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For example, consider the matrix $B$: \begin{bmatrix} 5 & -2\\ -2 & 2 \end{bmatrix} computation of the 2-norm of $B$ using its eigenvalues (which is $6$ and $1$) yields $\sqrt{6}$ (since 6 is the largest eigenvalue of $B$).

But Matlab norm(B,2) yields 6.0000

This is obviously the answer without the square root. What am I missing here? Is Matlab not square rooting the answer?

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    so should 6 be consider the 2-norm or $\sqrt{6}$? – 2017-02-16
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    @Moo you mean the *square root* of the largest eigenvalue of $A^HA$ (which is $6$, of course). – 2017-02-16
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    @Moo well, you did say "max eigenvalue of $A^HA$", which isn't correct, hence the correction. – 2017-02-16
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    Check the definition of $2$-norm of a matrix. $6$ looks correct to me. – 2017-02-17

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I hope it can help you: $$\bbox[yellow,5px] {\lVert B \rVert_{2}=\sqrt{\text {maximum eigenvalue of $B^TB$} }}$$

$$B=\begin{bmatrix} 5 & -2\\ -2 & 2 \end{bmatrix}$$ $$B^TB=\begin{bmatrix} 29 & -14\\ -14 & 8 \end{bmatrix}$$

\begin{aligned}|[B^TB]-\lambda I|&=\left|{\begin{bmatrix}29&-14\\-14&8\end{bmatrix}}-\lambda {\begin{bmatrix}1&0\\0&1\end{bmatrix}}\right|={\begin{vmatrix}29-\lambda &-14\\-14&8-\lambda \end{vmatrix}},\\&=\lambda ^{2}-37\lambda +36.\end{aligned} $$\lambda=\{1,36\}$$ $$\text{max $\lambda$}=36$$

$\lVert B \rVert_{2}=\sqrt{\text {maximum eigenvalue of $B^TB$} }=\sqrt{36}=6$