PCA on square matrix

Let's see an example:

 >>x = rand(4)

 x =

    0.8147    0.6324    0.9575    0.9572
    0.9058    0.0975    0.9649    0.4854
    0.1270    0.2785    0.1576    0.8003
    0.9134    0.5469    0.9706    0.1419

>>coeff1=pca(x)

 coeff1 =

    0.6374    0.1376   -0.1798
    0.1070    0.4092    0.9047
    0.6684    0.2762   -0.1656
   -0.3682    0.8587   -0.3490

 >>coeff2 = pcacov(cov(x))

 coeff2 =

    0.6374    0.1376   -0.1798    0.7365
    0.1070    0.4092    0.9047    0.0518
    0.6684    0.2762   -0.1656   -0.6705
   -0.3682    0.8587   -0.3490    0.0729

From my understanding, coeff1 and coeff2 should be the same. What is wrong here?

Answer 1

[Bish Erbas]

x=rand(4);
c1=pca(x,'algorithm','eig','economy',false)
d1=pcacov(cov(x))

If you use the eig algorithm and set economy to false, outputs will be equal.

Reference:

https://www.mathworks.com/help/stats/pca.html