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Found in: Page 267

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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?)

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

See the step by step solution

Step 1: 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.

Step 2: 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.