# Using inverse to find eigenvalues

• ChiralSuperfields
In summary: Maybe the OP thought the discussion was about a particular matrix, but that's a pretty big assumption.
ChiralSuperfields
Homework Statement
Relevant Equations
For this,

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

Many thanks!

#### Attachments

• 1684729399030.png
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$$A-2I =\begin{pmatrix}1&-2\\1&-2\end{pmatrix}-\begin{pmatrix}2&0\\0&2\end{pmatrix}=\begin{pmatrix}-1&-2\\1&-4\end{pmatrix}$$
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
$$(A-2I)^{-1}\cdot (A-2I)\cdot \begin{pmatrix}x\\y\end{pmatrix}= I\cdot \begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}x\\y\end{pmatrix} =(A-2I)^{-1}\cdot \begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$

ChiralSuperfields
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##.

Last edited:
ChiralSuperfields
fresh_42 said:
$$A-2I =\begin{pmatrix}1&-2\\1&-2\end{pmatrix}-\begin{pmatrix}2&0\\0&2\end{pmatrix}=\begin{pmatrix}-1&-2\\1&-4\end{pmatrix}$$
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.
fresh_42 said:
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
$$(A-2I)^{-1}\cdot (A-2I)\cdot \begin{pmatrix}x\\y\end{pmatrix}= I\cdot \begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}x\\y\end{pmatrix} =(A-2I)^{-1}\cdot \begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$

ChiralSuperfields
Mark44 said:
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!

ChiralSuperfields
fresh_42 said:
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.

ChiralSuperfields

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