In a first course of linear algebra it is usually establish that the space of linear operators on $\mathbb R^n$ is isomorphic to the space of $n \times n $ matrices over $\mathbb R^n $. If we let the operators have the operator norm $||T||_{op }:=\sup_{||x||=1 }||Tx|| $, where $|| \ ||$ is the euclidean norm, and the spcace of matrices the Frobenius norm $||(a_{ij })_{1 \le i,j \le n }||_{Fr }:=(\sum _i ^n \sum _j ^n a_{ij } ^2)^{1/2 } $ are the normed spaces isometric?
Suppose $n=3 $ the matric of the linear operator $$T(x_1,x_2,x_3)=(a_1x_1+a_2x_2+a_3x_3,b_1x_1+b_2x_2+b_3x_3,c_1x_1+c_2x_2+c_3x_3)$$ is given by $\begin{pmatrix}
a_1 & a_2 & a_3 \\
b_2 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{pmatrix} $ with norm $\sqrt {a_1 ^2 +a_2 ^2+…+c_2^2+c_3^2}$. How would I then show this is the least upper bound of the set of numbers $$\sqrt {(a_1x_1+a_2x_2+a_3x_3)^2 + (b_1x_1+b_2x_2+b_3x_3)^2+(c_1x_1+c_2x_2+c_3x_3)^2} $$
where $\sqrt {x_1^2+x_2^2+x_3^2 } =1$?
Thanks in advance!
Best Answer
Hint The Frobenius norm $||{\,\cdot\,}||_{Fr}$ is induced by the inner product $(A, B) \mapsto \operatorname{tr}(B^{\top} A)$ on the space of $n \times n$ matrices over $\Bbb R$, but it's straightforward to produce an example that shows that (for $n > 1$) the operatorname $||{\,\cdot\,}||_{op}$ does not satisfy the Parallelogram Law and hence does not arise from an inner product.