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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.
Find a linear transformation that maps each vector of the ordered set of vectors V, in the corresponding vector of the ordered set of vectors U.

V = {(4  1  2); (1  0  -3); (5  1  -2); (-1  0  2)}

U = {(2  -5  9  -6); (-2  -2  9  1); (-1  -5  0  -6); (1  4  -27  -2)}

How to solve this problem?
If V is a basis, there exists only one linear transformation that maps the vectors of V into the vectors of U. If the vectors of V are linearly independent but not a basis, there is an infinite number of linear transformations. If vectors of V are linearly dependent, then it could exist none, one, or an infinite number of linear transformations, depending on the vectors of U. All these problems can be solved algebraically or using matrices, but it will be used the second way, which is considered the easiest way.

- In practice, to solve the first case, multiply the matrix composed by the vectors of U as columns, by the inverse of the matrix composed by the vectors of V as columns. The resulting matrix is the matrix of the linear transformation relative to the standard bases. The matrix composed by the vectors of V as columns is always invertible; due to V is a basis for the input vector space. This practical way to find the linear transformation is a direct consequence of the procedure for finding the matrix of a linear transformation.

- To solve the second case, just expand the vectors of V into a basis, mapping additional vectors to null vector, and solve using the procedure of first case.

- To solve the third case, extract a linearly independent subset from V, let's call it A. Check for each vector no contained in A, that the coefficients when expressing this vector as a linear combination of the vectors of A are equal to the ones obtained when expressing its image as a linear combination of the images of the vectors of A, in the same order. If for one of these vectors the process fails, then, there is no linear transformation that can transform V in U. If this process is successful, then use the procedure for second case.

Step 1: Check if V is a linearly independent or dependent set of vectors.

- Write the matrix composed by the vectors as columns.
┌               ┐
│ 4   1   5  -1 │
│ 1   0   1   0 │
│ 2  -3  -2   2 │
└               ┘

- Transform the matrix to row echelon form.
r1 <---> r2
┌               ┐
│ 1   0   1   0 │
│ 4   1   5  -1 │
│ 2  -3  -2   2 │
└               ┘
r2 <---> r2 - 4•r1
r3 <---> r3 - 2•r1
┌               ┐
│ 1   0   1   0 │
│ 0   1   1  -1 │
│ 0  -3  -4   2 │
└               ┘
r3 <---> r3 + 3•r2
┌              ┐
│ 1  0   1   0 │
│ 0  1   1  -1 │
│ 0  0  -1  -1 │
└              ┘

- Determine the rank.
Rank(V) = 3

- Solution.
V is a linearly dependent set of vectors.

Step 2: Extract a linearly independent subset of vectors from V. Let's call it A.

- Select one column for each step.
Column indexes = {1, 2, 3}

- Solution.
A = {(4  1  2); (1  0  -3); (5  1  -2)}

Step 3: Determine if it is possible to find a linear transformation that transforms V in U.

- Select the subset of vectors composed by the vectors which are not in the subset of linearly independent vectors. Let's call it C.
C = {(-1  0  2)}

- Check for each vector in C, that the coefficients when expressing this vector as a linear combination of the vectors of A are equal to the ones obtained when expressing its image as a linear combination of the images of the vectors of A, in the same order.

f(-1  0  2) = (1  4  -27  -2)

(-1  0  2) =  - 1•(4  1  2) - 2•(1  0  -3) + 1•(5  1  -2)

f(-1  0  2) =  - 1•f(4  1  2) - 2•f(1  0  -3) + 1•f(5  1  -2)

(1  4  -27  -2) =  - 1•(2  -5  9  -6) - 2•(-2  -2  9  1) + 1•(-1  -5  0  -6)

(1  4  -27  -2) = (1  4  -27  -2)

Process was successful for all the vectors of C, therefore any linear transformation that transforms the vectors of A to their corresponding images, will transform the vectors of C to their corresponding images.

The set of vectors A is a basis for the input space. Let's call it B.

Step 4: Subtract from U the vectors corresponding to the images of the vectors of C.

U = {(2  -5  9  -6); (-2  -2  9  1); (-1  -5  0  -6)}

Step 5: Compute the inverse of the basis matrix of B.

- Calculate the determinant.
│ 4   1   5 │
│ 1   0   1 │ = 4•0•(-2) + 1•1•2 + 1•(-3)•5 - 5•0•2 - 1•1•(-2) - 1•(-3)•4 = 1
│ 2  -3  -2 │


- Calculate the matrix of cofactors.
         ┌              ┐
         │   3    4  -3 │
Cof(B) = │ -13  -18  14 │
         │   1    1  -1 │
         └              ┘

- Transpose the matrix of cofactors to obtain the adjugate matrix.
         ┌             ┐
         │  3  -13   1 │
Adj(B) = │  4  -18   1 │
         │ -3   14  -1 │
         └             ┘

- Divide each entry of the adjugate matrix by the determinant to obtain the inverse.
         ┌             ┐
         │  3  -13   1 │
Inv(B) = │  4  -18   1 │
         │ -3   14  -1 │
         └             ┘

Step 6: Multiply the matrix U by the inverse of the basis matrix of B.

           ┌              ┐
           │  1    -4   1 │
           │ -8    31  -2 │
U•Inv(B) = │ 63  -279  18 │
           │  4   -24   1 │
           └              ┘

Step 7: Final solution.

f(x, y, z) = (x-4y+z; -8x+31y-2z; 63x-279y+18z; 4x-24y+z)

  
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