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Let $A= a(i,j)$ and $B= b(i,j)$ be ($n\times n$) matrices that are positive definite such that $a(i,j) < b(i,j)$ .Let $C= c(i,j)= a(i,j) - b(i,j)$, then $C$ is also positive definite. Why or why not?

what i know is $x^T(C)x= x^T (A-B) x = x^T(A)x - x^T(B)x$. Now both terms on right-hand side are greater than $0$ as $A$ and $B$ are positive definite. Also $a(i,j) < b(i,j)$. But in such case the sign of $x^T(C)x$ being positive or negative will also depend on the value of elements of vector $x$. So, we can't assure that $x^T(C)x < 0$ always or that $C$ is not positive definite.

Am i right about this?

glS
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under root
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    Take $A=(1)$, $B=(2)$ two $1-1$ matrices. Then $A-B$ is obviously negative-definite. Then take $A=\begin{pmatrix}1&0 \0&1\end{pmatrix}$ and $B=\begin{pmatrix}2&1 \ 1&2\end{pmatrix}$. So $A-B=\begin{pmatrix}-1&-1\-1&-1\end{pmatrix}$ is negative semi-definite. So I think that the difference of two positive-definite matrices should not have a given fixed type. – awllower May 05 '13 at 02:37
  • $x^Ax > 0$ and $x^Bx>0$ imply that $x^*(A+B)x>0$, but not much can be said about $A - B$. – parsiad May 05 '13 at 02:42
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    Morover, if $A=\begin{pmatrix}2&0\0&2\end{pmatrix},$ $B=\begin{pmatrix}3&2\2&3\end{pmatrix}$, then $A-B=\begin{pmatrix}-1&-2\-2&-1\end{pmatrix}$ is indefinite.(it has eigen-values $-3$ and $1$.) (Let me try if an answer-form is possible.) – awllower May 05 '13 at 02:45
  • what i got is we cant decide if C is +ve definite or -ve definite .Even if a(i,j)>b(i,j) was given.As the value as a whole will also depend on the elements of the vector x.Let me know what you think about it. – under root May 05 '13 at 02:56
  • We cannot decide whether $C$ is defiite or not, but we can assure that $C$ is not positive-definite, nor positive semi-definite. – awllower May 05 '13 at 03:21
  • Related – Pedro Nov 20 '16 at 17:28

1 Answers1

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Proposition
If $A=(a_{ij})$, $B=(b_{ij})$ are two positive-definite matrices with $a_{ij}<b_{ij} \forall i,j$, then $A-B$ can only be indefinite or negative-definite or negative semi-definite.

Demonstration
This strange proposition says only that $A-B$ must take on negative values somewhere, and this is easy to prove: $A-B=(a_{ij}-b_{ij})=(c_{ij})$, so in particular $c_{1,1}<0$, and the form corresponding to $A-B$ contains the monomial $c_{1,1}x_1^2$, and hence must have negative values somewhere. Hence the proposition is proved.

Examples
The following are some illuminations showing each possibility:
negative-definite: $A=\begin{pmatrix}1\end{pmatrix}, B=\begin{pmatrix}2\end{pmatrix}, A-B=\begin{pmatrix}-1\end{pmatrix}$.
negative semi-definite: $A=\begin{pmatrix}1&0\\0&1\end{pmatrix}, B=\begin{pmatrix}2&1\\1&2\end{pmatrix}, A-B=\begin{pmatrix}-1&-1\\-1&-1\end{pmatrix}$.
indefinite: $A=\begin{pmatrix}2&0\\0&2\end{pmatrix}, B=\begin{pmatrix}3&2\\2&3\end{pmatrix}, A-B=\begin{pmatrix}-1&-2\\-2&-1\end{pmatrix}$.
Barring mistakes, and thanks for the attention.

awllower
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  • thanks for your feedback.... you are right on this one...need to add one more point...when a(ij)<b(ij), then A-B can never be positive definite according to sylvester's criterion.The determinant of the 1st principle minor matrix of C (i.e. a(11)-b(11) ) is -ve . Thus C can never be +ve definite in this case. – under root May 09 '13 at 08:59
  • @sunnyverma Thanks for sharing with us this point. :D – awllower May 09 '13 at 12:47