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 June 17th, 2019, 02:32 PM #1 Newbie   Joined: Jun 2019 From: London Posts: 13 Thanks: 0 proof of property Let a 3 × 3 matrix A be such that for any vector of a column v ∈ R3 the vectors Av and v are orthogonal. Prove that At + A = 0, where At is the transposed matrix.
 June 17th, 2019, 07:42 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 do you mean $A^TA=0$ Consider the 3x3 identity matrix. Clearly all the column vectors are orthogonal, and $Iv = v$ $I^T=I$ $I+I= 2I \neq 0$
 June 17th, 2019, 09:51 PM #3 Member     Joined: Oct 2018 From: USA Posts: 89 Thanks: 61 Math Focus: Algebraic Geometry I may have read this wrong, but I don't believe $I_{3 \times 3}$ can be $A$ since it would mean $Iv$ is orthogonal to $v$.
June 18th, 2019, 01:54 PM   #4
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Quote:
 Originally Posted by Greens I may have read this wrong, but I don't believe $I_{3 \times 3}$ can be $A$ since it would mean $Iv$ is orthogonal to $v$.
Yeah, A is definitely not an identity matrix. I think that maybe A is zero matrix. Or maybe I have forgotten some theorem or criteria.

Last edited by skipjack; July 1st, 2019 at 03:44 PM.

 June 18th, 2019, 03:34 PM #5 Member     Joined: Oct 2018 From: USA Posts: 89 Thanks: 61 Math Focus: Algebraic Geometry This was somewhat long so I wouldn't be surprised if there's an error in here somewhere so read critically. Let $A = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$ and $v= \begin{bmatrix} v_1 \\ v_2\\ v_3 \end{bmatrix}$ Two vectors are othogonal if their dot product is $0$, so we know $Av \cdot v = 0$ $Av = \begin{bmatrix} a_1 v_1 + a_2 v_2 + a_3 v_3 \\ b_1 v_1 + b_2 v_2 + b_3 v_3 \\ c_1 v_1 + c_2 v_2 + c_3 v_3 \end{bmatrix}$ $Av \cdot v = v_1 (a_1 v_1 + a_2 v_2 + a_3 v_3)\\ + v_2 (b_1 v_1 + b_2 v_2 + b_3 v_3)\\ + v_3 (c_1 v_1 + c_2 v_2 + c_3 v_3) = 0$ after a bit of simplifying: $Av \cdot v = a_1 v_{1}^{2} + b_2 v_{2}^{2} + c_3 v_{3}^{2} + v_1 v_2 (a_2 + b_1) + v_1 v_3 (a_3 + c_1) + v_2 v_3 (b_3 + c_2) = 0$ So , if $a_1 , b_2 , c_3 = 0$ and $b_1 = -a_2$ , $c_1 = -a_3$ , and $c_2 = -b_3$ then $Av \cdot v = 0$ So $A=\begin{bmatrix} 0 & a_2 & a_3 \\ -a_2 & 0 & b_3 \\ -a_3 & -b_3 & 0 \end{bmatrix}$ And you then have $A^{t} + A = 0$ Thanks from topsquark and mathodman25
 July 1st, 2019, 07:53 AM #6 Newbie   Joined: Jun 2019 From: London Posts: 13 Thanks: 0 But how do you get that a1 , b2 , c2= 0?
July 1st, 2019, 01:09 PM   #7
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Quote:
 Originally Posted by Greens $Av \cdot v = a_1 v_{1}^{2} + b_2 v_{2}^{2} + c_3 v_{3}^{2} + v_1 v_2 (a_2 + b_1) + v_1 v_3 (a_3 + c_1) + v_2 v_3 (b_3 + c_2) = 0$
If $a_1 , b_2 , c_3 = 0$ , then $a_1 v_{1}^{2} + b_2 v_{2}^{2} + c_3 v_{3}^{2} =0$ for any $v$

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