The book Introduction to Linear Algebra by Strang has an exercise that asks for the solution to $\frac{d\mathbf{u}}{dt}=-P\mathbf{u}$ (notice the minus sign, it says), where $P$ is the projection matrix onto the $45^\circ$ ($y = x$ in $\mathbb{R}^2$) line. I am not going to post the solution here (I could if needed). And then it asks to find the limit of $\mathbf{u}(t)$ at $t = \infty$, with the initial condition $\mathbf{u}(0) = (3, 1)$. For that I get $(1, -1)$ (one solution is vanishing the other is a constant).
What is the interpretation of this result? I don't understand where the author is pointing to. There seems to be a connecting thread starting from the choice of the matrix (projection), the minus sign in the DE, and the limit given that particular (maybe arbitrary) initial condition, but I don't get it.
The projection (orthogonal or not) will have two eigenvectors, one of them defined by the image of the projection and the other depending on the direction of the projection. In this case we can take ${\bf v}_1= (1,1)$; vector ${\bf v}_2$ is not given. We have $$ P{\bf v}_1 = {\bf v}_1, \qquad P{\bf v}_2 = 0$$
In any case, vectors ${\bf v}_1$ and ${\bf v}_2$ form a basis. Given any vector ${\bf u}$ you can decompose it in this basis $$ {\bf u} = \alpha {\bf v}_1 + \beta {\bf v}_2$$ If we have time dependnce, we have $$ {\bf u}(t) = \alpha(t) {\bf v}_1 + \beta(t) {\bf v}_2$$ Given our differential equation we have $$ \frac{d}{dt}\big(\alpha(t) {\bf v}_1 + \beta(t) {\bf v}_2\big) = -P\big(\alpha(t) {\bf v}_1 + \beta(t) {\bf v}_2\big)$$ $$ \frac{d\alpha(t)}{dt} {\bf v}_1 + \frac{d\beta(t)}{dt} {\bf v}_2 = -\alpha(t) {\bf v}_1$$ That is $$ \frac{d\alpha(t)}{dt} = -\alpha(t), \qquad \frac{d\beta(t)}{dt} = 0$$ $$ \alpha(t) = \alpha(0) e^{-t}, \qquad \beta(t) = \beta(0)$$ $$ {\bf u}(t) = \alpha(0)e^{-t} {\bf v}_1 + \beta(0) {\bf v}_2$$ $$ \lim_{t\rightarrow \infty} {\bf u}(t) = \beta(0) {\bf v}_2$$ As was said, vector ${\bf v}_2$ defines the direction of the projection. In this case we have $ \lim_{t\rightarrow \infty} {\bf u}(t) = (1,-1)$, which means that the projection is along the direction given by the vector $(1,-1)$, or in other words, along the lines parallel to the line $y=-x$.