4 Updated answer to address a question in the comments.
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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.
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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
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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.

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