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I've seen the pattern ABA', where A and B are matrices, and ' stands for the transpose, many times so I want to know if there is an intuition for this pattern.

I did some development to see what the resulting matrix would look like and I got this:

Let A be a matrix of dimensions (l,m),
    B be a matrix of dimensions (m,m) and
    R = ABA' with dimensions (l,l)
Then:

enter image description here

  • Is my development correct?
  • When does this pattern occur? (I've seen it a lot in machine learning).

One context where I've seen this pattern is in the equations of the Kaman filter (taken from wikipedia) :

enter image description here

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    $\begingroup$ Is A an orthogonal matrix in these contexts? $\endgroup$ Mar 5, 2018 at 22:36
  • $\begingroup$ @MatthewDrury I can't recall an example where it wasn't $\endgroup$ Mar 5, 2018 at 23:33
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    $\begingroup$ Your image says it all: the expression is a covariance matrix. It can always be interpreted as such when $(B+B^\prime)/2$ represents a nonnegative-definite quadratic form, regardless of the contents or structure of $A$. $\endgroup$
    – whuber
    Mar 6, 2018 at 0:07
  • $\begingroup$ @MatthewDrury $A$ is state transition matrix so it's square matrix but it not orthogonal and $H$ might not be square that all. this matrix map real system state into observable state. $\endgroup$
    – M lab
    Sep 6, 2021 at 10:32

2 Answers 2

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This pattern often occurs when there is, explicitly or implicitly, an orthonormal change of basis going on.

If we have a matrix $B$ and would like to re-express the transformation of multiplication by $B$ in another basis, standard linear algebra tells us that the matrix expression of the same transformation in the new basis is a similar matrix:

$$ B_{\text{new}} = A B A^{-1} $$

Here the matrix $A$ is called a change of basis matrix.

When we want to change between two orthonormal basies, a simple calculation shows that the change of basis matrix is an orthogonal matrix. Orthogonal matrices $B$ satisfy the identity:

$$ B^{-1} = B' $$

So this is how you end up with expressions like $B A B'$.

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  • $\begingroup$ I've added an example where I've seen this pattern. Could you comment if in this case, a change of basis is what is occurring? $\endgroup$ Mar 5, 2018 at 22:50
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    $\begingroup$ Since the question explicitly posits that $A$ might not be square, this answer--although insightful and correct--is not directly relevant. $\endgroup$
    – whuber
    Mar 6, 2018 at 14:18
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For short

This is covariance matrix's property $var(AX)=Avar(X)A^T$.

Also in scalar variance, $var(aX) = a^2var(X)$ when $a$ is constant. (Maybe) we can interprete as squred of matrix as well.

In other , we can interprete this and linear tranformation of ramdom variable inside covariance matrix. when we find $ABA^T$. and we know that $B=Var(X)$ so $ABA^T = Var(AX)$

For long answer.

I learning about Kalman filter recently, so I would like to explain with covariance perspective.

  1. From Kalman filter state transition equation.

$$x_k = Fx_{k-1}+Bu_k+w_k$$

Which define new state from last state and control input, we can predect next state (step k) given last step (step k-1) data by

$$\hat{x}_{k|k-1} = F\hat{x}_{k-1|k-1}+Bu_k$$

1.1 Then we find error covariance matrix of $\hat{x}_{k|k-1}$ called $P_{k|k-1}$ by $$ P_{k|k-1}=Var(x_k-\hat{k}_{x|k-1}) $$ $$ P_{k|k-1}=Var((Fx_{k-1}+Bu_k+w_k)-(F\hat{x}_{k-1|k-1}+Bu_k)) $$ $$ P_{k|k-1}=Var(F(x_{k-1}-\hat{x}_{k-1|k-1})+w_k) $$ 1.2 With covariance matrix property, $Var(A\pm B)=Var(A)+Var(B)$ when A and B are independent. we got $$P_{k|k-1}=Var(F(x_{k-1}-\hat{x}_{k-1|k-1}))+Var(w_k)$$ and we know that $Var(w_k) = Q_k$ $$P_{k|k-1}=Var(F(x_{k-1}-\hat{x}_{k-1|k-1}))+Q_k$$ 1.2 when $F$ is constant matrix, with caovaraince matrix's property $var(AX)=Avar(X)A^T$, then we got $$P_{k|k-1}=F Var(x_{k-1}-\hat{x}_{k-1|k-1})F^T+Q_k$$ 1.3 we know that $P_{k-1|k-1}=Var(x_{k-1}-\hat{x}_{k-1|k-1})$, we got $$P_{k|k-1}=F P_{k-1|k-1}F^T+Q_k$$

You see that $F P_{k-1|k-1}F^T$ pattern is just for update covariance matrix into new state coresponding to state update.

Example of $F$ and $H$

this is example value from wikipedia $$F= \begin{bmatrix} 1 & \Delta t\\ 0 & 1 \end{bmatrix} $$ $$H= \begin{bmatrix} 1 & 0\\ \end{bmatrix} $$

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