To show a function is bilinear symmetric non degenerate form Let $V$ be vector space of set of $n×n$ matrices over $R$. Define $\langle A,B \rangle = \mathrm{trace}(AB)$, $A$, $B$ in $V$. show that $\langle \ \ ,\  \rangle$ is a non degenerate symmetric bilinear form.
Now succeeded in showing that the function  $\langle \ \ ,\  \rangle$ is a symmetric bilinear form by checking the properties of bilinear form but now to show that it is non degenerate I need help.
I know a function is nondegenerate if left radical or right radical of  $\langle \ \ ,\  \rangle$ is zero.
That means for a bilinear form $B$ on $V$ , if 
I get $S=\{y\in V \ |\  B(x,y)=0, \mbox{for all } x\in V\}=\{0\}$ then $B$ is nondegenerate. 
So if I start with  $\langle A, B \rangle =0$
Then $\mathrm{trace}(AB)=0$ for all $A\in V$
My claim is to show $B=0$?
Am I going right?
 A: First let's check non-degeneracy. It is enough to check this on basis elements
Let $E_{ij}$ be the nxn matrix with a $1$ in the $(i,j)$ position and zero elsewhere.
Then $\langle E_{ij},E_{ji}\rangle=Tr(E_{ii})=1$
So the form is non-degenerate.
To be more expicit: We can write any nonzero matrix $A$ as
$A=\sum_{i,j} a_{ij}E_{ij}$
Since A is nonzero, at least one $a_{kl}$ is nonzero.
So that $Tr(AE_{kl})=a_{kl}$
Billinearity follows from the fact that if $X,Y,Z$ are nxn matrices and $\lambda\in\mathbb{R}$
$Tr(\lambda XY)=Tr(X\lambda Y)=\lambda Tr(XY)$
and 
$\langle X+Z,Y\rangle =Tr((X+Z)Y)=Tr(XY+ZY)=Tr(XY)+Tr(ZY)=\langle X,Y\rangle +\langle Z,Y\rangle$
Symmetry follows from
$Tr(XY)=Tr(YX)$
A: In the case of the proof you have started, you are assuming $B$ is in the radical, so $\mathrm{trace}(AB)=0$ for all $A$. In particular, if we take $A = B^{T}$ then we have that $\mathrm{trace}(B^{T}B) = 0$.
So let's consider what this tells us.  If we have $b_{1}$, $\ldots$, $b_{n}$ as the columns of $B$, then
$$B^{T}B = \begin{bmatrix} b_{1}^{T}b_{1} & b_{1}^{T}b_{2} & \cdots & b_{1}^{T}b_{n}\\
b_{2}^{T}b_{1} & b_{2}^{T}b_{2} & \cdots & b_{2}^{T}b_{n}\\
\vdots & \vdots & \ddots & \vdots\\
b_{n}^{T}b_{1} & b_{n}^{T}b_{2} & \cdots & b_{n}^{T}b_{n}
\end{bmatrix}$$
so $\mathrm{trace}(B^{T}B) = \sum_{i=1}^{n} b_{i}^{T}b_{i}$.  Each of these terms is nonnegative, and is equal to $0$ if and only if $b_{i} = 0$ (since this is the standard inner product on $V$). Therefore if this sum is $0$, then all of the $b_{i}=0$, and so $B = 0$.
A: That is indeed the goal. My first instinct is to use elementary matrices, which can add a multiple of one row to another (this is the realization of Gaussian elimination in terms of matrix multiplication). You can find an $A$ such that $AB$ consists entirely of $1$'s and $0$'s, with at least one $1$ on the diagonal. This gives us a positive trace.
