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I was looking for questions about linear algebra and I found the following statement:

If $A$ is a real positive definite symmetric matrix, then $A+A^{-1}-2I$ is positive definite matrix. Can we assume that $A+A^{-1}-2I>0$ entrywise? Or $\geq 0$? What if $A$ is not symmetric?.

Rodrigo de Azevedo
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User_09
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  • This is the matrix version of the statement $(x - y)^2 = x^2 + y^2 -2xy$, where $x = \sqrt A$ and $y = 1/x$. The important thing is that a matrix square-root will exist in the setting you describe. – guy Nov 12 '20 at 21:15
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    In fact, it seems that $A+A^{-1}-2I$ is positive semi-definite matrix but may not be positive definite matrix. (Take e.g. $A=I$.) –  Nov 12 '20 at 21:25

2 Answers2

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Since $A$ is real and symmetric, it has a decomposition $A=P\Lambda P^\top$ where $\Lambda$ is the diagonal matrix of eigenvalues and $P$ is orthogonal. Hence

$A+A^{-1} -2I = P(\Lambda + \Lambda^{-1} -2)P^\top$

Using the fact that $\lambda+\frac{1}{\lambda} \geq 2$, we have $A+A^{-1} -2I$ is positive semi-definite.

Dunham
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The property of being positive definite/semi-definite does not change if we switch from a matrix $A$ to a similar matrix $PAP^{-1}$ with an invertible matrix $P$. In particular, we can diagonalise $A$ using an orthogonal matrix $P$, i.e. $D=P^{-1}AP$ for some orthogonal matrix $P$ and diagonal matrix $D$. All the entries on the diagonal of $D$ are positive as $A$ is a positive-definite matrix.

Now, if $A$ is similar to $D$, then $A+A^{-1}-2I$ is similar to $D+D^{-1}-2I$ because:

$$P^{-1}(A+A^{-1}-2I)P=D+D^{-1}-2I$$

However, $D+D^{-1}-2I$ is also diagonal, and if the entries on the diagonal of $D$ are $d_i, i=1,\ldots,n$, then the entries on the diagonal of $D+D^{-1}-2I$ are $d_i+\frac{1}{d_i}-2=\frac{1}{d_i}(d_i-1)^2\ge 0$. Thus, the matrix $D+D^{-1}-2I$ is positive semi-definite, and so the matrix $A+A^{-1}-2I$ is positive semi-definite.

Therefore, the matrix $A+A^{-1}-2I$ is always positive semi-definite. It may not be positive definite, as the example $A=I$ shows.

  • But can we assume that all entries of $A+A^{-1}-2I$ are $\geq 0$ if $A$ is positive definite matrix? – User_09 Nov 12 '20 at 21:44
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    No. A positive definite matrix can have plenty of negative entries. $\begin{bmatrix}2&-1\-1&2\end{bmatrix}$ is a positive definite matrix, for example, and it is similar to a diagonal matrix $\begin{bmatrix}1&0\0&3\end{bmatrix}$. The entries on the diagonal of the diagonal matrix are the eigenvalues of the original matrix, and on positive definite/semi-definite symmetric matrix those eigenvalues are all positive/nonnegative. –  Nov 12 '20 at 21:55