# Where do matrix A and A transpose come from in calculating the predicted covariance matrix?

I don't understand where the matrices A and A transpose come from in the equation in this series. I have done a one-dimensional example in the series but I don't see how to derive these matrices in matrix form.

The $$A$$ matrix (sometimes $$F$$) comes from your equations of motion. There's a technical example on Wikipedia that goes over modeling position and speed of a truck.

You can model speed and position with the following equations:

$$x = x_0 + \Delta t \dot{x} + \frac{1}{2}\Delta t^2 \ddot{x} \\ \dot{x} = \dot{x}_0 + \Delta t \ddot{x} \\$$

Assuming your input $$u$$ is the acceleration ($$u = \ddot{x}$$), and being more explicit with the coefficients on everything, you could rewrite the above as:

$$x = \left(1\right)x_0 + \left(\Delta t\right) \dot{x} + \left(\frac{1}{2}\Delta t^2\right) u \\ \dot{x} = \left(0\right)x_0 + \left(1\right)\dot{x}_0 + \left(\Delta t\right)u \\$$

And then you convert to matrix form:

$$\left[\begin{matrix} x \\ \dot{x}\end{matrix}\right] = \left[\begin{matrix} 1 & \Delta t \\ 0 & 1 \end{matrix}\right] \left[\begin{matrix} x_0 \\ \dot{x}_0 \end{matrix}\right] + \left[\begin{matrix} \frac{1}{2}\Delta t^2 \\ \Delta t\end{matrix}\right]u$$

The current sample being: $$x_k = \left[\begin{matrix} x \\ \dot{x}\end{matrix}\right]$$ and the previous sample being: $$x_{k-1} = \left[\begin{matrix} x_0 \\ \dot{x}_0 \end{matrix}\right]$$

You can then define your "state matrix" $$A$$ or $$F$$ as:

$$A = \left[\begin{matrix} 1 & \Delta t \\ 0 & 1 \end{matrix}\right]$$ and your input matrix $$B$$ or $$G$$ as:

$$B = \left[\begin{matrix} \frac{1}{2}\Delta t^2 \\ \Delta t\end{matrix}\right]$$

And so you have:

$$x_k = Ax_{k-1} + Bu$$

And then those are used with the output equations:

$$y_k = Cx_k + Du$$

to form the Kalman filter. These are all standard matrices from the state space representation of a system.

• To expand on this. The value we are estimating is assumed to be a gaussian distribution with $x - \mathcal{N}(x,P)$. Chuck shows how $A$ works on the mean($x$). In your equation you are estimating the covariance. The transpose is used to keep the matrix symmetric, and because the covariance is the variance^2. You essentially need to square the $A$ matrix. Which is accomplished by multiplying by the transpose. A proof of this can be found here(math.stackexchange.com/questions/332441/…) – edwinem Jun 5 '20 at 15:31
• Thanks, @edwinem ! Op's question was about "where the matrices A and A transpose come from" in the equations, so I tried to answer just that aspect as directly as possible, but this is great context and excellent further reading :) – Chuck Jun 5 '20 at 15:34
• I needed the math part too. – heretoinfinity Jun 5 '20 at 17:39
• @heretoinfinity - What do you mean? You need the math part for what? What is missing from the answer I provided, or what can I explain better? – Chuck Jun 5 '20 at 19:32
• The linear algebra and how I ended up with the transpose. I see it in the link with the comment so that's solved. – heretoinfinity Jun 5 '20 at 20:08