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?

  • 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

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 $3$.

  • $\begingroup$ 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

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 $T$.


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