How do I find $\|T\|$ when given a matrix $T$? How do I find the norm $\|T\|$ of T: $\mathbb{R}^3 \rightarrow  \mathbb{R}^3$ is defined by $T(x) := Ax$, where $A:= \begin{pmatrix}
0 & 2 & 0 \\
1 & 0 & 0 \\
0 & 0 & 3 \end{pmatrix} $.
I was thinking of writing $T(x_1,x_2,x_3)=(2x_2,x_1,3x_3)$
and then $\|T\|$ =$((2x_2)^2+(x_1)^2+(3x_3)^2))^{1/2}$.
Do I find $\|T\|$ like this or not?
 A: No. By definition
$$\|T\|_{2\to2} = \max_{\|x\|_2 = 1} \|Tx\|_2$$
Now this special operator norm can be found by taking the square root of the largest eigenvalue of $T^H T$.
In this case, since $T$ is a permutation of a diagonal matrix, you can simply read it off as $3$.
A: First, $\|T\|$ is a single number describing the size of the $3\times3$ matrix.
It cannot depend on any coordinate $x_1,x_2,x_3$.
You can solve the problem by writing $|T(x)|=((2x_2)^2+(x_1)^2+(3x_3)^2))^{1/2}$ and playing with that, but let me give a more elaborate answer.
There are several possible norms you could use on the space of $3\times3$ matrices, and all of them are comparable to each other.
I assume you mean the operator norm, defined by
$$
\|T\|
=
\max_{|x|=1}|Tx|.
$$
There are many equivalent definitions out there.
(Usually the definition is given with the supremum rather than the maximum, but for a linear map between finite dimensional spaces the maximum is always reached.)
I also assume that you use the standard Euclidean norm for the vectors.
You could just take this constrained optimization problem and try to solve it, but there is a simpler way.
Instead of just stating the result, let me explain how to find it.
Your matrix is so simple that one could solve the problem faster with a different method, but a method that works just as easily for any $3\times3$ matrix might be more enlightening.
From the definition one sees that
$$
\|T\|^2
=
\max_{|x|=1}|Tx|^2
=
\max_{|x|=1}\langle x,T^\dagger Tx\rangle,
$$
where ${}^\dagger$ stands for transpose.
(The same argument goes through in the complex case as well if you use the conjugate transpose.)
The matrix $T^\dagger T$ is symmetric, so it can be orthogonally diagonalized.
Orthogonal transformations don't change norms of the vectors and therefore they don't change the operator norms of matrices.
Changing the basis suitably, we can assume $T^\dagger T$ to be diagonal, say $\text{diag}(a,b,c)$.
Now
$$
\langle x,T^\dagger Tx\rangle
=
ax_1^2+bx_2^2+cx_3^2
$$
and we have $x_1^2+x_2^2+x_3^2=1$.
I hope it is clear enough now that $\|T\|^2$ is the largest eigenvalue of $T^\dagger T$.
Therefore $\|T\|$ is the square root of the largest eigenvalue of $T^\dagger T$.
In your case we have $T^\dagger T=\text{diag}(1,4,9)$.
The largest eigenvalue of this is 9, so $\|T\|=3$.
I will leave it as an exercise to check that if the columns of $T$ are orthogonal (as they now are), then $T^\dagger T$ is diagonal and the eigenvalues are the squared norms of the column vectors.
Therefore $\|T\|$ is simply the largest norm of the column vectors.
Your matrix has orthogonal columns, so $\|T\|=3$.
Note:
I identified $A$ with $T$ in my notation above and considered $x$ to be a column vector.
If you find this confusing, let me know and I can try to make a distinction between $A$ and $T$.
