SOLUTION: Problem: Construct an orthogonal basis for R^4 that has no zeros in it and demonstrate that your basis is orthogonal.
I don't understand what it is asking and I have no idea how
Algebra.Com
Question 34257: Problem: Construct an orthogonal basis for R^4 that has no zeros in it and demonstrate that your basis is orthogonal.
I don't understand what it is asking and I have no idea how to get the problem going and solved.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
I AM GIVING BELOW ANSWER FOR R^4
X1 1 1 1 1
X2 1 -2 -1 1
X3 2 1 2 3
X4 1 0 0 0
Y1 1 1 1 1
SUM
X2.Y1 1 -2 -1 1 -1
Y1.Y1 1 1 1 1 4
Y2 1.25 -1.75 -0.75 1.25
X3.Y2 2.5 -1.75 -1.5 3.75 3
X3.Y1 2 1 2 3 8
Y2.Y2 1.5625 3.0625 0.5625 1.5625 6.75
Y3 -0.555555556 -0.222222222 0.333333333 0.444444444
X4.Y3 -0.555555556 0 0 0 -0.555555556
X4.Y2 1.25 0 0 0 1.25
X4.Y1 1 0 0 0 1
Y3.Y3 0.308641975 0.049382716 0.111111111 0.197530864 0.666666667
Y4 0.055555556 -0.111111111 0.166666667 -0.111111111
Y1.Y2 1.25 -1.75 -0.75 1.25 0
Y1.Y3 -0.555555556 -0.222222222 0.333333333 0.444444444 0
Y1.Y4 0.055555556 -0.111111111 0.166666667 -0.111111111 0
Y2.Y3 -0.694444444 0.388888889 -0.25 0.555555556 0
Y2.Y4 0.069444444 0.194444444 -0.125 -0.138888889 0
Y3.Y4 -0.030864198 0.024691358 0.055555556 -0.049382716 -9.02056E-17
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problem: Construct an orthogonal basis for R^4 that has no zeros in it and demonstrate that your basis is orthogonal.
I don't understand what it is asking and I have no idea how to get the problem going and solved.
OK..SHALL WE DO IT THIS WAY?I SHALL EXPLAIN THE MEANING AND METHODAND THE WORKING FOR R^3..YOU CONTINUE ON THAT FOR R^4 TO UNDERSTAND AND LEARN THE TEHNIQUE..OK!
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FIRSTLY WE ARE DEALING WITH VECTORS..YOU KNOW I THINK I,J,K AS UNIT VECTORS ALONG X,Y AND Z AXIS..
THEY CAN HAVE ANY DIMENSION ..THAT IS 1 OR 2 OR 3 OR 4 OR....N..IT MEANS FOR YOUR UNDERSTANDING
1 DIMENSION...ONLY ALONG A LINE SAY X AXIS..AS ALONG A CORNER LINE IN A ROOM...I IS THE EXAMPLE
2 DIMENSIONS...IN A PLANE SAY XY PLANE...AS IN A FLOOR OF A ROOM....I AND J ARE THE EXAMPLES
3 DIMENSIONS...IN A SPACE SAY SPANNED BY X,Y AND Z AXIS...AS IN A ROOM .....I,J AND K ARE THE EXAMPLES.
4 DIMENSIONS...IN A SPACE AND TIME SAY SPANNED BY X,Y,Z AND TIME AXIS...AS IN A ROOM AT DIFFERENT TIMES....I,J,K,T...COULD BE EXAMPLES
N DIMENSIONS....YOU HAVE TO ONLY IMAGINE THIS......
NOW NEXT IS BASIS...OR IN COMMON LANGUAGE SOME UNIT OF MEASUREMENT.AGAIN GOING BY DIMENSION...WE HAVE
1 DIMENSION..ALL POINTS OR SAY OUR HOUSES ARE ON ONE ROAD.WE ONLY NEED TO SAY
1 UNIT SAY 0.1 MILE FROM BEGINING OF THE ROAD,SEOND HOUSE IS 0.2 MILES FROM THE BEGINING ETC...WE DNOTE I AS UNIT VECTOR WHICH YOU MAY IMAGINE AS UNIT DISTANCE FROM START IN THAT DIRECTION.
2 DIMENSIONS...ALL HOUSES ARE NOW IN A RECTAGULAR ZONE..SO WE HAVE TO SPECIFY 2 DIRECTIONS ALONG MAIN ROAD AND CROSS ROAD.WE CAN SAY GO FROM START 0.1 MILE ALONG THE MAIN ROAD AND THEN GO 0.2 MILES ALONG A CROSS ROAD TO REACH A HOUSE...SO WE NEED 2 DIRECTIONS USUALLY CALLED X AND Y AXIS ..DENOTED BY I AND J VECTORS.
3 DIMENSIONS..HERE ALL HOUSES ARE IN APARTMENT BLOCKS SAY..SO WE HAVE TO REACH A BLOCK AS ABOVE AND THEN GO UP BY SOME DISTANCE TO REACH A PARTICULAR APARTMEN...SO WE NEED X,Y AND Z AXIS AND I,J,K AS 3 DIMENSIONED UNIT VECTORS
LIKE THIS WE CAN EXPAND THOUGH ,IT WOULD BE ONLY IMAGINATION AND IT IS DIFFICULT TO GIVE EXAMPLES AS ABOV.
SO THOSE MEASURING UNITS` IN DIFFERENT DIRECTIONS ARE CALLED BASIS...I,J,K..ETC.
OFCOURSE STRICTLY SPEAKING IN MATHS FOR THEM TO BE CALLED BASIS THEY SHOULD SATISFY 2 CRITERIA,...
1.THEY SHOULD BE INDEPENDENT....IT IS SO FOR I,J,K..SINCE A DISTANCE ALONG X AXIS CAN ONLY BE SHOWN BY I AND IT CANNOT BE REPRESENTED BY ANY COMBINATION OF J AND K..SIMILARLY FOR OTHER Y AND Z AXIS
2.TYHEY SHOULD BE SUFFIIENT TO SPAN ANY POINT / ECTOR IN THAT SPACE UNDER CONSIDERTATION....IT IS SO WITH I,J,K...FOR EXAMPLE TAKING THE EXAMPLE OF SPACE AS OUR ROOM..ANY POINT IN THAT CAN BE DESCRIBED AS GO FROM START SAY 2 UNITS ALONG X AXIS,THEN SAY 3 UNITS ALONG Y AXIS AND THEN SAY 5 UNITS ALONG Z AXIS ETC....
SO NOW WE HAVE DISCUSSED VECTORS,DIMENSIONS AD BASIS..NOW ON TO ORTHOGONAL BASIS..
IF BASIS VECTORS ARE ORTHOGONAL TO EACH OTHER WE SAY THEY FORM ORTHOGONAL BASIS..I,J,K ARE INDEED ORTHOGONAL SINCE THEY ARE ALOING X,Y,AND Z AXIS WHICH ARE MUTUALLY PERPENDICULAR TO EACH OTHER.AS YOU KNOW 2 VECTORS ARE ORTHOGONAL OR PE
RPENDICULAR IF THEIR DOT PRODUCT OR INNE OR SCALAR PRODUCT IS ZERO
WE FIND I.J=1*1*COS(90)=0..SO IS....J.K AND K.I ZERO
OK NOW WE GOT ALL THE DOUGH..SO LET US FIND HOW TO MAKE AN ORTHOGONAL BASIS IN 3 DIMENSIONS FROM ANY 3 GIVEN BASIS VECTORS IF THEY ARE NOT ORTHOGONAL..SUPPOSE I,J,K ARE GIVEN AS THE BASIS THEN THERE IS NOTHING ELSE TO BE DOBNE..THEY ARE ALREADY ORTHOGONAL AS ROVED ABOVE..SO LET US TAKE 3 VECTORS WHICH FORM A BASIS IN 3D (THAT IS THEY SATISFY THE 2 CONDITIONS MENTIONED ABOVE),BUT ARE NOT ORTHOGONAL AND HENCE WE HAVE TO MAKE A SET OF 3 ORTHONAL BAIS VECTORS FROM THEM..THAT IS AS A LINEAR COMBINATION OF THE 3 GIVEN VECTORS..OK..THE OBJECTIVE IS CLEAR I SUPPOSE!!
LET THE GIVEN BASIS BE...
X1=(1,1,1).............(THAT IS A VECTOR I+J+K..FOR YOUR UNDERSTANDING)
X2=(1,-2,1)
X3=(1,2,3)
WE TAKE IT THAT THEY ARE INDEPENDENT AND COVER ENTIRE SPAN OF R^3 AS THEY ARE GIVEN TO FORM BASIS..WE DONT PROVE IT..WE ONLY BUILD AN ORTHOGONAL BASIS WITH THESE 3 VECTORS
LET US SAY THE ORTHOGONAL BASIS BASED ON THESE IS Y1,Y2,Y3..WE HAVE TO FIND Y1,Y2,Y3 SO THAT Y1.Y2=Y2.Y3=Y3.Y1=0...AND THEY ARE ALL A LINEAR COMBINATION OF X1,X2 AND X3.
WE SAY FIRST ..Y1=X1...SO....Y1=*(1,1,1)
NOW FORMULA FOR Y2 IS
Y2=X2-Y1(Y1.X2)/(Y1.Y1)....
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{I SHALL SHOW THE PROOF HERE...THOUGH NOT ASKED FOR OR NEEDED...LET Y2=X2+AY1...SINCE Y1 AND Y2 ARE ORTHOGONAL...Y1.Y2=0...SO DOT BOTH SIDES WITH Y1...WE GET...
Y1.Y2=X2.Y1+AY1.Y1
0=X2.Y1+AY1.Y1...SO....A=-(X2.Y1)/(Y1.Y1)...OK...}
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Y2=(1,-2,1)-(1,1,1){(1,-2,1).(1,1,1)/(1,1,1).(1,1,1)}
=(1,-2,1)-(1,1,1){1*1+(-2*1)+1*1}/{1*1+1*1+1*1}
=(1,-2,1)-(12,1,1)*0/3=(1,-2,1)
NOW FORMULA FOR Y3 IS SIMILAR...
Y3=X3-Y2(Y2.X3)/(Y2.Y2) -Y1(Y1.X3)/(Y1.Y1).....YOU CAN WORK OUT AS ABOVE THE DOT PRODUCTS AND YOU WILL GET THE ANSWER AS
Y3=(1,2,3)-Y2(0/6)-(1,1,1)(6/3)
=(1,2,3)-(2,2,2)
=(-1,0,1)
SO THE ORTHOGONAL VECTORS CORRESPONDING TO THE GIVEN BASIS OF X1,X2 AND X3 ARE
Y1=(1,1,1)
Y2=(1,-2,1)
Y3=(-1,0,1)
YOU CAN CHECK THAT THEY ARE MUTUALLY ORTHOGONAL..THAT IS Y1.Y2=Y2.Y3=Y3.Y1=0
AND THEY ARE ALL OBTAINED AS LINEAR COMBINATIONS OF X1,X2 AND X3 AS THE FORMULAE FOR Y1 ,Y2 AND Y3 OBVIOUSLY PROVE...
NOW ON TO YOUR HOME WORK FOR R^4...HAPPY WORK..HOPE YOU WILL DO AND GET IT..IF STILL IN DOUBT OR DIFFICULTY PLEASE COME BACK....
venugopalramana
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