How to compute the induced $||\cdot||_{2} $ matrix norm of an SPD matrix Hi, I know they are related questions on the board but mine is more specific. Although the    answer for any non-singular matrix would be also interesting. Thanks!
UPDATE: I am sorry I though this was clear, but as I know it the $||\cdot||_{2}$ norm is defined as follow: Let be $A\in \mathbb{R}^{m\times n};\;||A||_{2} =  \displaystyle{max_{x \in \mathbb{R}^n} \frac{||Ax||_{2}}{||x||_{2}}}$
 A: To lay the question to rest, let me do two things: (i) restate it; (ii) answer it.
By $\|x\|$, we mean the Euclidean 2-norm throughout.

Show that the induced 2-norm $$\max_{\|x\|\not= 0} \frac{\|Ax\|}{\|x\|}$$ is given by $\sqrt{\lambda_{\max}(A^TA)}$

The proof is textbook material. For the lazy, here is an informal sketch.
Notice that since without loss of generality, we may rescale vector $x$, hence we may equivalently consider maximizing $\|Ax\|$ such that $\|x\|=1$.
Consider, $\|Ax\|^2 = x^TA^TAx$. The matrix $A^TA$ is SPD, so it has the eigendecomposition $V\Lambda V^T$, where $\Lambda$ is a nonnegative diagonal matrix. Thus, we have $x^TA^TAx = x^TV\Lambda V^Tx = y^T\Lambda y = \sum_i \lambda_i y_i^2$. This, implies that $\|Ax\|^2 \le \lambda_{\max}y^Ty = \lambda_{\max}x^TV^TVx=\lambda_{\max}$ because $V^TV=I$ and $x^Tx=1$.
To conclude the proof we now need to show that in fact $\|Ax\|^2 = \lambda_{\max}$. But this is trivial, because picking $x=v_{\max}$ (eigenvector corr to max eigenvalue), we attain this equality.
PS: Other proofs based on Lagrange multipliers etc. can also be given, but ultimately one needs to invoke something like $A^TAx=\lambda x$ at some point.
A: I fill bad about this but I found my answer thanks to the comment of Yemon Choi!  After looking for operator norm on Wikipedia I got that: $||A||_{2} = \sqrt{\lambda_{max}(A^*A)}$ where $A^*$ is the conjugate transpose of $A$ (but since in my question I asked only for the values of $A\in\mathbb{R}$ it's only the tranpose) and $\lambda_{max}(B)$ is the largest eigenvalue of the matrix $B$. If someone can give me a proof for the real case, I will vote for his answer as the correct one (if I am allowed to do that in the rules since I'm slightly changing the question).
A: For square matrices, isn't the induced 2-norm equivalent to the largest singular value of the matrix? Knowing this, you would use the optimal algorithm to find that value given knowledge that your matrix is SPD.
