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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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 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 $v$ Tags linear algebra, math, proof, property, prove Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Trinity Algebra 3 May 11th, 2014 07:12 PM WWRtelescoping Complex Analysis 2 January 27th, 2014 03:38 AM jozou Number Theory 2 July 22nd, 2011 11:02 AM elim Real Analysis 1 November 25th, 2010 12:39 PM

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