What does this matrix algebra mean?

Question

If $A$ is a matrix of size $R^{h\times w}$, then what's the meanning of $A^{\frac{1}{2}}$ and $A^{-\frac{1}{2}}$? Does it means $\sqrt{A_{ij}}$ or $-\sqrt{A_{ij}}$?

There is a formula as follows: $$ f(x)=x^T D x,x\in R^{d\times 1}, D\in R^{d\times d} $$ then, we could let the following formulas $$ y=D^{\frac{1}{2}}x $$ to get the following formula $$ f(y)=y^Ty. $$ Is there any difference between the domain of $f(x)$ and $f(y)$?

Answer 1

This is only possible when $D$ is positive semi-definite. Since $D$ is related to a quadric, it is implicitly assumed that it is symmetric. A symmetric matrix $D$ can be diagonalized by orthonormal matrices $M$: $$ D=M\Lambda M^T,\quad \Lambda=\text{diag}(\underbrace{\lambda_1,\dots,\lambda_n}_{\text{eigenvalues}}) $$ In the case $D$ is positive semi-definite, meaning its eigenvalues are all non-negative, $D^{1/2}$ could only mean: $$ D^{1/2}=M\Lambda^{1/2} M^T,\quad \Lambda^{1/2}=\text{diag}(\lambda_1^{1/2},\dots,\lambda_n^{1/2}) $$ It is easy to check that $D^{1/2}D^{1/2}=D$: $$ D^{1/2}D^{1/2}=M\Lambda^{1/2} \underbrace{M^TM}_{\text{cancel out}}\Lambda^{1/2} M^T=M\Lambda M^T $$ If $D^{-1/2}$ is needed, $D$ should be positive definite.

There should be no difference between the domain of $x$ and $y$ if $D$ is positive definite (non-singular). When $D$ is only positive semi-definite, meaning it does not have a full rank, the domain of $y$ is restricted to its image.

Answer 2

$A^{\frac12}$ should be a matrix $X$ with the property that $X^2=A$. As with real numbes, taking the square root is not always possible and requires a a choice otherwise - where the natural choice and convention for $a\ge 0$ is that $\sqrt a$ is the unique nonnegative solution of $x^2=a$. Such a choice is less obvious with matrices. If $A=D$ is a diagonal matrix, then a matrix square root can be taken simply entry-wise on the diagonal - provided these are nonnegative; and in that case we use the convention again to pick the nonnegative square roots. Thus $\sqrt D:= (\sqrt{d_{i,j}})$ has the desired property $(\sqrt D)^2=D$ if $D$ is positive semidefinite diagonal. If $A$ is conjugate to a diagonal, $A=BDB^{-1}$, then we notice that $(B\sqrt D B^{-1})^2=A$, so here we can define a square root as well (but is it unique? Notice that $B,D$ are not unique! Check!)