# Upper bound of a norm of an inverse of a matrix

I'm looking for a way to bound the norm of an inverse of a matrix (specifically considering infinity-norm). I know that the norm of an inverse of a matrix has the relationship,

$||A^{-1}|| \geq \left(||A|| \right)^{-1}$ (Eq. 1)

but this provides a lower bound. Are there any useful equations or identities that could form the upper bound on this norm instead? (For example, in which situations does the equality for Eq. 1 hold?)

I do not have knowledge of eigenvalues/singular values of the matrix (that's why I'm using the infinity-norm instead). The matrix is complex and has the form $A := \hat{I} - {\mathrm i} \alpha B$, where $\hat{I}$ is the identity, ${\mathrm i}$ is the imaginary unit, $\alpha$ is a real constant, and $B$ is a matrix with real elements and a symmetric structure.

Attempt: I know we can split up the norm of a sum of a matrix into the sum of norms of matrices, i.e.

$||A+B|| \leq ||A|| + ||B||$. (Eq. 2)

While this provides an upper bound on $(||A||)^{-1}$ if we use Eq. 1 to decompose the norm of an inverse matrix, I don't know how it relates to the upper bound on $||A^{-1}||$.

• Related problems will involve estimation of the condition number of a matrix, usually defined as $||A|| ||A^{-1}||$ for a convenient matrix norm. Note that your Eq. 1 amounts to the fact that the condition number is at least $1$, giving that lower bound, and conversely any upper bound on the condition number of a matrix would give an analogous upper bound on the norm of the inverse. – hardmath Apr 27 '18 at 5:29

Let $$\Delta=B-A$$, where $$A$$, $$B$$ and $$\Delta$$ are complex square matrices. In this scenario I want to compute a bound on $$\|A^{-1}-B^{-1}\|$$ as a function of $$\|\Delta\|=\|A-B\|$$.

Since $$A^{-1}\Delta B^{-1}=A^{-1}\left(B-A\right) B^{-1}=A^{-1} - B^{-1}$$, it follows that

\begin{align} \| A^{-1} - B^{-1} \| &= \|A^{-1}\Delta B^{-1}\| \\ &\leq \|A^{-1}\Delta\| \| B^{-1}\| \ , \tag 1 \label{eq:1} \end{align}

due to the fact that matrix norms are submultiplicative. Furthermore, since $$B^{-1}=A^{-1}-A^{-1}\Delta B^{-1}$$ we can use the triangle inequality and submultiplicative property (once again) to obtain

\begin{align} \|B^{-1}\| &\leq \|A^{-1}\|+\|A^{-1}\Delta B^{-1}\| \\ &\leq \|A^{-1}\|+\|A^{-1}\Delta\| \|B^{-1}\| \ , \end{align}

which is equivalent to

\begin{align} \|B^{-1}\| &= \|\left(A+\Delta \right)^{-1}\| \\ &\leq \frac{\|A^{-1}\|}{1-\|A^{-1}\Delta\|} \ . \tag 2 \label{eq:2} \end{align}

Finally, if we plug \eqref{eq:2} back in \eqref{eq:1} we get the desired upper bound

\begin{align} \|A^{-1}-B^{-1}\|&\leq\frac{\|A^{-1}\|\|A^{-1}\Delta\|}{1-\|\|A^{-1}\Delta\|} \\ &\leq \frac{\|A^{-1}\|\|A^{-1}\|\|\Delta\|}{1-\|A^{-1}\Delta\|} \\ &\leq \frac{\|A^{-1}\|^2\|A-B\|}{1-\|A^{-1}\Delta\|} \ . \end{align}

Full disclosure: This answer is based on the beginning of section "5.8 Condition numbers: inverses and linear systems" from "Matrix Analysis" by Roger A. Horn and Charles R. Johnson (2nd Edition, p. 381).

• This argument requires an additional assumption: namely that $\|A^{-1} \Delta \| < 1$. Without this condition the inequality in (2) is reversed. See (5.8.0) from Horn & Johnson (2012). – inhuretnakht Oct 2 '20 at 16:08