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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.
Determine if the subspaces spanned by the sets of vectors V and U are equal.

V = {(2  3  -2  -2); (0  4  8  0); (1  3  3  0); (-4  -1  12  2)}

U = {(1  3  3  0); (0  3  8  2); (-1  1  5  0)}

How to solve this problem?
This problem can be solved using different ways, but generally on the premise that the dimensions of the two subspaces are equal. Once this has been proved, a practical way is to demonstrate that the vectors of a spanning set for the first subspace belong to the second subspace. Another way is to demonstrate that the sum or the intersection of both subspaces has the same dimension of the subspaces. Here it will be used the second way, calculating the dimension of the sum of both subspaces, because it is known the spanning sets for both subspaces.

Step 1: Calculate the dimension of the subspace spanned by the set of vectors V.

- Write the matrix composed by the vectors of V as columns.
┌              ┐
│  2  0  1  -4 │
│  3  4  3  -1 │
│ -2  8  3  12 │
│ -2  0  0   2 │
└              ┘

- Transform the matrix to row echelon form.
r2 <---> 2•r2
┌              ┐
│  2  0  1  -4 │
│  6  8  6  -2 │
│ -2  8  3  12 │
│ -2  0  0   2 │
└              ┘
r2 <---> r2 - 3•r1
r3 <---> r3 + r1
r4 <---> r4 + r1
┌             ┐
│ 2  0  1  -4 │
│ 0  8  3  10 │
│ 0  8  4   8 │
│ 0  0  1  -2 │
└             ┘
r3 <---> r3 - r2
┌             ┐
│ 2  0  1  -4 │
│ 0  8  3  10 │
│ 0  0  1  -2 │
│ 0  0  1  -2 │
└             ┘
r4 <---> r4 - r3
┌             ┐
│ 2  0  1  -4 │
│ 0  8  3  10 │
│ 0  0  1  -2 │
│ 0  0  0   0 │
└             ┘

- The dimension of the subspace spanned by the set of vectors V is the rank of the matrix.
dim([V]) = 3

Step 2: Calculate the dimension of the subspace spanned by the set of vectors U.

- Write the matrix composed by the vectors of U as columns.
┌          ┐
│ 1  0  -1 │
│ 3  3   1 │
│ 3  8   5 │
│ 0  2   0 │
└          ┘

- Transform the matrix to row echelon form.
r2 <---> r2 - 3•r1
r3 <---> r3 - 3•r1
┌          ┐
│ 1  0  -1 │
│ 0  3   4 │
│ 0  8   8 │
│ 0  2   0 │
└          ┘
r2 <---> r4
┌          ┐
│ 1  0  -1 │
│ 0  2   0 │
│ 0  8   8 │
│ 0  3   4 │
└          ┘
r4 <---> 2•r4
┌          ┐
│ 1  0  -1 │
│ 0  2   0 │
│ 0  8   8 │
│ 0  6   8 │
└          ┘
r3 <---> r3 - 4•r2
r4 <---> r4 - 3•r2
┌          ┐
│ 1  0  -1 │
│ 0  2   0 │
│ 0  0   8 │
│ 0  0   8 │
└          ┘
r4 <---> r4 - r3
┌          ┐
│ 1  0  -1 │
│ 0  2   0 │
│ 0  0   8 │
│ 0  0   0 │
└          ┘

- The dimension of the subspace spanned by the set of vectors U is the rank of the matrix.
dim([U]) = 3

Step 3: Calculate the dimension of the subspace spanned by the vectors of both sets: V and U.

- Write the matrix composed by the vectors of V and U as columns.
┌                        ┐
│  2  0  1  -4  1  0  -1 │
│  3  4  3  -1  3  3   1 │
│ -2  8  3  12  3  8   5 │
│ -2  0  0   2  0  2   0 │
└                        ┘

- Transform the matrix to row echelon form.
r2 <---> 2•r2
┌                        ┐
│  2  0  1  -4  1  0  -1 │
│  6  8  6  -2  6  6   2 │
│ -2  8  3  12  3  8   5 │
│ -2  0  0   2  0  2   0 │
└                        ┘
r2 <---> r2 - 3•r1
r3 <---> r3 + r1
r4 <---> r4 + r1
┌                       ┐
│ 2  0  1  -4  1  0  -1 │
│ 0  8  3  10  3  6   5 │
│ 0  8  4   8  4  8   4 │
│ 0  0  1  -2  1  2  -1 │
└                       ┘
r3 <---> r3 - r2
┌                       ┐
│ 2  0  1  -4  1  0  -1 │
│ 0  8  3  10  3  6   5 │
│ 0  0  1  -2  1  2  -1 │
│ 0  0  1  -2  1  2  -1 │
└                       ┘
r4 <---> r4 - r3
┌                       ┐
│ 2  0  1  -4  1  0  -1 │
│ 0  8  3  10  3  6   5 │
│ 0  0  1  -2  1  2  -1 │
│ 0  0  0   0  0  0   0 │
└                       ┘

- The dimension of the subspace [V] + [U], where [V] and [U] are the subspaces spanned by V and U respectively, is the rank of the matrix.
dim([V] + [U]) = 3

Step 4: Solution.

The subspace spanned by V and the subspace spanned by U are equal, because their dimensions are equal, and equal to the dimension of the sum subspace too.

  
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