4 Updated answer to address a question in the comments. edited Nov 4 '13 at 20:05 ramhiser 1,3351212 silver badges1313 bronze badges I find it helpful to consider the singular value decomposition for questions like this with the assumption that $$X$$ is a real matrix. Writing $$X = UDV^T$$, we can see that $$XX^T = UD^2U^T$$ and $$X^TX = VD^2V^T$$. As we can see, the eigenvalues of both $$XX^T$$ and $$X^TX$$ are contained in the diagonal matrix $$D^2$$ and are indeed equal. Also, we see that the matrix of eigenvectors of $$XX^T$$ is $$U$$, while the matrix of eigenvectors of $$X^TX$$ is $$V$$. Because $$X^TU = VDU^TU = VD,$$ we have the relationship $$X^TU = VD$$, similar to that pointed out by whuber. Assuming $$D$$ is nonsingular, two additional properties that prove to be quite useful are: \begin{align} U &= XVD^{-1}\\ V &= X^TUD^{-1}. \end{align} I find it helpful to consider the singular value decomposition for questions like this with the assumption that $$X$$ is a real matrix. Writing $$X = UDV^T$$, we can see that $$XX^T = UD^2U^T$$ and $$X^TX = VD^2V^T$$. As we can see, the eigenvalues of both $$XX^T$$ and $$X^TX$$ are contained in the diagonal matrix $$D^2$$ and are indeed equal. Also, we see that the matrix of eigenvectors of $$XX^T$$ is $$U$$, while the matrix of eigenvectors of $$X^TX$$ is $$V$$. Because $$X^TU = VDU^TU = VD,$$ we have the relationship $$X^TU = VD$$, similar to that pointed out by whuber. I find it helpful to consider the singular value decomposition for questions like this with the assumption that $$X$$ is a real matrix. Writing $$X = UDV^T$$, we can see that $$XX^T = UD^2U^T$$ and $$X^TX = VD^2V^T$$. As we can see, the eigenvalues of both $$XX^T$$ and $$X^TX$$ are contained in the diagonal matrix $$D^2$$ and are indeed equal. Also, we see that the matrix of eigenvectors of $$XX^T$$ is $$U$$, while the matrix of eigenvectors of $$X^TX$$ is $$V$$. Because $$X^TU = VDU^TU = VD,$$ we have the relationship $$X^TU = VD$$, similar to that pointed out by whuber. Assuming $$D$$ is nonsingular, two additional properties that prove to be quite useful are: \begin{align} U &= XVD^{-1}\\ V &= X^TUD^{-1}. \end{align} 3 Improved clarity of last equation to answer comment. edited Jun 19 '13 at 21:20 ramhiser 1,3351212 silver badges1313 bronze badges I find it helpful to consider the singular value decomposition for questions like this with the assumption that $$X$$ is a real matrix. Writing $$X = UDV^T$$, we can see that $$XX^T = UD^2U^T$$ and $$X^TX = VD^2V^T$$. As we can see, the eigenvalues of both $$XX^T$$ and $$X^TX$$ are contained in the diagonal matrix $$D^2$$ and are indeed equal. Also, we see that the matrix of eigenvectors of $$XX^T$$ is $$U$$, while the matrix of eigenvectors of $$X^TX$$ is $$V$$. Hence, weBecause $$X^TU = VDU^TU = VD,$$ we have the relationship $$X'U = VD$$$$X^TU = VD$$, similar to that pointed out by whuber. I find it helpful to consider the singular value decomposition for questions like this with the assumption that $$X$$ is a real matrix. Writing $$X = UDV^T$$, we can see that $$XX^T = UD^2U^T$$ and $$X^TX = VD^2V^T$$. As we can see, the eigenvalues of both $$XX^T$$ and $$X^TX$$ are contained in the diagonal matrix $$D^2$$ and are indeed equal. Also, we see that the matrix of eigenvectors of $$XX^T$$ is $$U$$, while the matrix of eigenvectors of $$X^TX$$ is $$V$$. Hence, we have the relationship $$X'U = VD$$, similar to that pointed out by whuber. I find it helpful to consider the singular value decomposition for questions like this with the assumption that $$X$$ is a real matrix. Writing $$X = UDV^T$$, we can see that $$XX^T = UD^2U^T$$ and $$X^TX = VD^2V^T$$. As we can see, the eigenvalues of both $$XX^T$$ and $$X^TX$$ are contained in the diagonal matrix $$D^2$$ and are indeed equal. Also, we see that the matrix of eigenvectors of $$XX^T$$ is $$U$$, while the matrix of eigenvectors of $$X^TX$$ is $$V$$. Because $$X^TU = VDU^TU = VD,$$ we have the relationship $$X^TU = VD$$, similar to that pointed out by whuber. 2 added 1 characters in body edited Jun 3 '13 at 15:39 ramhiser 1,3351212 silver badges1313 bronze badges I find it helpful to consider the singular value decomposition for questions like this with the assumption that $$X$$ is a real matrix. Writing $$X = UDV^T$$, we can see that $$XX^T = UD^2U^T$$ and $$X^TX = VD^2V^T$$. As we can see, the eigenvalues of both $$XX^T$$ and $$X^TX$$ are contained in the diagonal matrix $$D^2$$ and are indeed equal. Also, we see that the matrix of eigenvectoreigenvectors of $$XX^T$$ is $$U$$, while the matrix of eigenvectors of $$X^TX$$ is $$V$$. Hence, we have the relationship $$X'U = VD$$, similar to that pointed out by whuber. I find it helpful to consider the singular value decomposition for questions like this with the assumption that $$X$$ is a real matrix. Writing $$X = UDV^T$$, we can see that $$XX^T = UD^2U^T$$ and $$X^TX = VD^2V^T$$. As we can see, the eigenvalues of both $$XX^T$$ and $$X^TX$$ are contained in the diagonal matrix $$D^2$$ and are indeed equal. Also, we see that the matrix of eigenvector of $$XX^T$$ is $$U$$, while the matrix of eigenvectors of $$X^TX$$ is $$V$$. Hence, we have the relationship $$X'U = VD$$, similar to that pointed out by whuber. I find it helpful to consider the singular value decomposition for questions like this with the assumption that $$X$$ is a real matrix. Writing $$X = UDV^T$$, we can see that $$XX^T = UD^2U^T$$ and $$X^TX = VD^2V^T$$. As we can see, the eigenvalues of both $$XX^T$$ and $$X^TX$$ are contained in the diagonal matrix $$D^2$$ and are indeed equal. Also, we see that the matrix of eigenvectors of $$XX^T$$ is $$U$$, while the matrix of eigenvectors of $$X^TX$$ is $$V$$. Hence, we have the relationship $$X'U = VD$$, similar to that pointed out by whuber. 1 answered Jun 3 '13 at 15:27 ramhiser 1,3351212 silver badges1313 bronze badges