# How do you deduce that matrices are equal

i wanted to ask for a clarification.

I was looking around in my linear algebra text when i reached this justification:

Considered two coloumn vectors $$X$$ and $$Y$$, and assume $$X ^tC Y = X^t C^t Y$$ for every $$X,Y \in V$$, with $$V$$ an $$n$$-dimensional vectorial space, with $$X^t, Y^t$$ being the transposed of $$X, Y$$.

My book says that because of this is valid for every $$X,Y$$, we can deduce: $$C^t = C$$

Now, it is intuitively true, but i was wondering if ,maybe the general sum (it's a bilinear form) could equals without needing $$C^t = C$$. I've seen this type of justification also in other theorems, but i want to know if there is a way to prove it formally, beacuse i'm not satisfied.

• See what happens if $X=e_i$ and $Y=e_j$ ($i$-th and $j$-th standard basis vectors of $V$). Oct 23 at 0:32
• What does this have to do with the “null” in the title? Oct 23 at 0:40
• The transpose of a linear transformation doesn’t make sense over a general vector space. But it does make sense if the vector space is given an inner product. Oct 23 at 0:43
• Sorry, language translations Oct 23 at 0:45

since these are column vectors, evidently $$V=\mathbb F^n$$

select $$n$$ linearly independent vectors $$\mathbf x_k$$ and $$n$$ linearly independent $$\mathbf y_k$$ (if you like you can set $$\mathbf y_k:=\mathbf x_k$$)

Then for $$k,j \in \big\{1,2,...,n\big\}$$
$$\mathbf x_k^T C\mathbf y_j=\mathbf x_k^T C^T\mathbf y_j$$
$$\implies\mathbf x_k^T\big( C-C^T\big)\mathbf y_j = \mathbf 0$$
$$\implies \mathbf X^T\big( C-C^T\big)\mathbf Y = \mathbf 0$$
where $$\mathbf X$$ and $$\mathbf Y$$ have column $$j$$ given by $$\mathbf x_j$$ and $$\mathbf y_j$$ respectively. But $$\mathbf X$$ and $$\mathbf Y$$ are invertible,
$$\implies \big( C-C^T\big) = \mathbf 0$$
$$\implies C=C^T$$

Setting $$A = C-C^t$$ you have $$X^tAY=0$$ for all $$X,Y$$. Let $$e_j$$ denote the $$j$$-th standard basis vector. The $$(i,j)$$-th entry of $$A$$ can be expressed as $$e_i^tAe_j$$. But this is zero, hence $$A=0$$, i.e., $$C^t = C$$.