(A−λI)x=0 and x≠0 iff det(A−λI)=0: Why [[1,1],[1,1]][[2],[3]] = [[5],[5]] ≠ 0 when det([[1,1],[1,1]]) = 0?

Question

As refered to Why non-trivial solution only if determinant is zero,

I wonder why \begin{gather} \begin{bmatrix} 1 & 1 \\1 & 1 \end{bmatrix} \begin{bmatrix} 2 \\3 \end{bmatrix} = \begin{bmatrix} 5 \\5 \end{bmatrix} ≠ \begin{bmatrix} 0 \\0 \end{bmatrix} \end{gather} when \begin{gather} \det \begin{pmatrix} \begin{bmatrix} 1 & 1 \\1 & 1 \end{bmatrix}\end{pmatrix} = 0 ?\end{gather} Above is conflict to the fact that "$(A−\lambda I)x=0$ has a nontrivial solution (a solution where $x\neq 0$) if and only if $\det(A−\lambda I)=0$ " to me.

Answer 1

The determinant of that matrix $A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$is indeed $0$ and there is indeed a non-trivial solution for $Ax=0$ , namely $x=\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ and its multiples.

Why you get $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ into it, I don't know. Your computation shows that $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ is not an eigenvector, and mine shows that $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ is.

Answer 2

First, the determinant of the transformation matrix has little to do with finding the eigenvalues or eigenvectors. It is the determinant of $A - \lambda I$ that matters. Here, you apply the theorem that a homogeneous system $AX = 0$, where $A$ is an $n \times n$ matrix, has a non-trivial solution if and only if the determinant of $A$ is $0$. Note that using 0-determinant as a criteria works for homogeneous system only. For in-homogeneous system $AX=b$, where $A$ is an $n \times n$ matrix and $b$ is not $0$, the system has a unique solution if the determinant of $A$ is not $0$. The criteria for homogeneous and in-homogeneous systems are exactly opposite.

Going back to your matrix

$$A=\begin{bmatrix} 1 \ 1 \\ 1 \ 1 \end{bmatrix}$$

there is indeed no contradiction with the theorem that the homogeneous system $AX=0$ has a non-trivial solution if and only if the determinant of the $n \times n$ matrix $A$ is $0$.

In the context of eigenvalues, this is your homogeneous system:

$$(A - \lambda \cdot I)X = 0$$

$X$ is non-trivial if and only if $det(A-\lambda I)$ is $0$. You actually have two eigenvalues by solving this homogeneous system: $\lambda_1 = 0$ and $\lambda_2 = 2$.

Substituting $\lambda_1=0$ back into the homogeneous system $(A - \lambda I) X = 0$, you have

$$\begin{bmatrix} {1-0} \ 1 \\ 1 \ {1-0} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$

As long as $x_1$ is the opposite of $x_2$, this system will be true. So, $\lambda_1=0$ gives you infinitely many eigenvectors $\begin{bmatrix} t \ -t \end{bmatrix}$, i.e., $\begin{bmatrix} 1 \ -1 \end{bmatrix}$ or $\begin{bmatrix} -1 \ 1 \end{bmatrix}$ or scalar multiples of them.

Substituting $\lambda_2=2$ back into the homogeneous system as above, you got infinitely many eigenvectors as well $\begin{bmatrix} t \ t \end{bmatrix}$, i.e., $\begin{bmatrix} 1 \ 1 \end{bmatrix}$ or scalar multiples.

As shown, given the transformation matrix $\begin{bmatrix} 1 \ 1 \\ 1 \ 1 \end{bmatrix}$, the theorem for solutions of homogeneous square matrix system is not contradicted.