tf_svd_gradient.py

def mmul(*tensors):
    return tf.foldl(tf.matmul, tensors)

def msym(X):
    return (X + tf.matrix_transpose(X)) / 2

def mdiag(X):
    return tf.matrix_diag(tf.matrix_diag_part(X))

@tf.RegisterGradient('Svd')
def gradient_svd(op, dL_ds, dL_dU, dL_dV):
    s, U, V = op.outputs
    # NOTE: based on https://arxiv.org/pdf/1509.07838.pdf
    # this version works for square matrices only
    # in practice it means that only U_1 part of (17) is used
    assert U.shape == V.shape, U.shape[1] == U.shape[2]
    I = tf.eye(tf.shape(s)[1])        
    S = tf.matrix_diag(s)    
    dL_dS = tf.matrix_diag(dL_ds)
    V_T = tf.matrix_transpose(V)
    U_T = tf.matrix_transpose(U)
    s_2 = tf.square(s)
    K = 1.0 / (s_2[:,tf.newaxis,:] - s_2[:,:,tf.newaxis] + I) - I
    D = mmul(dL_dU, tf.matrix_diag(1.0 / s))
    D_T = tf.matrix_transpose(D)
    return (mmul(D, V_T) + 
            mmul(U, mdiag(dL_dS - mmul(U_T, D)), V_T) + 
            2 * mmul(U, S, msym(K * mmul(V_T, dL_dV - mmul(V, D_T, U, S))), V_T))

Thanks for sharing the code! I have a similar version of your code, but I runs very slow... I am not sure whether such customized gradient runs on GPU? THX!!

Also, one more question in the paper it says when dU is (m,n) did you assume that m and n are equal?
And in https://i.imgur.com/Bzyb1My.png I notice that the second term is not considered, I guess since m and n are equal we are only considering the first term?

Oh oh nvm you set the n to be m - hence the second part of block decomposition never gets to be considered gotcha