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Linear Algebra Decoded
 

Linear Algebra Decoded Example

Linear Algebra Decoded
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This example has been taken directly from the solution given by Linear Algebra Decoded to the formulated problem.
Compute the change of basis matrix from the basis A to the basis B.

A = {(-1  0  -1); (-4  8  2); (1  -3  -1)}

B = {(0  1  -1); (-1  1  0); (6  -4  -1)}

How to solve this problem?
The change of basis matrix (or transition matrix) C[A->B] from the basis A to the basis B, can be computed transposing the matrix of the coefficients when expressing the vectors of A as the linear combination of the vectors of B. Another way, the one that will be used, it is multiplying the inverse of the basis matrix of B by the basis matrix of A, where the basis matrix is a matrix that represents the basis, because its columns are the components of the vectors of the basis. The basis matrices are always invertible due to their rank coincides with its order. This calculation method is based on the following formula: C[A->B] = C[N->B]•C[A->N] where N is the standard basis, and C[N->B] = inv(C[B->N]). The change of basis matrix from any basis B to the standard basis N is equal to the basis matrix of B.

Step 1: Write the change of basis matrix from the basis B to the standard basis N (It is the basis matrix of B).

          ┌            ┐
          │  0  -1   6 │
C[B->N] = │  1   1  -4 │
          │ -1   0  -1 │
          └            ┘

Step 2: Invert the matrix C[B->N].

│  0  -1   6 │
│  1   1  -4 │ = 0•1•(-1) + (-1)•(-4)•(-1) + 1•0•6 - 6•1•(-1) - (-1)•1•(-1) - (-4)•0•0 = 1
│ -1   0  -1 │

               ┌          ┐
               │ -1  5  1 │
Cof(C[B->N]) = │ -1  6  1 │
               │ -2  6  1 │
               └          ┘
               ┌            ┐
               │ -1  -1  -2 │
Adj(C[B->N]) = │  5   6   6 │
               │  1   1   1 │
               └            ┘
                         ┌            ┐
                         │ -1  -1  -2 │
Inv(C[B->N]) = C[N->B] = │  5   6   6 │
                         │  1   1   1 │
                         └            ┘

Step 3: Write the change of basis matrix from the basis A to the standard basis N (It is the basis matrix of A).

          ┌            ┐
          │ -1  -4   1 │
C[A->N] = │  0   8  -3 │
          │ -1   2  -1 │
          └            ┘

Step 4: Multiply the matrices C[N->B] and C[A->N].


C[A->B]11 = (-1)•(-1) + (-1)•0 + (-2)•(-1) = 3
C[A->B]12 = (-1)•(-4) + (-1)•8 + (-2)•2 = -8
C[A->B]13 = (-1)•1 + (-1)•(-3) + (-2)•(-1) = 4

C[A->B]21 = 5•(-1) + 6•0 + 6•(-1) = -11
C[A->B]22 = 5•(-4) + 6•8 + 6•2 = 40
C[A->B]23 = 5•1 + 6•(-3) + 6•(-1) = -19

C[A->B]31 = 1•(-1) + 1•0 + 1•(-1) = -2
C[A->B]32 = 1•(-4) + 1•8 + 1•2 = 6
C[A->B]33 = 1•1 + 1•(-3) + 1•(-1) = -3
          ┌              ┐
          │   3  -8    4 │
C[A->B] = │ -11  40  -19 │
          │  -2   6   -3 │
          └              ┘

  
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