PropositionSolutions of linear matrix equations

Let \(\Mtrx{A}\) and \(\Mtrx{C}\) be matrices of size \((m,n)\) and \((m,1)\), respectively. For an unknown \((n,1)\)-matrix \(\Mtrx{X}\), consider the matrix equation

\[(E)\qquad \Mtrx{A}\cdot \Mtrx{X} = \Mtrx{C}\]

together with the associated homogeneous equation

\[(E_0)\qquad \Mtrx{A}\cdot \Mtrx{X} = \ZMtrx{(m,1)}.\]

Let \(Z\) be the set of solutions of \((E_0)\), and let \(\Mtrx{U}\) be a particular solution of \((E)\). Then \(\Mtrx{X}\) solves \((E)\) if and only if

\[\Mtrx{X} = \Mtrx{U} + \Mtrx{Y}\]

for some \(\Mtrx{Y}\) in \(Z\).

Proof

Let us show first that every matrix \(\Mtrx{U}\) in \((E_0)\) yields a solution \(\Mtrx{X}\DefEq \Mtrx{U}+\Mtrx{Y}\)of \((E)\). Indeed,

\[\Mtrx{A}\cdot (\Mtrx{U}+\Mtrx{Y}) = \Mtrx{A}\cdot \Mtrx{U} + \Mtrx{A}\cdot \Mtrx{Y} = \Mtrx{C} + \ZMtrx{(m,1)} = \Mtrx{C}\]

Conversely, if \(\Mtrx{X}\) is a solution of \((E)\), then \(\Mtrx{Y}\DefEq \Mtrx{X}-\Mtrx{U}\) is a solution of \((E_0)\), because

\[\Mtrx{A}\cdot (\Mtrx{X}-\Mtrx{U}) = \Mtrx{A}\cdot \Mtrx{X} - \Mtrx{A}\cdot \Mtrx{U} = \Mtrx{C}-\Mtrx{C} = \ZMtrx{(m,1)}\]

But then \(\Mtrx{X} = \Mtrx{U} + \Mtrx{Y}\) is of the required form.