Consider $AX + XA = B$, how many equations and how many unknowns?

Question

Here, $A$, $X$, and $B$ are real $n\times n$ matrices. A and B are given, but X is unknown.

Also, A is symmetric and positive definite, so $A^T = A$ and $z^TAz >0$ for all nonzero $z$ in $\mathbb R^n$.

My question is: I can easily show that $AX + XA$ is linear in $x$.

but...how many equations and unknowns does it involve?

I wrote out some scratch-work and agree with the solution that it involves $n^2$ unknowns. However, the solution also says it involves $n^2$ equations. I'm pretty sure that's wrong, and that there are only $n$ equations involved.

Thanks,

Edit:

How could I show that F(x) = AX + XA is a bijection? I feel like I should be using what's given in the question: that A is real symmetric, positive definite, so then A has real, positive eigenvalues. Also, A is diagonalizable - orthogonally diagonalizable, in fact - so it has n eigenvectors that form a basis for R^n. Not sure how to proceed to prove the bijection...

Answer 1

Two matrices are equal if and only if their corresponding elements are equal. So the matrix equation $AX + XA = B$ yields $n^2$ equations, one for each entry of the two $n \times n$ matrices $AX + XA$ and $B$.

Answer 2

The $n^2$ eigenvalues of the linear application $F:X\in M_n(\mathbb{R})\rightarrow AX+XA\in M_n(\mathbb{R})$ are the $(\lambda_i+\lambda_j)_{i,j}$ where $spectrum(A)=(\lambda_i)_i$. Since $A$ is symmetric $>0$, $F$ is a bijection.

Moreover one may write the equation in this stable form: $-AX-XA=-B$ and the unique solution is $X=\int_0^{+\infty}e^{-tA}Be^{-tA}dt$.

EDIT .(answer to Lebron James). 1 $M_n(\mathbb{R})$ has dimension $n^2$; then $f$ has $n^2$ eigenvalues. We may assume that $A=diag((\lambda_i)_i)$. $f$ admits the $n^2$ elements $(E_{i,j})_{i,j}$ of the canonical basis of $M_n(\mathbb{R})$ as eigenvectors ($f(E_{i,j})=(\lambda_i+\lambda_j)E_{i,j})$. Since the $(\lambda_i)_i$ are $>0$, $f$ has positive eigenvalues and is a bijection and $AX+XA=B$ has a sole solution $X_0$.

2. $X_1=\int_0^{+\infty}e^{-tA}Be^{-tA}dt$ exists because $||e^{-tA}Be^{-tA}||=O(e^{-2\lambda t})$ where $\lambda$ is a minimal eigenvalue of $A$ (the matrix $-A$ gives the stability to infinity). If $Y=e^{-tA}Be^{-tA}$, then $Y'=-AY-YA$ and, after integration, $-B=-AX_1-X_1A$, that implies $X_1=X_0$.