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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?

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    $\begingroup$ That's right. $\,$ $\endgroup$
    – D_S
    Apr 16, 2017 at 18:46
  • $\begingroup$ Then I am just stuck there. I don't know how to show $B=0$ given that trace$AB=0$ for all $A\in V$ $\endgroup$
    – Kavita
    Apr 17, 2017 at 2:34
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    $\begingroup$ If you're stuck, try an example. Work it out in the two by two case, then the three by three case. $\endgroup$
    – D_S
    Apr 17, 2017 at 2:37

3 Answers 3

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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

$Tr(XY)=Tr(YX)$

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  • $\begingroup$ Edited to address your concern. I guess my proof was a bit terse after consideration. But the above is what I meant by "It is enough to check this on basis elements". $\endgroup$
    – user433011
    Apr 17, 2017 at 20:53
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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 $B = 0$.

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  • $\begingroup$ Then trace $AA^{T}$=0 which means $A$ must be zero matrix? Is it right what I M saying? $\endgroup$
    – Kavita
    Apr 17, 2017 at 2:44
  • $\begingroup$ That means I need to use now inner products . I guess the steps are easy here but it's a kind request for you to write your steps you are describing. Thanks $\endgroup$
    – Kavita
    Apr 17, 2017 at 2:49
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    $\begingroup$ Yes I got it. I am getting trace = sum of squares of each element in A =0 that means each entry should be zero $\endgroup$
    – Kavita
    Apr 17, 2017 at 2:57
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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.

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