Understanding Eigenvalues of a Matrix

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Homework Statement
Please see below
Relevant Equations
Please see below
For this,
1684540765113.png

I am confused by the second line. Does someone please know how it can it be true since the matrix dose not have an inverse.

Many thanks!
 
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Why do you think it is true?
 
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ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For this,
View attachment 326802
I am confused by the second line.
Me, too. I don't know what this means. ##A\cdot 0=0## no matter whether ##A## is invertible or not.
ChiralSuperfields said:
Does someone please know how it can it be true since the matrix dose not have an inverse.

Many thanks!
The matrix doesn't have an inverse. If we want to prove this by contradiction then we assume it has an inverse. Say the matrix is ##A.## Then ##A## maps the entire space ##\mathbb{R}^2## onto itself: ##A\cdot v= w.## Now, if we set ##v=\begin{pmatrix}x\\ y\end{pmatrix}## then ##A\cdot v=\begin{pmatrix}-x+2y\\-x+2y\end{pmatrix}.## But this means that both coordinates of ##w## are the same and we have no chance to get any other vector with different coordinates. So the image of ##A## is one-dimensional, not two-dimensional, so it cannot be invertible.

You can also argue with a vector in the kernel of ##A##. Can you name one for which ##A\cdot v=0## while ##v\neq 0?##
 
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The statement in the first two lines is vacuously true: if a singular matrix has an inverse, then the equality holds. The equality has no meaning because ##^{-1}## doesn't exist for this matrix.
 
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The part in the lower half of the image doesn't make much sense in the context of what you're asking.

You have ##\begin{bmatrix} 1 & -2 \\ 1 & -2\end{bmatrix} = A - 2I##
If you work this out, you find that ##A = \begin{bmatrix} -1 & -2 \\ 1 & -4\end{bmatrix}##.

Since you're asking about eigenvalues for a matrix (presumably A, above), it turns out that the eigenvalues are -2 and -3. This means that for one eigenvector ##x_1##, ##Ax_1 = -2x_1##, and for the other eigenvector ##x_2##, ##Ax_2 = -3x_2##.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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