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Suppose $A$ is symmetric, positive semidefinite and all its diagonal entries are strictly positive (real coefficients - even integer if it helps). Suppose that the first $r$ rows of $A$ are linearly independent.

Is it true that the $r$-th leading principal submatrix is full rank (and therefore positive definite)?

(The $r$-th leading principal submatrix is the top-left $r\times r$ submatrix of $A$ - it is certainly positive semidefinite and symmetric)

What if all the entries are strictly positive?

I asked this on math.stackexchange last week but I did not receive any suggestion: https://math.stackexchange.com/questions/2221146/full-rank-submatrices-of-positive-semidefinite-matrix-with-positive-entries

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Yes, this is true. To see this note that for $A$ positive semidefinite, $v^T A v = 0$ if and only if $Av = 0$. For the less obvious direction, write $A = B^TB$ for a real matrix $B$. Then $0 = v^TAv = v^TB^TBv = \lVert Bv\rVert^2$, so $Bv = 0$ and therefore $Av = B^TBv = 0$.

Now let $R$ be the principal submatrix of $A$ in question. Let $u$ be such that $Ru = 0$. Let $v$ be $u$ padded with zeros up to the dimension of $A$. Then $0 = u^TRu = v^TAv$, so $Av = 0$. Since the first $R$ columns (equiv. rows) of $A$ are linearly independent, $v$ must be zero, so $u$ must be zero and $R$ is full rank.

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  • $\begingroup$ Then one does not even need positive diagonal or positive entries. Is it correct? The statement is true for every positive semidefinite matrix: If $r$ rows are linearly independent, then the corresponding principal minor is positive definite! $\endgroup$
    – fulges
    Commented Apr 10, 2017 at 21:21
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    $\begingroup$ Yes, the only positivity constraint needed is that $A$ be positive semidefinite. $\endgroup$
    – Noah Stein
    Commented Apr 10, 2017 at 21:35
  • $\begingroup$ The answer is no. The issue is that the linear independence of the first r rows needs to take place in the first r columns. If there is a later linear dependence, then the matrix may not be full rank. For example, consider $\begin{matrix} [ 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 2 & 2 ] \end{matrix}$. The leading matrix is not of full rank. $\endgroup$
    – MDR
    Commented Aug 11, 2022 at 14:41
  • $\begingroup$ @MDR : Your example is not a positive semidefinite matrix. The proof above shows that indeed, as you say correctly, the linear independent occurs in the 'principal' part of the matrix $\endgroup$
    – fulges
    Commented Feb 9, 2023 at 16:20

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