I'm trying to work through how to calculate eigenvalues and eigenvectors.
I start with
$$Ax=\lambda x$$
Where $A$ is a $p \times p$ matrix, $\lambda$ is the eigenvalue and $x$ is the eigenvector.
This is the same as:
$$Ax=I\lambda x$$
$$Ax-I\lambda x=0$$
$$(A-I\lambda) x=0$$
We define the matrix $A$ as a $2 \times 2$ matrix:
$\begin{bmatrix}4 & -2\\-3 & 6\end{bmatrix}$
Thus this -$I\lambda$ equals
$\begin{bmatrix}4-\lambda & -2\\-3 & 6-\lambda\end{bmatrix}$
$$Det(A-I\lambda)=(4-\lambda(6-\lambda)-(-3)*-2)$$
$$Det(A-I\lambda)=24-10\lambda +\lambda^2 -6$$
$$Det(A-I\lambda)=18 – 10\lambda + \lambda^2 $$
Then, out of the blue my textbook claims that
$$0=30 – 10\lambda + \lambda^2 $$
How do I justify setting the determinant to $0$?
(I do "not" have an advanced knowledge in linear algebraic analysis, I only know how the determinant is used to calculate the inverse matrix)
Best Answer
The text is not claiming that the determinant is $0$. The text is saying "Let's find out for which values of lambda the determinant is $0$!"
So the determinant is $\lambda^2 - 10\lambda + 30$, and you want to find the $\lambda$ such that it is equal to zero. What do you do? You set it equal to zero and solve for $\lambda$. That is, you solve the equation
$$\lambda^2 - 10\lambda + 30 = 0$$
As for why you are interested in the values of $\lambda$ that make the determinant equal to $0$, remember that
$$rank(A-\lambda I) = n \iff det(A - \lambda I) \neq 0$$
So, if $det(A-\lambda I) \neq 0$, you will find that the only solution to $(A - \lambda I)x = 0$ is $x = 0$ (due to the fact that the rank of the matrix is full, hence the kernel only contains the $0$ vector). This means that the only $x$ such that $Ax = \lambda x$ is $x=0$, which means that $x$ is not an eigenvector.
So the only way to have eigenvectors is to have the determinant of $A - \lambda I$ be equal to zero, so that's why to find eigenvalues you look for the values of $\lambda$ that make $det(A - \lambda I) = 0$