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Suppose we have $A,B \in \operatorname{Mat}(n, \mathbb{R})$, and let $\alpha(A,B) = \text{Tr}(AB)$ and $\beta(A,B) = \text{Tr}(AB^T)$. Then is it possible to show that both $\alpha$ and $\beta$ cannot be bilinear forms? If not then what can be shown as a counter-example?

Furthermore, how can I show that only $\beta$ gives an inner product space on $\operatorname{Mat}(n, \mathbb{R})$ but $\alpha$ doesn't? How can it be shown that β(A,A)≥0 but that this is not true for α ? I assume this is the condition for the inner product which α does not satisfy but β does?

My attempts

I have tried to use the definition of bilinear forms paired with proof by contradiction, I first assumed that both α and β are bilinear forms and then have tried to use the definition to show this cannot be the case at the same time, however I am not sure how I can do this in a general form.

I did this for both α and β and found that they both satisfy the conditions for being bilinear forms, but I fail to understand how one of α or β being a bilinear form contradicts the other being bilinear too.

Anne Bauval
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John Doe
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  • What have you tried? – user469053 Mar 28 '24 at 11:41
  • @user469053 I have tried to use the definition of bilinear forms paired with proof by contradiction, I first assumed that both $\alpha$ and $\beta$ are bilinear forms and then have tried to use the definition to show this cannot be the case at the same time, however I am not sure how I can do this in a general form. – John Doe Mar 28 '24 at 11:47
  • For $\gamma:\text{Mat}(n,\mathbb{R})^2\to \mathbb{R}$, to check that $\gamma$ is a bilinear form, we want to check that for a fixed $B$, $A\mapsto \gamma(A,B)$ is linear, and for a fixed $A$, $B\mapsto \gamma(A,B)$ is linear.

    For a fixed $B$, linearity of $A\mapsto \gamma(A,B)$ means that for any $A,A_1,A_2$ and any scalar $a$, $$\gamma(aA,,B)=a\gamma(A,B)$$ and $$\gamma(A_1+A_2,B)=\gamma(A_1,B)+\gamma(A_2,B).$$ Similar for a fixed $A$. How much of this did you accomplish?

     – user469053 Mar 28 '24 at 11:56
  • I did this for both $\alpha$ and $\beta$ and found that they both satisfy the conditions for being bilinear forms, but I fail to understand how one of $\alpha$ or $\beta$ being a bilinear form contradicts the other being bilinear too – John Doe Mar 28 '24 at 12:10
  • Whether or not one of them is a bilinear form doesn't affect whether or not the other is.

    What about the inner product part?

     – user469053 Mar 28 '24 at 12:24
  • @user469053 I am getting that $\alpha$ is an inner product while $\beta$ isn't, because $\beta(A,B) = Tr(AB^T) \neq Tr(BA^T) = \beta(B,A)$. Is this a correct reasoning? – John Doe Mar 28 '24 at 12:38
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    Note that $(AB^T)^T=BA^T$, so these two matrices have the same main diagonal and the same trace. Therefore $\beta$ is symmetric. – user469053 Mar 28 '24 at 12:53
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    We should get that $\alpha,\beta$ are both symmetric. $\beta$ is an inner product but $\alpha$ is not. – user469053 Mar 28 '24 at 12:53
  • How can it be shown that $\beta(A,A) \geq 0$ but that this is not true for $\alpha$? I assume this is the condition for the inner product which $\alpha$ does not satisfy but $\beta$ does? – John Doe Mar 28 '24 at 14:39
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    That's correct. It helps if we write out the definitions in terms of coordinates. Let $A^i_j$ be the row $i$, column $j$ entry of $A$, and similar for $B$. Then $$[AB^T]^i_k=\sum_{j=1}^n A^ij [B^T]^j_k = \sum_{j=1}^n A^i_j B^k_j.$$ The trace is $$\sum_{i=1}^n [AB^T]^i_i=\sum_{i=1}^n\sum_{j=1}^n A^i_j B^i_j,$$ which is just like a dot product. If $A=B$, it's the sum of squares, so it can't be negative. – user469053 Mar 28 '24 at 15:20
  • If we know that $AA^T>0$ for $A\neq 0$, then if $A^T=-A$, $$AA=-A(-A)=-AA^T<0.$$ So this should tell us how to find a candidate to show that $\alpha$ is not an inner product. – user469053 Mar 28 '24 at 15:22

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