general. It does hold for symmetric matrices but symmetric matrices have real eigenvalues and are diagonalizable so there is no need to calculate anything to deduce your matrix is diagonalizable.
You can say that $A$ is diagonalizable because it is real and symmetric without calculating anything. Alternatively, you can compute the eigenvalues (they will be real) and then compute the geometric multiplicity of each eigenvalue and then conclude $A$ is diagonalizable. It seems that whoever wrote the solution mixed both approaches.
Best Answer
For a $2\times 2$ matrix with repeated eigenvalue, and only one linearly independent eigenvector, one can obtain a second solution as $\vec y_2(t)=te^{\lambda t}\vec v_1+e^{\lambda t}\vec v_2$, where $\vec v_1$ is the eigenvector for the matrix, and $\vec v_2$ is a "generalized eigenvector", which satisfies $(A-\lambda I)\vec v_2=\vec v_1$, or equivalently satisfies $(A-\lambda I)^2\vec v_2=0$. So the general solution becomes $$\vec y(t)=c_1e^{\lambda t}\vec v_1+c_2(te^{\lambda t}\vec v_1+e^{\lambda t}\vec v_2)$$ If you are familiar with the matrix exponential, then you could also solve this problem using $e^{At}$ as a fundamental matrix. Notice here that $A$ can be written as the sum of the identity matrix, and a nilpotent matrix (i.e. $N^k=0$ for some $k\in\Bbb N$), so $A=I+N$, and $e^{At}=e^{It}e^{Nt}$.