By filtering the given vectors into a matrix we have
B = "\\begin{bmatrix}\n 1 & 1 &2\\\\\n 0 & 1&2\\\\\n-1&1&3\n\\end{bmatrix}"
B-1 = "\\begin{bmatrix}\n 1 & -1 &0\\\\\n -2 & 5&-2\\\\\n1&-2&1\\\\\n\\end{bmatrix}"
Now the product B-1 B = I
Hence we see that the the rows of B-1 multiplied by the columns of B satisfy the condition for duality.
Hence, defining
"\\gamma_1" = [1, -2, 1]
"\\gamma_2" = [-1, 5, -2]
"\\gamma_3" = [0, -2, 1]
form the dual basis for {v1, v2, v3} .
Comments
Leave a comment