And finally $\boldsymbol{X}\boldsymbol{X}^T$

For $\boldsymbol{X}\boldsymbol{X}^T$ we found

$$\boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T\boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{U}^T=\boldsymbol{U}\boldsymbol{\Sigma}^T\boldsymbol{\Sigma}\boldsymbol{U}^T.$$

Since the matrices here have dimension $n\times n$, we have

$$\boldsymbol{\Sigma}\boldsymbol{\Sigma}^T = \begin{bmatrix} \tilde{\boldsymbol{\Sigma}} \\ \boldsymbol{0}\\ \end{bmatrix}\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} \boldsymbol{0}\\ \end{bmatrix}=\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0}\\ \end{bmatrix},$$

leading to

$$\boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{U}\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0}\\ \end{bmatrix}\boldsymbol{U}^T.$$

Multiplying with $\boldsymbol{U}$ from the right gives us the eigenvalue problem

$$(\boldsymbol{X}\boldsymbol{X}^T)\boldsymbol{U}=\boldsymbol{U}\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0}\\ \end{bmatrix}.$$

It means that the eigenvalues of $\boldsymbol{X}\boldsymbol{X}^T$ are again given by the non-zero singular values plus now a series of zeros. The column vectors of $\boldsymbol{U}$ are the eigenvectors of $\boldsymbol{X}\boldsymbol{X}^T$ and measure how much correlations are contained in the rows of $\boldsymbol{X}$.

Since we will mainly be interested in the correlations among the features of our data (the columns of $\boldsymbol{X}$, the quantity of interest for us are the non-zero singular values and the column vectors of $\boldsymbol{V}$.

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