Can a 2x2 Matrix Have a Cube Root Near a Given Matrix?

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Homework Statement

Prove or Disprove: A 2x2 matrix $$\epsilon A+I$$ has a cube root near $$\left( \begin{array}{cc} 2 & 0 \\ 0 & -5 \end{array} \right)$$.

Homework Equations

Inverse Function Theorem

The Attempt at a Solution

I'm just confused about the "near $$\left( \begin{array}{cc} 2 & 0 \\ 0 & -5 \end{array} \right)$$" bit. Is this the matrix $$A$$? Or is it the matrix $$H$$ in the derivative $$Df(A)(H)$$ of the function $$f(A)=A^3$$? Otherwise I know what to do.

 

[mighty2000]

The phrase "near \left( \begin{array}{cc} 2 & 0 \\ 0 & -5 \end{array} \right)" is referring to the desired cube root of the 2x2 matrix. In other words, the question is asking if there exists a matrix that, when cubed, will result in a matrix close to \left( \begin{array}{cc} 2 & 0 \\ 0 & -5 \end{array} \right).

To prove or disprove this, you can use the Inverse Function Theorem. First, define the function f(A) = A^3. Then, take the derivative of f(A) at the point A = \left( \begin{array}{cc} 2 & 0 \\ 0 & -5 \end{array} \right). This will give you Df(A) = 3A^2.

Next, use the Inverse Function Theorem to check if there exists a matrix H such that Df(A)(H) = I. If such a matrix H exists, then it can be considered a cube root of A.

If you are able to find a matrix H that satisfies this condition, then you can conclude that a cube root of A exists near \left( \begin{array}{cc} 2 & 0 \\ 0 & -5 \end{array} \right). Otherwise, you can disprove the statement.