Matrices - homogeneous system

Question

Hello I’m taking the course 18.02 multivariate calculus of MIT. I don’t understand the thinking behind the answer of a problem.

The question is (with Julia notation):

For what c-value(s) will $$\left(\begin{matrix}2& 1\\ 0 &-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right) = c\left(\begin{matrix}x\\y\end{matrix}\right)$$ have a non-trivial solution? (Write it as a system of homogeneous equations)

The answer is:

(1): $\left(\begin{matrix}(2-c)x+y\\(-1-c)y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$ has a non-trivial solution if

(2): $\left|\begin{matrix}(2-c)&1\\0&(-1-c)\end{matrix}\right|=0$ i.e., if:

(3): $(2-c)(-1-c) = 0$ or $c=2, c=-1$

I don’t understand how we come from the question to (1) or (2). Which theorem is used?

Thank you very much in advance

Answer 1

$A\mathbf x = c\mathbf x\\ A\mathbf x - c\mathbf x = 0\\ A\mathbf x - cI\mathbf x = 0\\ (A\mathbf - cI)\mathbf x = 0$

Either $\mathbf x = \mathbf 0$ (the trivial solution) or $A - cI$ is singular, and $\det (A - cI) = 0$

Answer 2

We have (1), which is $$\left(\begin{matrix}(2-c)x+y\\(-1-c)y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$ This can be rewritten as $$\left(\begin{matrix}(2-c)&1\\0&(-1-c)\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$ There are now two ways to understand why the determinant of this matrix must be zero:

(a) - Suppose it does not have zero determinant. Then the matrix is invertible, so you can multiply by its inverse on both sides of this equation. Then you would get $(x,y)=(0,0)$, which is not a non-trivial solution (which is what you are asked for). So our assumption that it has non-zero determinant must be false.

(b) - You are looking for a non-trivial vector multiplied by a matrix to give the zero vector. This vector must then be an eigenvector of this matrix, with eigenvalue $0$. The determinant of a matrix is the product of the eigenvalues of the matrix, so if we have a $0$ eigenvalue, then the determinant must also be zero.

Therefore we get (2) $$\left|\begin{matrix}(2-c)&1\\0&(-1-c)\end{matrix}\right|=0$$

Answer 3

The theorem behind is Rouché–Capelli theorem which states that

A system of linear equations with n variables has a solution if and only if the rank of its coefficient matrix $A$ is equal to the rank of its augmented matrix $[A|b]$.If there are solutions, they form an affine subspace of $\mathbb {R} ^{n}$ of dimension $n − rank(A)$.

In particular:

• if $n = rank(A)$, the solution is unique,

• otherwise there is an infinite number of solutions.

In this case we are looking for no trivial solutions of $(A-cI)x=0$, that is for solutions with $x\neq 0$, and that require that $$rank(A)<n \iff \det(A-cI)=0$$