Using inverse to find eigenvalues

[ChiralSuperfields]

Homework Statement

Pleases ee below

Relevant Equations

Please see below

For this,

I don't understand how if $(A - 2I_2)^{-1}$ has an inverse then the next line is true.

Many thanks!

 

[fresh_42]

$$
A-2I =\begin{pmatrix}1&-2\\1&-2\end{pmatrix}-\begin{pmatrix}2&0\\0&2\end{pmatrix}=\begin{pmatrix}-1&-2\\1&-4\end{pmatrix}
$$
and thus $(A-2I)^{-1}=\dfrac{1}{6}\begin{pmatrix}-4&2\\-1&-1\end{pmatrix}$

So $A-2I$ is invertible. Since we have $A\cdot \begin{pmatrix}x\\y\end{pmatrix}=2I\cdot \begin{pmatrix}x\\y\end{pmatrix},$ we get
$$
A\cdot \begin{pmatrix}x\\y\end{pmatrix}-2I\cdot \begin{pmatrix}x\\y\end{pmatrix}= (A-2I)\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}
$$
Applying $(A-2I)^{-1}$ on both sides results in
$$
(A-2I)^{-1}\cdot (A-2I)\cdot \begin{pmatrix}x\\y\end{pmatrix}= I\cdot \begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}x\\y\end{pmatrix} =(A-2I)^{-1}\cdot \begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}
$$

 

[Mark44]

The thread title, "Using inverse to find eigenvalues," doesn't make sense to me. When your goal is to find the eigenvalues and eigenvectors for a given matrix A, the matrix expression you work with is by definition noninvertible. The process of finding eigenvalues involves a matrix expression whose determinant is zero; i.e., $|A - \lambda I| = 0$. The determinant of a invertible matrix is always nonzero.

The matrix expression $A - 2I_2$ in this thread turns out to be invertible precisely because 2 is not an eigenvalue of A. If the author of the material in the picture you uploaded has a point, it's not clear to me what it is.

The matrix A - 2I that fresh_42 shows is the same as the matrix A I found in your thread from yesterday, namely $A = \begin{bmatrix}-1 & -2 \\ 1 & -4 \end{bmatrix}$. I mentioned yesterday that the eigenvalues for this matrix expression happen to be -2 and -3.

The matrix $A + 2I_2 = A - (-2)I_2 = \begin{bmatrix}1 & -2 \\ 1 & -2 \end{bmatrix}$. Because the determinant of $A + 2I_2 = 0$, $A + 2I_2$ does not have an inverse. The same is true for the matrix $A + 3I_2 = A - (-3)I_2$.

The process of finding an eigenvalue $\lambda$ for a matrix A is this:

1. Write the matrix $A - \lambda I_n$, with n = 2 for 2 x 2 matrices, n = 3 for 3 x 3 matrices, and so on.
2. Set the determinant of $A - \lambda I_n$ to zero, and solve the resulting polynomial involving powers of $\lambda$.

 

[Mark44]

$$
A-2I =\begin{pmatrix}1&-2\\1&-2\end{pmatrix}-\begin{pmatrix}2&0\\0&2\end{pmatrix}=\begin{pmatrix}-1&-2\\1&-4\end{pmatrix}
$$
and thus $(A-2I)^{-1}=\dfrac{1}{6}\begin{pmatrix}-4&2\\-1&-1\end{pmatrix}$

The attached image in post 1 of this thread doesn't specify any particular matrix, so it's not possible to determine the entries of A - 2I. You might be confusing what was in the hand-drawn sketch of the previous thread from the OP, which itself was confused.

So $A-2I$ is invertible. Since we have $A\cdot \begin{pmatrix}x\\y\end{pmatrix}=2I\cdot \begin{pmatrix}x\\y\end{pmatrix},$ we get
$$
A\cdot \begin{pmatrix}x\\y\end{pmatrix}-2I\cdot \begin{pmatrix}x\\y\end{pmatrix}= (A-2I)\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}
$$
Applying $(A-2I)^{-1}$ on both sides results in
$$
(A-2I)^{-1}\cdot (A-2I)\cdot \begin{pmatrix}x\\y\end{pmatrix}= I\cdot \begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}x\\y\end{pmatrix} =(A-2I)^{-1}\cdot \begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}
$$

 

[fresh_42]

The attached image in post 1 of this thread doesn't specify any particular matrix, so it's not possible to determine the entries of A - 2I.

Occam's razor. Why complicate things? I do not debate typos.

However, the nice part is: it does not even matter! The line of the argument remains the same if $A$ has a different form as long as $A-2I$ remains regular!

 

[Mark44]

Occam's razor. Why complicate things?

The OP is already sufficiently confused as evidenced in the thread title, in thinking that finding the inverse of a matrix plays any role in finding eigenvalues. Muddying up the water by tossing in a specific matrix where none was given doesn't help alleviate that confusion.