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If I have a matrix $X \in R^{n \times n} $ and an index set $ I \subseteq \{1,\dots,n\} $,

Is $X_I$ also positive-semidefinite $\forall \ \ I $? Why ?

$X_I $ is the submatrix that is formed by choosing all rows and columns from index-set $I $

Edit : How would you prove that the determinant is product of eigenvalues for $X$ ?

Alice
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  • What are your thoughts on the question? – Sudarsan Nov 19 '13 at 18:17
  • On an intuition I think $X_I$ is positive-semidefinite. But I don't know how to prove that it is. – Alice Nov 19 '13 at 18:19
  • @Sudarsan - what are your thoughts? – Alice Nov 19 '13 at 18:20
  • I presume $\mathbf{X}_I$ is the matrix formed by removing all rows and columns from index-set $I$. Think of the definition of a PSD matrix. $y^T\mathbf{X}y \geq 0,\forall y$. If we let the positions of $y$ corresponding to the index set to vanish, then what'll happen? – Sudarsan Nov 19 '13 at 18:20
  • Then would the elements of $X$ would be $\geq$ 0 corresponding to those positions of y that vanished? Therefore giving $X_I$ , where all the elements are $\geq$ 0? – Alice Nov 19 '13 at 18:27
  • $y^T \mathbf{X}y \geq 0$ even if those positions vanished. This means that if you formed a Matrix $\mathbf{X}_I$ with those rows and columns from the index-set $I$ removed, then $y_I^T\mathbf{X}_Iy_I \geq 0$ according to our first statement where you can think of $y_I$ as the vector $y$ with the positions corresponding to index-set $I$ forced to $0$. – Sudarsan Nov 19 '13 at 18:30
  • I understand. Therefore proving the $X_I \geq 0$. Thank you. – Alice Nov 19 '13 at 18:32

1 Answers1

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(I presume your question has a statement error). $\mathbf{X}_I$ should be the Matrix formed by removing the rows and columns from index-set $I$. Yes, the new matrix $\mathbf{X}_I$ is also positive semi-definite.
Proof: (already shown in comments).

Sudarsan
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