1) Could you please give me more details on how we obtain this result : (edit : figured it out)
I think that considering that V is orthogonal, V^T @ v_j should be equal to a vector where v_j = 1 else 0 because v_i @ v_j = 0 if i != j else 1.
Thus, I come up with a different result.
2) I also don't understand the second part of the solution.
We have a way to compute efficiently s_j² but we are losing the sign of s_j. Is it important ?
Also, I don't get what is u_j and how we obtain the equality u_j = X_vj. Is it a column or a row of U ?
My first guess is that :
X^T @ X @ v_j = s_j² v_j
=>
X @ X^T @ X @ v_j = s_j² X @ v_j
= s_j² U @ S @ V^T @ v_j
= s_j² U @ S @ V'j where V'j is a vector equal to 1 for i = j else 0
= s_j² U @ S'j where S'j is a vector equal to s_j for i = j else 0
= s_j^3 u_j where u_j is the j-th column of U.
We have an extra s_j that we need to remove in this case but I might be wrong somewhere.
Let me answer the second question. \(u_j\) is the \(j\)-th column of \(U\), and noticing \(XX^\top X v_j = s_j^2 X v_j\) means that \(X v_j\) is an eigenvector of \(XX^\top\) of eigenvalue \(s_j^2\), we have \(u_j = X v_j\).
We can also formally prove this as follows: define \(w_j = X v_j\). Then according to the screenshot you've posted, \(XX^\top w_j = s_j^2 w_j\). We only need to prove \(w_j = u_j\). Noticing \(X = USV^\top\), we have
\(USS^\top U^\top w_j = s_j^2 w_j\), which means \(SS^\top U^\top w_j = s_j^2 U^\top w_j\), expanding this equation and compare each element gives the result.
Lab 12 Problem 1
Hi,
1) Could you please give me more details on how we obtain this result : (edit : figured it out)
I think that considering that V is orthogonal, V^T @ v_j should be equal to a vector where v_j = 1 else 0 because v_i @ v_j = 0 if i != j else 1.
Thus, I come up with a different result.
2) I also don't understand the second part of the solution.
We have a way to compute efficiently s_j² but we are losing the sign of s_j. Is it important ?
Also, I don't get what is u_j and how we obtain the equality u_j = X_vj. Is it a column or a row of U ?
My first guess is that :
X^T @ X @ v_j = s_j² v_j
=>
X @ X^T @ X @ v_j = s_j² X @ v_j
= s_j² U @ S @ V^T @ v_j
= s_j² U @ S @ V'j where V'j is a vector equal to 1 for i = j else 0
= s_j² U @ S'j where S'j is a vector equal to s_j for i = j else 0
= s_j^3 u_j where u_j is the j-th column of U.
We have an extra s_j that we need to remove in this case but I might be wrong somewhere.
Hi,
Let me answer the second question. \(u_j\) is the \(j\)-th column of \(U\), and noticing \(XX^\top X v_j = s_j^2 X v_j\) means that \(X v_j\) is an eigenvector of \(XX^\top\) of eigenvalue \(s_j^2\), we have \(u_j = X v_j\).
We can also formally prove this as follows: define \(w_j = X v_j\). Then according to the screenshot you've posted, \(XX^\top w_j = s_j^2 w_j\). We only need to prove \(w_j = u_j\). Noticing \(X = USV^\top\), we have
\(USS^\top U^\top w_j = s_j^2 w_j\), which means \(SS^\top U^\top w_j = s_j^2 U^\top w_j\), expanding this equation and compare each element gives the result.
2
Add comment