Miscellaneous
4.4 Coordinate Systems
Remark. What this theorem means is that for a transformation \(T: V \to \mathbb{R}^n\) which is defined as \(\mathbf{x} \mapsto [\mathbf{x}]_{\mathcal{B}}\), \(T\) is a linear transformation and is both one-to-one and onto.
Remark. One way to solve this is to equate the coefficients of the equation \[c_1(1+t) + c_2(1+t^2) + c_3(t+t^2)=6+3 t-t^2\] and solve for \(c_1,c_2,c_3\).
However, by observation, we can also first write \(1+t,1+t^2,t+t^2,6+3t-t^2\) as \(\begin{bmatrix}1\\ 1\\ 0\end{bmatrix}, \begin{bmatrix}1\\ 0\\ 1\end{bmatrix}, \begin{bmatrix}0\\ 1\\ 1\end{bmatrix}, \begin{bmatrix}6\\ 3\\ -1\end{bmatrix}\). Then, as for subspaces in \(\mathbb{R}\), we solve the coefficients \(\mathbf{c}\) using row reduction.
Let \(T: \mathbf{x} \mapsto [\mathbf{x}]_\mathcal{B}\), so \(T\) is a bijective linear transformation.
\[\begin{align} T(c_1\mathbf{b}_1 + \cdots + c_n\mathbf{b}_n) &= c_1 T(\mathbf{b}_1) + \cdots + c_n T(\mathbf{b}_n) \\ &= T(\mathbf{p}) \end{align}\]
where \(T(\mathbf{b}_1),\dots,T(\mathbf{b}_n)\) are the coordinate vectors of the basis. Since \(T\) is bijective, we can apply \(T^{-1}\) to both side of the equation. Reversing the steps gives the same coefficients.
Isomorphism
In some problems, we are given a basis of a vector space other than a subspace of \(\bb{R}^n\). One of the most frequent ones is about polynomials, such as Exercise 1.
A formalized version of this type of problems: Given basis \(\cal{B}=\{\bf{b}_1, \dots, \bf{b}_n\}\) and \(\bf{v}\) in vector space \(V\), find \([\bf{v}]_\cal{B}\).
Proof. Define transformation \(T: \bf{x} \mapsto [\bf{x}]_\cal{E}\), where \(\cal{E}\) is another basis of vector space \(V\) (usually \(\{1,x,x^2,\dots\}\) for polynomial spaces).
Solve \(\mat{[\bf{b}_1]_\cal{E} & \cdots & [\bf{b}_n]_\cal{E}}\bf{x} = [\bf{v}]_\cal{E}\) for \(\bf{x}\). Then,
\[x_1[\bf{b}_1]_\cal{E} + \dots + x_n[\bf{b}_n]_\cal{E} = [\bf{v}]_\cal{E}\]
Because \(T\) is a linear transformation, \([x_1\bf{b}_1+\dots+x_n\bf{b}_n]_\cal{E}=[\bf{v}]_\cal{E}\).
Since \(T\) is one-to-one, \([\bf{u}]_\cal{E}=[\bf{v}]_\cal{E}\) if and only if \(\bf{u}=\bf{v}\). Therefore, \(\bf{v}=x_1\bf{b}_1+\dots+x_n\bf{b}_n\), \[[\bf{v}]_\cal{B}=\bf{x}\]
Solving for Bases Using Column Operations
The standard method to obtain a basis for either column spaces, row spaces, or null spaces is performing row reductions.
Let \(B\) be the RREF of \(A\). The columns in \(A\) that correspond with a pivot column in \(B\) form a basis for \(\opn{Col}A\), the non-zero rows of \(B\) form a basis for \(\opn{Row}A\), and the parametric vector form of its solution set we derive from \(B\) indicates a basis for \(\opn{Nul}A\).
My friend: Why should I perform row reductions? Why not column operations?
Me: If you clearly know what you are doing, you can solve them using column operations.
— during a linear algebra class.
Using column operations to solve for the column space and the row space is simply doing row reduction on the transpose of the original matrix. Therefore, the non-zero columns of the reduced column echelon form (RCEF) give a basis for the column space, and the pivot rows in the original matrix constitute a basis for the row space.
However, null spaces are different because they are defined as \(\{\bf{x} : A\bf{x}=\bf{0}\}\) rather than explicitly the span of some vectors. The idea of transpose doesn’t work here because its counterpart is \(\opn{Nul}A^T\), which even complicates the problem.
Similar to row operations, column operations can also be expressed by right-multiplying elementary matrices. Let the RCEF of a matrix \(A\) with \(n\) columns, \(B=A E_1 \cdots E_p = AM\), where \(E_1,\dots,E_p\) are the elementary matrices corresponding to the column operations. Observe that \(B\) has the following form: \[\mat{1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \ast & \ast & 0 & 0 \\ 0 & 0 & 1 & 0}\]
Let \(r=\opn{rank}A\), then \(x_1=\cdots=x_r=0\) for equation \(B\bf{x}=(AM)\bf{x}=\bf{0}\), while \(x_{r+1},\dots,x_n\) are free variables completely independent of \(x_1,\dots,x_r\). Therefore, we can write a basis for \(\opn{Nul}(AM)\)—\(\{\bf{e}_{r+1},\dots,\bf{e}_n\}\).
Since elementary matrices are invertible, \(M\) is also invertible. We show that \(\{M\bf{e}_{r+1},\dots,M\bf{e}_n\}\) is a basis for \(\opn{Nul}A\).
- Span: For all \(\bf{v} \in \opn{Nul}A\), \(A\bf{v}=(AM)(M^{-1}\bf{v})=\bf{0}\), so \(M^{-1}\bf{v}\in\opn{Nul}(AM)\)
\[\begin{align} \bf{v}&=M(M^{-1}\bf{v}) \\ &=M(c_{r+1}\bf{e}_{r+1}+\dots+c_n\bf{e}_n) \\ &=c_{r+1}(M\bf{e}_{r+1})+\dots+c_n(M\bf{e}_n) \end{align}\]
- Linear Independence: By contradiction, suppose it is linearly dependent. There exists \(\bf{x}\neq\bf{0}\) such that \[\mat{M\bf{e}_{r+1} & \cdots & M\bf{e}_n}\bf{x}=\bf{0}\] Left-multiply the equation by \(M^{-1}\), \[\begin{align} M^{-1}\mat{M\bf{e}_{r+1} & \cdots & M\bf{e}_n}\bf{x}&=\mat{M^{-1}M\bf{e}_{r+1} & \cdots & M^{-1}M\bf{e}_n}\bf{x} \\ &=\mat{\bf{e}_{r+1} & \cdots & \bf{e}_n}\bf{x} \\ &=M^{-1}\bf{0}=\bf{0} \end{align}\] Since \(\bf{x}\neq\bf{0}\), this leads to that \(\{\bf{e}_{r+1},\dots,\bf{e}_n\}\), a basis for \(\opn{Nul}(AM)\), is linearly dependent.
Therefore, \(\{M\bf{e}_{r+1},\dots,M\bf{e}_n\}\) forms a basis for \(\opn{Nul}A\). Specifically, these vectors are the \(r+1\)-th column to the \(n\)-th column of matrix \(M\).
One problem remains: how to obtain \(M\)? Using the same logic of finding the inverse of a matrix, we augment the original matrix with an identity matrix below \(A\). \[\mat{A \\ I}\sim\mat{B \\ M}\] As \(A\) is reduced to the reduced column echelon form, the identity matrix is transformed to \(M\).
An Example
\[\mat{1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1}\]
Solution (Row Operations). \[\mat{1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1}\sim \mat{1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0}\]
Therefore, \(\left\{\mat{1 \\ -2 \\ 1}\right\}\) is a basis for \(\opn{Nul}A\).
Solution (Column Operations). \[\mat{1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}\sim\mat{1 & 0 & 0 \\ 2 & 0 & 0 \\ 1 & -1 & -2 \\ 1 & -2 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 1}\sim\mat{1 & 0 & 0 \\ 2 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & -1 & -2 \\ 0 & 0 & 1}\]
The third column \(\mat{1 \\ -2 \\ 1}\) gives the exactly same result as what we obtain using row operations.
Remark. In fact, we may only reduce \(A\) to the column echelon form since all operations required to transform a matrix from its column echelon form to its reduced column echelon form do not change the \(r+1\)-th to the \(n\)-th columns of \(M\).