How to find the condition of having a non-trivial solution for a non-square matrix equation?

Question

I have this matrix equation. $$ M\pmatrix{a\\b\\c\\d}=\pmatrix{0\\0}\ , $$ $$M=\pmatrix{5e^{\frac{n}{2}}&-(n-1) e^{-\frac{n}{2}}&-5n&3m\\ -5e^{-\frac{(n-4)}{2}}&3e^{\frac{n+4}{2}}&5e^{mn}&-3e^{-mn}}.$$ I need to obtain the condition of having a non-trivial solution, i.e. the condition when $\operatorname{det} (M)=0$. But the matrix is non-square. Is there a way to find the determinant or solvability condition in this case?

Answer 1

We can use $\det(M)$ only for square matrix to check the existence of not trivial solutions. More in general $Mx=0$ has a not trivial solutions when column vectors of $M$ are not independent. Since in this case we have $4$ column vectors $\in \mathbb R^2$, for any fixed pair $(m,n)$ we always have a subspace with at least dimension $2$ of not trivial solutions.

Answer 2

When a matrix is non-square and the number of rows (equations) is less than the number of columns (variables), the equation $A \vec{x} = \vec{0}$ has infinitely many solutions (and at least $k$ linearly independent solutions), where $k$ is the number of columns minus the number of rows.

In the exact case you have, in fact there are non-trivial solutions to the equation $A \vec{x} = \vec{b}$, no matter your choice of $\vec{b}$. This is because $A \vec{x}$ returns a vector in the column space (space spanned by $A$'s columns) of $A$, and $A$ has $2$ linearly independent columns (meaning $A$'s column space is $\mathbb{R}^2$ itself, because any two linearly independent vectors in $\mathbb{R}^2$ span it). We can go further and observe that $A \vec{x} = \vec{b}$ has infinitely many solutions for all choices of $\vec{b} \in \mathbb{R}^2$, because if $\vec{x}^*$ is a solution to $A \vec{x} = \vec{b}$, then $A(\vec{x}^* + \lambda \vec{z}) = \vec{b}$, for any scalar $\lambda$ and nontrivial $\vec{z}$ such that $A \vec{z} = \vec{0}$ (which in the previous paragraph we showed must exist).