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First of all, I've read through all of the answers here and here, but neither of those threads was able to give completely satisfying answers.

Now, I understand that, if $A$ is symmetric and positive definite, then both the columns of $U$ and the columns of $V$ are eigenvectors of the strictly positive eigenvalues of $A$. Thus, if the eigenspace is one-dimensional, because of orthogonality and linear dependency, they only differ by a factor $|\lambda| = 1$, which necessarily implies $\lambda = 1$ since $A$ is positive definite.

The part I don't get is that, what if the corresponding eigenspace is multi-dimensional? Why wouldn't it be able for the columns of $U$ to be a (true) rotation of the columns of $V$, or even just a permutation? I unfortunately wasn't able to see any proofs or counter-examples for this proposition, so I was hoping to get some answers here.

kalkuluss
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2 Answers2

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You already know that columns of $U$ and $V$ are both eigenvectors associated to the eigenvalues of $A$ (strictly positive, as $A$ is symmetric, positive-definite). Furthermore, as they are orthogonal matrices, $V^T = V^{-1}$. Now, let the columns of $U$ be $\{u_1, \cdots, u_n\}$ and the columns of $V$ be $\{v_1, \cdots, v_n\}$. $\Sigma$ is a diagonal matrix with $\lambda_1, \cdots, \lambda_n$ on the diagonal, which are (strictly positive) eigenvalues of $A$. Then $Ve_i = v_i$, so $V^Tv_i = V^{-1}v_i = e_i$. Thus, $Av_i = U\Sigma V^Tv_i = U\Sigma e_i = U(\lambda_i e_i) = \lambda_i u_i$. $v_i$ is supposed to be an eigenvector of $A$, so $\lambda_i u_i = Av_i = \kappa_i v_i$ for some $\kappa_i > 0$. Then $u_i = \frac{\kappa_i}{\lambda_i}v_i$, so $u_i$ is a strictly positive multiple of $v_i$. But both $u_i$ and $v_i$ are supposed to be unit vectors, so the only possibility is $u_i = v_i$ for all $i$. Thus, $U = V$.

David Gao
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  • Wow, I feel pretty dumb now, I've definitely played around with the equations $Av_i = \lambda_i u_i$ and $Au_i = \lambda_i v_i$. Thank you so much! – kalkuluss Jul 19 '24 at 08:50
  • @kalkuluss No problem - if you find the answer satisfactory, you can accept the answer by clicking the checkmark on the left-hand side. – David Gao Jul 20 '24 at 05:01
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This is where the positive definiteness of $A$ is essential. That $A$ is symmetric already implies that $V^TU$ commutes with $\Sigma$. If $A$ is also positive definite, then so are $U^TAU=\Sigma V^TU$ and $V^TU$. Hence $V^TU$ is a positive definite orthogonal matrix and it must be equal to $I$. Consequently, $U=V$.

user1551
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