Let N(.) be a norm on $\mathbb{R}^n$. For a $n \times n$ matrix, A nor is defined as $$||A|| = \underset{x \neq 0}{sup}\ \frac{N(Ax)}{N(x)}$$

Assume $$||A|| < 1$$. Show that $I-A$ and $I+A$ are non singular. Also show that $$\frac{1}{1+||A||} \leq ||(I-A)^{-1}|| \leq \frac{1}{1-||A||}$$

My Approach: I have already proved that the given matrix norm is the norm along with some other results like

  • $N(Ax) \leq ||A||N(x)$ for all $x$.
  • $||AB|| \leq ||A||\ ||B||$

Let $B = (I-A)^{-1}$ and assume it exists.

$$||B|| \geq \frac{1}{||I-A||}$$ $$||B|| \geq \frac{N(x)}{N((I-A)x)}$$ $$||B|| \geq \frac{N(x)}{N(x-Ax)}$$ $$||B|| \geq \frac{N(x)}{N(x) + N(Ax)}$$ $$||B|| \geq \frac{1}{1 + ||A||}$$

For II inequality $$B = (I-A)^{-1}$$ $$(I-A)B = (I-A)(I-A)^{-1}$$ $$B - AB = I$$ $$B = I + AB$$ Take matrix norm both side $$||B|| = ||I + AB||$$ $$||B|| \leq ||I|| + ||AB||$$ $$||B|| \leq 1 + ||A||\ ||B||$$ $$||B|| \leq \frac{1}{1-||A||}$$ I am able to prove the bounds of the norm. How can I prove that $(I-A)$ and $(I+A)$ are non-singular?


1 Answer 1


Because $$N(Ax) \le \|A\| N(x) < N(x)$$ it is impossible for $Ax=x$ or $Ax=-x$ to occur, so $A-I$ and $A+I$ are nonsingular.

There is a lot of "hand-waving" in your proof of the norm bounds.

By the sub-multiplicative property $||AB\| \le \|A\| \|B\|$, we have $$\|(I - A)^{-1}\| \ge \frac{1}{\|I-A\|} \ge \frac{1}{1 + \|A\|}$$ where the last inequality is due to the triangle inequality for the matrix norm ($\|A+I\| \le \|A\| + \|I\|$).

For the other direction, for any $x \ne 0$, $$N(x) = N((I-A)(I-A)^{-1} x) \ge N((I-A)^{-1} x) - \|A\| N((I-A)^{-1} x) = (1-\|A\|)N((I-A)^{-1} x).$$ Dividing both sides by $N(x)$ and taking the supremum over $x \ne 0$ proves the other inequality.

  • $\begingroup$ In which step (or steps) do you think I need to specify a theorem or which are too direct? $\endgroup$
    – AxyuS
    Aug 12, 2020 at 6:33
  • $\begingroup$ A few issues: a) $\|B\| \ne \frac{1}{\|I-A\|}$, b) where did the supremum go? c) What is $x$ and why is it appearing sometimes and disappearing other times? $\endgroup$
    – angryavian
    Aug 12, 2020 at 6:35
  • $\begingroup$ agree with your first point, I had to use the inequality. $x$ is appearing because I am using the definition of the matrix norm defined in the first few lines of the question. I $\endgroup$
    – AxyuS
    Aug 12, 2020 at 6:44

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