如何放大每个向量的值

Question

我有100个向量的特征向量矩阵。我遇到的问题是向量的值非常低，几乎为零(平均e-10)！下面是一个向量的示例：

0.000000245023212264631
-0.00000000928673421733024
-0.00000000702744206187778
-0.0400022257147745
0.000715047161691289
0.000178299491594312
0.000000000000000111022302462516
-0.000000000000000444089209850063
-0.0000311452059972295
0.000000928320962476081
0.00025873217452001
-0.0000073255984986786
0.000226706852668388
-0.00000637215108697076
-0.00000741641656228371
0.000000000466042723561714
0.000000000279056699407741
-0.00000000926312312226756
0.00000000000000000000000000323117426778526
0.000000000000000000000000000403896783473158
-0.0366691252097418
-0.0305829466924761
-0.0300306491122465
0.99163522436172
-0.0000000033133072963138
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000331297544425818
-0.0000100525189687
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
0.000000100784783009722
-0.00000000339576544864114
-0.0000000033133072963138
-0.0000000033133072963138
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
-0.00000000339576544864114
0.000000000128704479996649
0.000000000128704479996649
0.000000000128704479996649
0.000000000128704479996649
0.000000000128704479996649
0.000000000128704479996649
0.000000000128704479996649
0.000000000128704479996649
0.000000000229274956160153
0.000000000128704479996649
0.000000000128704479996649
0.000000000229274956160153
0.000000000128704479996649
0.000000000128704479996649
0.000000000128704479996649
0.000000000128704479996649
0.000000000128704479996649

我想知道是否可以以某种方式放大每个向量，以便在不破坏特征向量的属性的情况下使值尽可能接近1(例如，0.01而不是0.0000001)。下面是我使用的代码：

library(igraph)
M <- data.matrix(test)
G <- graph.data.frame(M, directed = FALSE)
L <- laplacian_matrix(G, normalized = FALSE, weights = NULL, sparse = FALSE)
options(scipen=999)
df <- data.frame(eigen(L)) 
df[,1] <- NULL
write.table(df[,1], file = "~/test.csv", row.names=FALSE, col.names=FALSE, sep=",")

Answer 1

我不知道这是否会打乱属性，但是像这样的.

test <- rnorm(100, mean = 0.00000000001, sd = 0.00000000002) # fake small set
test <- ifelse(test<0, test*-1, test) # removed negatives
scaledtest <- test/(max(test)+test[which.max(test)]/10)

Answer 2

矩阵A的特征向量x的定义是，对于某个特征值c，Ax = cx，所以如果你放大x到kx，那么你的特征值必须是c/k才能满足方程。因此，您可以通过标量任意缩放特征向量，但相应的特征值将乘以逆值。

我不建议这样做，因为你引入的误差可能会很大。