Are there any known results on how big the gap between the absolute value of the largest Eigen value of matrix and the induced norm can be?

More formally, let the induced norm of A is given by $\|A\| = max_{\|x\| = 1}\|Ax\|$ and let $\lambda_{max}$ denote the largest Eigenvalue. I am interested in the quantity $\|A\| - |\lambda_{max}|$. Since the norm bounds the absolute value of the Eigenvalues, the quantity $\|A\| - |\lambda_{max}|$ is always positive. I also know that for positive-definite matrices, the quantity $\|A\| - |\lambda_{max}|$ is zero. But are there any results known for a generic matrix $A$?

p.s: I am working on a problem where I am trying to compute a bound on the norm of a matrix. I have bounds on the Eigenvalues of my matrix via Gresghorin's circle theorem and I am trying to see whether I can use that in some way to obtain a bound on the matrix norm....

Edit: To clarify, A is a square matrix over the field of reals and I am using the standard 2-norm on $R^n$

  • $\begingroup$ Is $A$ square, and is the vector norm the standard 2-norm for vectors? $\endgroup$
    – rajb245
    Apr 30, 2014 at 20:41
  • $\begingroup$ Yup, A is a square matrix and it is the standard 2-norm on the vectors.. $\endgroup$ Apr 30, 2014 at 20:42
  • $\begingroup$ In this case then they are the same. The norm is the largest eigenvalue (up to a $\pm$) $\endgroup$
    – rajb245
    Apr 30, 2014 at 20:48
  • $\begingroup$ You say $A$ is a real matrix - but you do take complex eigenvalues into account, right? Because if you don't, you can e.g. find a transform $T$ in $\mathbb{R}^3$ whose largest eigenvalue is $1$ but whose norm is arbitrarily large. Just let $T$ be the identity on some line through the origin, and a rotation plus scaling on the plane orthogonal to that line... $\endgroup$
    – fgp
    Apr 30, 2014 at 20:51
  • $\begingroup$ I don't think so. It is only true when $A$ is a symmetric matrix. The norm can also be expressed as the square root of the largest singular value which is the square root of the largest Eigenvalue of the matrix $A^TA$. This is not the same as the largest Eigenvalue of $A$... $\endgroup$ Apr 30, 2014 at 20:52

2 Answers 2


Presumably you are talking about the induced 2-norm, i.e. the largest singular value. When the spectral radius of $A$ is fixed, the 2-norm can be unbounded, as illustrated by the example $A=\pmatrix{1&n\\ 0&1}$. We have $\rho(A)=1$ but $\|A\|_2$ (and in turn the gap $\|A\|_2-\rho(A)$) approaches infinity when $n\to\infty$.

  • $\begingroup$ That was a very useful example. Thanks. I am going to leave the question up in the air just in case someone posts something interesting. If nothing shows up, I will accept yours as an answer after a while. $\endgroup$ Apr 30, 2014 at 20:58

Your two notions of norm are equal. If the maximum absolute value eigenvalue of $A$ is $\lambda$ and you take any vector $w$ of length $1$, then you will have $||Aw|| \leq |\lambda| ||w|| = |\lambda|$, and you will have equality when you take $w$ to be an eigenvector of $A$ with eigenvalue $\lambda$.

  • 1
    $\begingroup$ No, not true. The error is, that $\|Aw\|\leq |\lambda|\|w\$ is not true (why should it?). $\endgroup$
    – Dirk
    Jan 24, 2017 at 22:54
  • $\begingroup$ yes, @Dirk is correct. What you have written user is only true for eigenvectors and eigenvalues, but not true in geenral. $\endgroup$
    – makansij
    Oct 13, 2017 at 15:56

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