4
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ there are several different norms on $\mathbb{R}^9$ you can use. Which one do you want to compute? $\endgroup$ – demitau Jun 23 '15 at 12:28
2
$\begingroup$

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 $3$.

$\endgroup$
  • $\begingroup$ I did not quite get it @AlexR and it seems more like an algebra then analysis solution $\endgroup$ – abcdef Jun 23 '15 at 12:26
  • $\begingroup$ @user123 Well, the spectral norm is defined as I said and it's the naturally induced norm. Maybe you are supposed to find the Frobenius norm $\|T\|_F$? Also it would help me if you say what part you don't understand so I can try to clarify. $\endgroup$ – AlexR Jun 23 '15 at 12:28
1
$\begingroup$

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 $T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.