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Suppose that $A$ is a $3 \times 3$ real matrix such that for each $u=(u_1, u_2, u_3)’ \in \mathbb{R}^3, \: u’Au=0$ where $u’$ stands for the transpose of $u$. Then which one of the following is true?

  1. $A’=-A$
  2. $A’=A$
  3. $AA’=I$
  4. None of these
in Linear Algebra
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1 Answer

3 votes
Is the answer (A) ?

$\begin{bmatrix} u1 &u2 &u3 \end{bmatrix} \begin{bmatrix} a1 &a2 &a3 \\ a4&a5 &a6 \\ a7& a8 & a9 \end{bmatrix} \begin{bmatrix} u1\\ u2\\ u3 \end{bmatrix}$

$(u1a1 + u2a4 + u3a7)*u1 + (u1a2 + u2a5 + u3a8)*u2 + (u1a3 + u2a6 + u3a9)*u3 = 0$

 

$u1^{2}a1 + u1u2(a4 + a2) + u1u3(a7 + a3) + u2^{2}a5 + u2u3(a8 + a6) + u3^{2}a9 = 0$

This can only be possible when the matrix A is skew symmetric. Because all the diagonal elements $(a1,a5,a9)$ will be 0. And all the non diagonal elements will be negation of each other.$(a2 = -a4, a3= -a7, a6 = -a8)$
0
Nice !

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