How to go from $U^TS^TAC+C^TASU+2BSU+$ to $2(C^TAS+BS)U$?

Question

I'm reading Eves: Elementary Matrix Theory. Here:

I am trying to perform these computations to check it. I did the following:

$X^TAX+2BX+c=0$ $X:=C+SU$

$(C+SU)^TA(C+SU)+2B(C+SU)+c=0$

$(C^T+U^TS^T)A(C+SU)+2B(C+SU)+c=0$

$(C^TA+U^TS^TA)C+(C^TA+U^TS^TA)SU+2B(C+SU)+c=0$

$\require{cancel} \cancel{U^T(S^TAS)U}+U^TS^TAC+C^TASU+2BSU+\cancel{C^TAC+2BC+c}=0$

But I don't know how he gets to $2(C^TAS+BS)U$, (By the way, $2':=2^T$). I guess I should do: $(U^TS^TAC)^T=C^TA^TSU$ but it doesn't get the same. What did I do wrong or what did I miss?

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Answer 1

You did nothing wrong. Note that $C^TASU$ is a scalar (i.e., it's $1\times 1$), so it equals its own transpose. Finally, $A$ is a symmetric matrix, so $A^T=A$.