Hw3 - Exercise 2

Hello,
I don't really understand how to find the matrices \(U, V, \ Sigma \) without any calculation i.e. just based on the X matrix and the mappings \(Xa, Xb, Xc \). Could you explain to me how it works and what are the different steps ?

Thank you and have a good weekend,
Alessio

Top comment

Hi Alessio,

Sorry for my late reply.
For this exercise, you can refer to the Page 13 of Lecture 6. We can see that a,b,c vectors are orthogonal and firstly we can use V^t to change the xyz basis to the new abc basis. Then use the Sigma matrix to scale each axis (Hint: |Xa|/a, etc). Finally, you can infer the U matrix based on X[a b c] and SigmaV^t[a b c] to find the X without calculations.

Best,
Junze

Hi,
Sorry I didn't see your reply. Thanks a lot for the explanation. I manage to finish the exercise following your hints.

There is just one step I didn't fully understand. Why should we use \(V^T \) to change the xyz basis to the new abc basis ? Isn't the goal of \( V^Ta \) to express \(a\) in the the \( V \)'s basis ?

Best,
Alessio

Hi, I'm sorry but I’m trying to redo this exercise and I still don’t understand why we should compute V this way.

From the SVD decomposition, there are two « change of basis » (the matrices U and V) and I don’t see what is the reason (or the intuition) to compute V^T the way you described it. If you can explain this step in more detail, it will be really great.

Thank you for your help,
Best,
Alessio

Hi Alessio,

Your understanding of SVD is correct, U and V change the basis and \(\Sigma\) changes the scale. And the solution is based on this special case: a, b and c are orthogonal. If we do not have this condition, it is really hard for us to find X without calculation (hard to determine V and U respectively). If we know that a, b and c are orthogonal, then we can just use the abc basis to find V matrix (the 'new abc basis' in the solution means that we should keep \(VV^{T}=I\)). And the next steps are basic matrix multiplication. Overall, this special setting in this question is critical for us to find the starting point :)

Best,
Junze

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