I have a linear system such that:


It is easy to understand that if $\mathbf{X}$ is arbitrary, i.e. it has the structure: $$\mathbf{X}=\begin{bmatrix}x_1&x_2\\x_3&x_4\\x_5&x_6\\x_7&x_8\end{bmatrix}$$

The system is solvable if $\mathbf{y}$ belongs to the column-space of $\mathbf{A}$. So I verify that according to Rouchè-Capelli theorem:

$$rank(\mathbf{A})=rank([\mathbf{A} | \mathbf{y}])$$

Now, I am confused, in the case $\mathbf{X}$ has the prescribed structure: $$\mathbf{X}=\begin{bmatrix}x_1&0\\0&x_2\\x_3&0\\0&x_4\end{bmatrix}$$

which is the condition for the system being solvable? I am looking for something similar to the first case of matrix $\mathbf{X}$.

Note: $\mathbf{A}$ has 2 rows and 4 columns while $\mathbf{y}$ has 2 rows and 1 column and $\mathbf{X}$ is my unknown so I can't use its entries in the computation of such condition.


1 Answer 1


First note that $(I-AX)y = 0$. So, either $AX=I$ or $AX = yy^T / (y^T y)$. In the prescribed structure case $AX = [a_1 x_1 + a_3 x_3 ~~~~ a_2 x_2 + a_4 x_4]$ where $a_i$ are the columns of $A$. So a solution exists if and only if

$$\left( [1 ~~ 0]^T \in \operatorname{Im} A_{13} ~~\text{and}~~ [0 ~~ 1]^T \in \operatorname{Im} A_{24} \right) ~~\text{or}~~ \left( y \in \operatorname{Im} A_{13} ~~\text{and}~~ y \in \operatorname{Im} A_{24} \right)$$

where $A_{13} := [a_1 ~~ a_3]$ and $A_{24} := [a_2 ~~ a_4]$.

  • $\begingroup$ Nice idea! Do you have some references or more insight about the last part, i.e. "So a solution exists if and only if...." it is not clear to me why do you use the "and" $\endgroup$
    – me8
    Commented Nov 1, 2021 at 10:42
  • 1
    $\begingroup$ This is because we need to find $X$ such that $AX=I$ or $AX=yy^T/(y^Ty)$. Looking at the structure of $AX$ we can see that this is only possible with the given condition. Because the first column of $AX$ can be selected any point on the span of vectors $a_1$, $a_3$. Similarly, the second column. $\endgroup$
    – obareey
    Commented Nov 1, 2021 at 10:49

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