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_jX @ v_j=>

X @ X^T @ X @ v_j = s_j²

= s_j²

U @ S @ V^T @ v_jU @ S @ V'j where V'j is a vector equal to 1 for i = j else 0= s_j²

= s_j²

U @ S'j where S'j is a vector equal to s_j for i = j else 0u_j where u_j is the j-th column of U.= s_j^3

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

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