SOLUTION: I need to find the basis for the null space, the range of the matrix, and the orthogonal basis using Gram-Schmidt. The matrix is: 1 -2 1 -5 2 1 7 5 1 -1 2 -2

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 Question 61084This question is from textbook Introduction to Linear Algebra : I need to find the basis for the null space, the range of the matrix, and the orthogonal basis using Gram-Schmidt. The matrix is: 1 -2 1 -5 2 1 7 5 1 -1 2 -2This question is from textbook Introduction to Linear Algebra Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!I WAS WAITING FOR YOUR RESPONSE..ANY WAY I MADE MY OWN ASSUMPTIONS OF THE REQUIREMENTS AND WORKED OUT THE SOLUTION BELOW ---------------------------------- I need to find the basis for the null space, NULL SPACE? WHY?ARE YOU SURE? the range of the matrix, RANGE? ARE YOU SURE? OR IS IT RANK? PLEASE CONFIRM REQUIREMENTS AS THEY LOOK DOUBTFULL! and the orthogonal basis using Gram-Schmidt. SOLUTION:NULL SPACE , RANK ,ORTHONORMAL BASI The matrix is: 1 -2 1 -5 2 1 7 5 1 -1 2 -2 V1= 1 -2 1 -5 V2= 2 1 7 5 V3= 1 -1 2 -2 NR2=R2-2*R1 ; NR3=R3-R1 1 -2 1 -5 0 5 5 15 0 1 1 3 NR2=R2/5 1 -2 1 -5 0 1 1 3 0 1 1 3 NR3=R3-R2 1 -2 1 -5 0 1 1 3 0 0 0 0 NR1=R1-R2 1 -3 0 -8 0 1 1 3 0 0 0 0 RANK = 2 CONSIDERING THE SET OF EQNS. AX = 0,WE HAVE ONLY 2 INDEPENDENT EQNS. X1-2X2+X3-5X4=0…………….1 X1-X2+2X3-2X4=0……………2 EQN.2-EQN.1 GIVES X2+X3+3X4=0…..3 HENCE ONE SET OF SOLUTION COULD BE X2=1,X3=2,X4=-1..AND X1=-5…..THAT IS (-5,1,2,-1) = N1 AND ANOTHER SET OF SOLUTION COULD BE ..X2=2,X3=1,X4=-1 AND X1=-2..THAT IS (-5,2,1,-1)=N2 N1 AND N2 BEING 2 INDEPENDENT SOLUTIONS WILL FORM THE BASIS FOR NULL SPACE. HENCE THE 3 VECTORS HAVE 2 INDEPENDENT VECTORS. HENCE WE CAN CHOSE ANY 2 INDEPENDENT VECTORS TO FORM BASIS LET US TAKE V1 AND V3 WHICH ARE INDEPENDENT TO FORM THE BASIS. Y1=V1 = (1,-2,1,-5) Y3 =V3+AV1…DOT WITH Y1 Y3.Y1 = V3.Y1+AV1.Y1…….SINCE Y1 AND Y3 ARE ORTHOGONAL,Y1.Y3=0 A=-(V3.Y1)/(V1.Y1) = -[1*1+(-1)*(-2)+2*1+(-5)*(-2)]/[1*1+(-2)*(-2)+1*1+(-5)*(-5)]=-15/31 Y3 = (1,-1,2,-2)-(15/31)(1,-2,1,-5) = (1/31)*(16,-1,47,13) HENCE THE 2 ORTHOGONAL VECTORS ARE OBTAINED. NORMALISING THEM WE GET THE ORTHONORMAL BASIS Y1'=(1,-2,1,-5)/SQRT(31) = (1/5.568)(1,-2,1,-5) Y3'= [16,-1,47,13]= (1/51.33)(16,-1,47,13)