Solving for a matrix in an equation with trace

Question

I want to solve the following equation for the $m\times m$ matrix $X$: $$2X^T(A^TA)=-2\mathrm{trace}(X)I+V,$$ where $I$ is the identity matrix, $A$ is a $t\times m$ known matrix, and $V$ is an $m\times m$ known matrix. Assume that $t$ is less than $m$ so, in this case, $A^TA$ is a singular matrix.

I looked at the question in Solving matrix equation involving trace but it did not help me because of the singularity of $A^TA$.

Also, in the same link, they that said this kind of equations can be solve with vectorization. I searched the internet but could not find anything about this topic: could someone help me with that please?

Thank you.

Answer 1

Define the variables $$\eqalign{ B &= A^TA \cr W &= \tfrac{1}{2}V^T \cr \tau &= {\rm tr}(X) \cr }$$ Divide your equation by $2$ and transpose it $$\eqalign{ BX &= W - \tau I \cr }$$ The general solution of this linear equation can be written in terms of the Moore-Penrose inverse $B^+$ and the nullspace projector applied to an arbitrary matrix $Y$. $$\eqalign{ X_s &= B^+(W-\tau I) + (I-B^+B)Y \cr }$$ Be aware that your equation might not have any solutions.
To see this, multiply each side by the transposed projector. $$\eqalign{ (I-BB^+)(BX) &= (I-BB^+)(W - \tau I) \cr 0 &= (I-BB^+)W - \tau (I-BB^+) \cr (I-BB^+)W &= \tau (I-BB^+) \\ V(I-B^+B) &= (2\tau) (I-B^+B) \cr }$$ A solution only exists when $W$ or $V$, satisfies the projected equation for some $\tau.\,$ So in some sense, the choice of $A$ restricts your choice of $V$; they are not independent of each other.

Another way of looking at it: every column of $(I-B^+B)$ must be a linear combination of the eigenvectors of $V$ associated with a single eigenvalue, $\lambda=2\tau$.

$\tau$ can be calculated by equating the traces. $$\eqalign{ {\rm tr}(X) &= {\rm tr}(X_s) \\ \tau &= B^+:W^T - \tau B^+:I + (I- B^+B)^T:Y \\ \tau\,(1+I:B^+) &= W^T:B^+ + (I-B^+B):Y \\ \tau &= \frac{W^T:B^+ + (I-B^+B):Y}{1+I:B^+} \\ }$$ Substitute this value into the expression for $X_s\\$

NB: In the above, a colon is used as a convenient product notation for the trace, i.e. $$A:B = {\rm tr}(A^TB)$$ Properties of the trace allow terms in a colon product to be rearranged in lots of ways. $$\eqalign{ A:B &= B:A \\ A:B &= A^T:B^T \\ A:BC &= B^TA:C = AC^T:B \\ }$$

Answer 2

Presumably the matrices are real. Let $r$ be the rank of $A$. Orthogonally diagonalise $A^TA$ as $Q(D\oplus0)Q^T$, where $D$ is a $r\times r$ positive diagonal matrix. Let also $$ X^T=Q\pmatrix{X_1&X_2\\ X_3&X_4}Q^T,\quad V=Q\pmatrix{V_1&V_2\\ V_3&V_4}Q^T, $$ where both $X_1$ and $V_1$ are $r\times r$. The equation can then be rewritten as $$ 2\pmatrix{X_1D&0\\ X_3D&0}=-2\operatorname{tr}(X)I+\pmatrix{V_1&V_2\\ V_3&V_4}, $$ i.e. $$ \pmatrix{X_1&0\\ X_3&0}=\pmatrix{\frac12\left(-2\operatorname{tr}(X)I_r+V_1\right)D^{-1}&V_2\\ \frac12V_3D^{-1}&-2\operatorname{tr}(X)I_{m-r}+V_4}. $$ Therefore, it is solvable if and only if $V_2=0$ and $V_4=2tI_{m-r}$ for some number $t$. If this is the case, the general solution is given by $$ X^T=Q\pmatrix{\frac12\left(-2tI_r+V_1\right)D^{-1}&X_2\\ \frac12V_3D^{-1}&X_4}Q^T $$ where $X_2$ is arbitrary and $X_4$ is any matrix whose trace is $\alpha=t-\frac12\operatorname{tr}\left(\left(-2tI_r+V_1\right)D^{-1}\right)$ (so that the trace of $X$ is $t$).

If you don't want to use partitioned matrices, you may rewrite the above solution as follows. Let $P=A^TA$. Then the equation is solvable if and only if $V(I-P^+P)=2t(I-P^+P)$ for some $t\in\mathbb R$. If this is the case, the general solution is given by $$ X^T=\frac12(-2tI+V)P^+ + \left(\alpha I + M - \frac1n\operatorname{tr}(M(I-P^+P))I\right)(I-P^+P) $$ where $M$ is an arbitrary matrix and $$ \alpha = t-\frac12\operatorname{tr}\left((-2tI+V)P^+\right). $$