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Given a symmetric positive semi-definite matrix $A\in\mathbb{R}^{n\times n}$ and denote by $rank(A)$ the rank of $A$. Let $\bar A=Q^TAQ$, where $Q\in\mathbb{R}^{n\times n}$ is an arbitrary invertible matrix. Then it is clear that $\bar A$ is also symmetric positive semi-definite and $rank(\bar A)=rank(A)$. My question is: By choosing arbitrary invertible matrix $Q$, can $\bar A$ take all the symmetric positive semi-definite matrices with $rank(A)$? I think it is true when $A$ is symmetric positive definite. But I cannot prove for the semi-definite case, either right or wrong. Any help will be appreciated!

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The answer is yes. With

$\text{rank}\big(A) = d$
What Sylvester's Law of Intertia tells you is that your matrix is congruent to the $\text{n x n}$ matrix

$\begin{bmatrix} I_d & \mathbf 0\\ \mathbf 0 & \mathbf 0_{n-d}\mathbf 0_{n-d}^T\\ \end{bmatrix}$

More generally, any $\text{n x n}$ real symmetric matrix with rank of $d$ and $m$ positive eigenvalues is congruent to the $\text{n x n}$ matrix

$B =\begin{bmatrix} I_{m} & \mathbf 0&\mathbf 0\\ \mathbf 0 & -I_{d-m}&\mathbf 0\\ \mathbf 0 & \mathbf 0 & \mathbf 0\\ \end{bmatrix}$

where $(m,d-m)$ (or sometimes $(m,d-m, n-d)$) is the signature of a real symmetric matrix. Equivalently any $\text{n x n}$ real symmetric matrix with signature $(m,d-m)$ is an orbit of $B$.

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  • $\begingroup$ Typo in matrix $B$: $d\to m$. $\endgroup$ – sas Apr 13 '20 at 23:04
  • $\begingroup$ @sas fixed this $\endgroup$ – user8675309 Apr 13 '20 at 23:54
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Yes, it's true. If $\ A\ $ and $\ B\ $ are any non-negative semi-definite real $\ n\times n\ $ matrices with the same rank, $\ r\ $, then there exist orthogonal matrices $\ \Omega_A\ $, $\ \Omega_B\ $ such that \begin{align} \Omega_A^TA\,\Omega_A &= \text{Diag}\left(a_1,a_2,\dots,a_r,0,\dots,0\right)\ \text{ and}\\ \Omega_B^TB\,\Omega_B &= \text{Diag}\left(b_1,b_2,\dots,b_r,0,\dots,0\right)\ , \end{align} where $\ a_i, b_i>0\ $ are the non-zero eigenvalues of $\ A, B\ $, respectively. Now if \begin{align} D&=\text{Diag}\left(\sqrt{\frac{b_1}{a_1}}, \sqrt{\frac{b_2}{a_2}}, \dots, \sqrt{\frac{b_r}{a_r}},1,\dots,1 \right)\ ,\text{ and}\\ Q&=\Omega_A D\,\Omega_B^T\ ,\text{ then}\\ B&=Q^TA\,Q\ . \end{align}

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  • $\begingroup$ There is a problem in the answer. Notice that $Q$ is required to be invertible. However, the $Q$ constructed above is not invertible if $A$ is not invertible. If $Q$ is restricted to be invertible, then is it still true? Thanks a lot!! $\endgroup$ – user147687 Apr 17 '20 at 5:58
  • $\begingroup$ Sorry, I missed that condition. Yes, it's still true. Just replace the zeroes in the diagonal of $\ D\ $ with ones (or any other non-zero entry). The key identity you need is $$ \text{Diag}\left(b_1,b_2,\dots,b_r,0,\dots,0\right)=D^T \text{Diag}\left(a_1,a_2,\dots,a_r,0,\dots,0\right)D\ , $$ and this will be true no matter what you put in the last $\ n-r\ $ entries of the diagonal of $\ D\ $. $\endgroup$ – lonza leggiera Apr 17 '20 at 6:44
  • $\begingroup$ I see. Thanks!! $\endgroup$ – user147687 Apr 23 '20 at 2:16

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