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If a matrix $A_{n\times n}$ is invertible, then

$\left\|A^{-1}\right\|_2 = \dfrac{1}{\min\limits_{i} \sigma_i}$

where $\sigma_i$ is the $i$-th singular value of $A$

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    $\begingroup$ If this is a homework question (and it looks like something that appears verbatim in every linear algebra textbook ever written), you must tag it as such and show that you have put in reasonable effort. $\endgroup$
    – Jed Brown
    Jan 2, 2014 at 18:58

2 Answers 2

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Singular value decomposition of $A$ can be written as $A=USV^T$ with $S$ being a diagonal matrix of singular values of $\sigma_i$.

Euclidean norm of a matrix can be written as : $||A||_2=\sigma_{max}(A)$, meaning that the norm is the maximum singular value.

If $A$ is invertible, then $A^{-1}=(USV^T)^{-1}$.

From here it follows:

$$A^{-1}=(USV^T)^{-1} =(V^T)^{-1}S^{-1}U^{-1} $$

If $A$ is a real valued matrix, then from orthogonality the expression further reduces to: $$A^{-1}=VS^{-1}U^T.$$

This new matrix $A^{-1}$ now has singular values $S^{-1}$, and its norm would be $\max(\text{diag}(S^{-1}))$, where $\text{diag}$ takes the diagonal of the matrix. Because $S$ is a diagonal matrix, its inverse is computed by simply inverting every single element. Therefore,

$||A^{-1}||_2=\sigma_{max}(A^{-1})=\max\limits_{i} \text{diag}(S^{-1})_i = \max\limits_{i}\frac{1}{\sigma_i}=\frac{1}{\min\limits_{i}{\sigma_i}}$ and this completes the proof.

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    $\begingroup$ Of course, you've started with $\| A \|_{2}=\sigma_{max}(A)$. It isn't clear whether the original poster is aware of that fact or is willing to assume it. $\endgroup$ Jan 2, 2014 at 15:35
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    $\begingroup$ Well, considering the structure of the question, I guess this is a fair starting point as it is not explicitly stated. If desired it could also be interpreted either geometrically (cs.princeton.edu/courses/archive/spring12/cos598C/…) or algebraically using the induced norm $||A||_2=\max{\frac{||Ax||}{||x||}}$, where $x$ is real and non-zero. This is only a special case of $L_p$ norm. If $||x||_2=1$ is assumed, the maximizer is the largest singular value. $\endgroup$ Jan 2, 2014 at 15:55
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I think I'd rather see it done this way. Since $A$ is invertible, then $\min_i\sigma_i>0$, $$ \min_i\sigma_i = \inf_{x\neq 0}\frac{\|Ax\|_2}{\|x\|_2} \quad\Longleftrightarrow \frac{1}{\min_i\sigma_i} = \sup_{x\neq 0}\frac{\|x\|_2}{\|Ax\|_2}. $$ Then we have $$ \frac{1}{\min_i\sigma_i} = \sup_{x\neq 0}\frac{\|x\|_2}{\|Ax\|_2} = \sup_{A^{-1}y\neq 0}\frac{\|A^{-1}y\|_2}{\|y\|_2} = \sup_{y\neq 0}\frac{\|A^{-1}y\|_2}{\|y\|_2} = \|A^{-1}\|_2. $$ where we have made the substitution $Ax=y$, and utilized the fact that $A^{-1}y=0\Longleftrightarrow y=0$, again since $A$ is invertible.

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