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Show that \(I - A\) is invertible when all the eigenvalues of \(A\) are less than 1 in magnitude. (Hint: What would be true if \(I - A\) were not invertible?)

Short Answer

Expert verified

It is proved that \(I - A\) is invertible when all the eigenvalues of \(A\) are less than 1 in magnitude.

Step by step solution

01

Definition of magnitude

The eigenvalue \(\lambda \) is the real or complex number of a matrix \(A\) which is a square matrix that satisfies the following equation

\(\det \left( {A - \lambda I} \right) = 0\).

This equation is called the characteristic equation.

02

Showing that \(I - A\) is invertible

If\(I - A\)is not invertible, then the equation\(\left( {I - A} \right){\rm{x}} = 0\)would have a non-trivial solution of\({\rm{x}}\).

Then\({\rm{x}} - A{\rm{x}} = 0\)and\(A{\rm{x}} = 1 \cdot {\rm{x}}\), which shows that\(A\)would have\(1\)as an

eigenvalue.

This cannot happen if all the eigenvalues are less than\(1\)in magnitude.

So\(I - A\)must be invertible.

It is proved that\(I - A\)is invertible when all the eigenvalues of\(A\)are less than 1 in magnitude.

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