I think there is some confusion in the derivation of ii) -> iii), if p is a vector that can be right multiplied by the controllability gramian, it should have a dimension of (n by 1), then b' * e^{-A't} * p will become a scalar because it's (1 by n)*(n by n)*(n by 1). And it's the same for the " v' ", which will also become a scalar instead of a vector. In this case, would "v" and " v' " become two same scalars so the term in the integral will be a scalar's square so it has to be zero? I tried some examples in MATLAB by defining vector p(n by 1), vector B(n by 1), and matrix A(n by n). When I calculate B' * A" * p and p' * A* B, it turns out to be two same scalars but I am not sure if this is true for all cases. If this is true, then p' B = 0(scalar), p'AB = 0 (scalar) and so on, and the conclusion will be the same.
Hello! Very good observation - you are correct, for a system with one input, v and v' are scalars. I should have made this clearer. As a side note though, we actually end up proving something more general. In systems with more than one input, the B matrix will by n by m, where m is the number of inputs. Then the vector v will not be a scalar anymore - but the argument in the video still works! So actually we are seeing a proof in the general case for systems with multiple inputs!
I think you did a mistake at the end when you gave us u(t) to show that iii) -> i) it should be u(t) = -(B^T)* e^(-(A^T) *t)*(W(T))^1 *x0 But thank you anyway, its the first time I understand this proof
Yes, you're right (but with a ((W(T))^(-1)?)! Thank you, and thank you also for the encouragement! I see also that I made another mistake - in the bottom left for i)->ii) expressions for x(t), I forgot to add in the B in front of the u(tau). There could be others, but I'm extremely happy that you stuck with this all the way to the end and were able to follow the ideas! These are quite advanced topics, but it is always satisfying to know what the justification for these results are.
Thank you
I think there is some confusion in the derivation of ii) -> iii), if p is a vector that can be right multiplied by the controllability gramian, it should have a dimension of (n by 1), then b' * e^{-A't} * p will become a scalar because it's (1 by n)*(n by n)*(n by 1). And it's the same for the " v' ", which will also become a scalar instead of a vector.
In this case, would "v" and " v' " become two same scalars so the term in the integral will be a scalar's square so it has to be zero? I tried some examples in MATLAB by defining vector p(n by 1), vector B(n by 1), and matrix A(n by n). When I calculate B' * A" * p and p' * A* B, it turns out to be two same scalars but I am not sure if this is true for all cases.
If this is true, then p' B = 0(scalar), p'AB = 0 (scalar) and so on, and the conclusion will be the same.
Hello! Very good observation - you are correct, for a system with one input, v and v' are scalars. I should have made this clearer. As a side note though, we actually end up proving something more general. In systems with more than one input, the B matrix will by n by m, where m is the number of inputs. Then the vector v will not be a scalar anymore - but the argument in the video still works! So actually we are seeing a proof in the general case for systems with multiple inputs!
I think you did a mistake at the end when you gave us u(t) to show that iii) -> i) it should be u(t) = -(B^T)* e^(-(A^T) *t)*(W(T))^1 *x0
But thank you anyway, its the first time I understand this proof
Yes, you're right (but with a ((W(T))^(-1)?)! Thank you, and thank you also for the encouragement! I see also that I made another mistake - in the bottom left for i)->ii) expressions for x(t), I forgot to add in the B in front of the u(tau). There could be others, but I'm extremely happy that you stuck with this all the way to the end and were able to follow the ideas! These are quite advanced topics, but it is always satisfying to know what the justification for these results are.