张量流中的频谱范数2x2矩阵 [英] Spectral norm 2x2 matrix in tensorflow

问题描述

我有一个2x2矩阵，该矩阵由来自其他输入的变量J00, J01, J10, J11定义.由于矩阵很小，因此我能够通过先计算迹线和行列式来计算光谱范数

I've got a 2x2 matrix defined by the variables J00, J01, J10, J11 coming in from other inputs. Since the matrix is small, I was able to compute the spectral norm by first computing the trace and determinant

J_T = tf.reduce_sum([J00, J11])
J_ad = tf.reduce_prod([J00, J11])
J_cb = tf.reduce_prod([J01, J10])
J_det = tf.reduce_sum([J_ad, -J_cb])

然后求解二次方

L1 = J_T/2.0 + tf.sqrt(J_T**2/4.0 - J_det)
L2 = J_T/2.0 - tf.sqrt(J_T**2/4.0 - J_det)
spectral_norm = tf.maximum(L1, L2)

这有效，但是看起来很丑陋，并且不能推广到更大的矩阵.有没有更简洁的方法(也许是我所缺少的方法调用)来计算spectral_norm?

This works, but it looks rather ugly and it isn't generalizable to larger matrices. Is there cleaner way (maybe a method call that I'm missing) to compute spectral_norm?

推荐答案

矩阵J的频谱范数等于最大的矩阵的奇异值.

The spectral norm of a matrix J equals the largest singular value of the matrix.

因此，您可以使用 tf.svd() 进行奇异值分解，并采用最大的奇异值:

Therefore you can use tf.svd() to perform the singular value decomposition, and take the largest singular value:

spectral_norm  = tf.svd(J,compute_uv=False)[...,0]

其中J是您的矩阵.

注意:

• 我使用compute_uv=False是因为我们只对奇异值感兴趣，而对奇异矢量不感兴趣.
• J不必是正方形.
• 此解决方案也适用于J具有任意数量的批次尺寸(只要最后两个尺寸是矩阵尺寸)的情况.
• 省略号...操作与NumPy中的操作相同.
• 我选择0索引是因为我们只对最大的奇异值感兴趣.

• I use compute_uv=False since we are interested only in singular values, not singular vectors.
• J does not need to be square.
• This solution works also for the case where J has any number of batch dimensions (as long as the two last dimensions are the matrix dimensions).
• The elipsis ... operation works as in NumPy.
• I take the 0 index because we are interested only in the largest singular value.

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