# Thread: Bilinear Transformations

1. ## Bilinear Transformations

$\displaystyle Let f : F^{nxn} x F^{nxn} -> F$ as follows : $\displaystyle F(A,B)=tr(AB).$

Prove that there is no matrix A (other than the 0 matrix) for which$\displaystyle f(A,B)=0$for every B.

2. Originally Posted by adam63
Let $\displaystyle f : F^{nxn} x F^{nxn} -> F$ as follows : $\displaystyle f(A,B)=tr(AB).$

Prove that there is no matrix A (other than the 0 matrix) for which $\displaystyle f(A,B)=0$ for every B.
let $\displaystyle A=[a_{ij}]$ and suppose that $\displaystyle \{e_{ij} : \ 1 \leq i,j \leq n \}$ is the standard basis for the vector space of all $\displaystyle n \times n$ matrices over $\displaystyle F.$ now $\displaystyle \text{tr}(AB)=0,$ for all $\displaystyle B,$ if and only if $\displaystyle \text{tr}(Ae_{ij})=0,$ for all

$\displaystyle 1 \leq i,j \leq n.$ but $\displaystyle Ae_{ij}=\sum_{k=1}^n a_{ki}e_{kj}$ and thus $\displaystyle \text{tr}(Ae_{ij})=a_{ji}.$ therefore $\displaystyle \text{tr}(Ae_{ij})=0,$ for all $\displaystyle 1 \leq i,j \leq n,$ if and only if $\displaystyle A=0.$

3. Well, thank you very much!

BTW, are you sure there is no other way, more 'elegant'?

4. Originally Posted by NonCommAlg
let $\displaystyle A=[a_{ij}]$ and suppose that $\displaystyle \{e_{ij} : \ 1 \leq i,j \leq n \}$ is the standard basis for the vector space of all $\displaystyle n \times n$ matrices over $\displaystyle F.$ now $\displaystyle \text{tr}(AB)=0,$ for all $\displaystyle B,$ if and only if $\displaystyle \text{tr}(Ae_{ij})=0,$ for all

$\displaystyle 1 \leq i,j \leq n.$ but $\displaystyle Ae_{ij}=\sum_{k=1}^n a_{ki}e_{kj}$ and thus $\displaystyle \text{tr}(Ae_{ij})=a_{ji}.$ therefore $\displaystyle \text{tr}(Ae_{ij})=0,$ for all $\displaystyle 1 \leq i,j \leq n,$ if and only if $\displaystyle A=0.$
Originally Posted by adam63
Well, thank you very much!

BTW, are you sure there is no other way, more 'elegant'?
Now, I am not the most knowledgeable linear algebraist (? :S) in the world, but I now an "elegant" proof when I see one. That was an elegant proof.

5. Originally Posted by Drexel28
Now, I am not the most knowledgeable linear algebraist (? :S) in the world, but I now an "elegant" proof when I see one. That was an elegant proof.
Thanks !

Just thought there would be something more 'verbal', or an assumption that would lead to a contradiction.

6. Originally Posted by adam63
Well, thank you very much!

BTW, are you sure there is no other way, more 'elegant'?
it's math, there's always another way! besides, looking for (possible) more elegant proofs is tempting when the problem is not straightforward.