Vector spaces: Coordinates
Basis and echelon form
In theory Finding bases we discussed two methods for finding a basis of a vector space. One of the two methods does not need a set of spanning vectors; this method uses of the Growth criterion for independence. The other method does use a spanning set of vectors and works with thinning out, a process that requires a dependence calculation at each step. By switching to coordinates, we can find a basis efficiently in this case by use of row reduction.
We recall that a matrix is in echelon form if it has the following two characteristics:
- all of the elements in the rows below a leading element, in the column in which this leading element lies as well as in the columns to the left of it, are zero;
- all null rows are at the bottom.
The leading element of a row in a matrix is the first element (from the left) of the row that is not zero. The general form of a matrix in echelon form is:
\[
\left(\,\begin{array}{cccccccccccccccccc}
0 & \ldots & 0 & 1 & \ast & \ldots & \ast & \ast & \ast & \ldots & \ast & \ast & \ast & \ldots & \ast & \ast & \ldots & \ast\\
0 & \ldots & 0 & 0 & 0 & \ldots & 0 & 1 & \ast & \ldots & \ast & \ast & \ast & \ldots & \ast & \ast & \ldots & \ast\\
0 & \ldots & 0 & 0 & 0 & \ldots & 0 & 0 & 0 & \ldots & 0 & 1 & \ast & \ldots & \ast & \ast & \ldots & \ast\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & \ldots & 0 & 0 & 0 & \ldots & 0 & 0 & 0 & \ldots & 0 & 0 & 0 & \ldots & 1 & \ast & \ldots & \ast\\
0 & \ldots & 0 & 0 & 0 & \ldots & 0 & 0 & 0 & \ldots & 0 & 0 & 0 & \ldots & 0 & 0 & \ldots & 0\\
\vdots & & \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots & & \vdots & \vdots & & \vdots\\
0 & \ldots & 0 & 0 & 0 & \ldots & 0 & 0 & 0 & \ldots & 0 & 0 & 0 & \ldots & 0 & 0 & \ldots & 0
\end{array}\,\right)
\]
We view the rows of a #(m\times n)#-matrix as vectors of #\mathbb{R}^n#.
Echelon Form and independence
The rows of a matrix in echelon form that are distinct from the null row are independent.
In particular, the rank of a matrix is equal to the dimension of the span of the row vectors.
The proposition Standard operations with spanning sets and the above theorem show how to find a basis of a subspace spanned by vectors in the coordinate space #\mathbb{R}^n#:
Finding bases by coordinates Let #\vec{a}_1,\ldots,\vec{a}_m# be a set of vectors in #\mathbb{R}^n#. A basis of #\linspan{\vec{a}_1,\ldots,\vec{a}_m}# can be found by use of the following three steps:
- Set up the matrix #M# whose rows are the vectors #\vec{a}_1,\ldots,\vec{a}_m#.
- Row reduce #M# to an echelon form #T#.
- The set of non-zero row vectors of #T# is a basis of #\linspan{\vec{a}_1,\ldots,\vec{a}_m}#.
From the theorem Standard operations with spanning sets it follows that the span of the rows of #M# is equal to the span of the rows of #T#. Therefore, the rows of #T# that are not equal to the null row, span #\linspan{\vec{a}_1,\ldots,\vec{a}_m}#. Because of the above statement, they are independent. They thus form a basis of #\linspan{\vec{a}_1,\ldots,\vec{a}_m}#.
If a vector space #W# is not a coordinate space, we can apply the same techniques after transition to coordinates with respect to a given basis #\basis{\vec{b}_1,\ldots,\vec{b}_n}# of #W#. If #W# is described by a spanning set of #m# vectors, we first determine the coordinates with respect to #\basis{\vec{b}_1,\ldots,\vec{b}_n}# of the spanning vectors, then apply the above theorem Finding bases by coordinates, and finally write the basis vectors found in #\mathbb{R}^n# as linear combinations of #\vec{b}_1,\ldots,\vec{b}_n#. This way we find a basis of #W#.
If we apply this process to the vector space of linear polynomial functions on #\mathbb{R}^n#, with basis #\basis{x_1, x_2,\ldots,x_n,1}#, then we wind up using the row reduction process of the augmented coefficient matrix pertaining to a system of linear equations. This idea is further discussed in one of the examples below.
We form the matrix
\[ \left(\begin{array}{cccc}
1&0&2&0 \\
1&1&2&1\\
2&-1&4&-1 \\
1&3&2&3
\end{array} \right) \] Row reduction does not change the space spanned by the rows. We reduce the matrix to reduced echelon form (reducing to echelon form would suffice) and find:
\[\left(\begin{array}{cccc}
1&0&2&0\\
0&1&0&1\\
0&0&0&0\\
0&0&0&0
\end{array} \right)\] We conclude that the vectors #\rv{1,0,2,0}# and #\rv{0,1,0,1}# are coordinate vectors of a basis for #V#. Therefore, a good answer is \[\basis{\rv{1,0,2,0},\rv{0,1,0,1}}\]
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