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)}
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.
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
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
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
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.