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can we say if $\|AB - I\|<1$ then $\|BA - I\|<1$ for some arbitrary norm. I am trying to make counter example but I stuck please help me.

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Note that if we let $B=\left(\begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array} \right)$, $\ \ A=\left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$, then: $$BA= \left(\begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array} \right)\cdot \left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)=\left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right) $$ while: $$AB= \left(\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)\cdot \left( \begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array} \right)=\left( \begin{array}{cc} 0 & 2 \\ 0 & 0 \end{array} \right) $$ The existence of zero divisors, allows $||BA-I||=||I||$. So, if you set a norm with $||I||=1$ and $||AB-I ||< 1$, then your relation is not true.

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Consider the matrices: $A = \begin{bmatrix} 0.8 & 0.2 \\ 0.3 & 0.5 \end{bmatrix}$ and $B = \begin{bmatrix} 0.2 & 0.1 \\ 0.3 & 0.7 \end{bmatrix}$.

Then, we have: $\| AB - I\|_1 = 0.99$ and $\|BA - I\|_1 = 1.26$.

PS. I was just testing some random matrices.

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For given matrices $A$ and $B$, it depends on the norm you use (see this for examples of matrix norms), so it is important to specify your norm. We have already seen in other answers that for some norms this is false. For the Frobenius norm, namely $\|A\|_F \overset{def}= \sqrt{\mathrm{tr}(A^{\top}A)}$, this is true for symmetric matrices $A$ and $B$ ; this follows from the fact that the Frobenius norm is symmetric, i.e. $\|A^{\top}\|_F = \|A\|_F$, and $(AB-I)^{\top} = BA-I$. For any symmetric norm the result is true on symmetric matrices. Other instances of symmetric norms include the $L_{p,q}$ norms with $q=p$ (see this again, it is described there).

Otherwise producing counter examples is just a matter of crunching numbers ; you don't need to be smart, the identity will generally fail. Just try a bunch of numbers until it works.

Hope that helps,

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As the other answers nicely show we cannot say that your claim holds for all norms. But if we set $\| A \| = \max\{|\lambda_i|\ \ | \ \lambda_i \in \sigma(A) \}$, $\sigma(A)$ stands for the spectrum of $A$ , we can show that $\|AB \| = \|BA\|$. This is simply because $\sigma(AB) = \sigma(BA)$. Subtracting identity is just shift of the spectrum so $\sigma(AB-I) = \sigma(BA-I)$ and also $\| AB - I \| = \| BA - I \|$.

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  • $\begingroup$ Note that your answer is a special case of mine, because your norm is symmetric! $\endgroup$ – Patrick Da Silva Mar 5 '16 at 3:37
  • $\begingroup$ @PatrickDaSilva Not quite a special case. Yes, the norm Korf uses here is symmetric, but the equality here holds for arbitrary $A$ and $B$, not just for symmetric matrices. $\endgroup$ – user1551 Mar 5 '16 at 6:54
  • $\begingroup$ @user1551 : True! I missed that. +1 $\endgroup$ – Patrick Da Silva Mar 5 '16 at 7:10

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