1. ## Bilinear Transformations

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

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

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

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

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

3. Well, thank you very much!

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

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

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