Linear algebra: the theory of vector spaces and linear transformations

Computing coordinate vectors relative to one of our standard bases for \(\R^n\text{,}\) \(M_{mn}\text{,}\) or \(P_{n}\) amounts to just listing the coefficients or entries used to specify the given vector. The examples below serve to illustrate the general method in this setting.

1. Let \(V=\R^3\) and \(B=(\bolde_1, \bolde_2, \bolde_3)\text{.}\) For any \(\boldv=(a,b,c)\in \R^3\) we have \([\boldv]_{B}=(a,b,c)\text{,}\) since \((a,b,c)=a\bolde_1+b\bolde_2+c\bolde_3\text{.}\)

2. Let \(V=M_{22}\) and \(B=(E_{11}, E_{12}, E_{21}, E_{22})\text{.}\) For any \(A=\begin{amatrix}[rr]a\amp b\\ c\amp d \end{amatrix}\) we have \([A]_B=(a,b,c,d)\) since

   \begin{equation*} A=aE_{11}+bE_{12}+cE_{21}+dE_{22}\text{.} \end{equation*}

If we choose an alternate ordering of one of the standard bases, the entries of the coordinate vector are reordered accordingly, as illustrated by the examples below.

1. Let \(V=\R^3\) and \(B=(\bolde_2, \bolde_1, \bolde_3)\text{.}\) Given \(\boldv=(a,b,c)\in \R^3\) we have \([\boldv]_B=(b,a,c)\text{,}\) since

   \begin{equation*} \boldv=b\bolde_2+a\bolde_1+c\bolde_3\text{.} \end{equation*}

2. Let \(P=P_3\) and \(B=(1,x,x^2, x^3)\text{.}\) Given \(p(x)=ax^3+bx^2+cx+d\) we have \([p(x)]_B=(d, c, b, a)\text{,}\) since

   \begin{equation*} p(x)=d(1)+cx+bx^2+ax^3\text{.} \end{equation*}

For a nonstandard ordered basis, we compute coordinate vectors by solving a relevant system of linear equations, as the examples below illustrate.

1. Let \(V=\R^2\text{,}\) \(B=((1,2),(1,1))\text{,}\) and \(\boldv=(3,3)\text{.}\) To compute \([\boldv]_B\) we must find the unique pair \((c_1, c_2)\) satisfying

   \begin{equation*} (3,3)=c_1(1,2)+c_2(1,1)\text{.} \end{equation*}

   By inspection, we see that

   \begin{equation*} (3,3)=3(1,1)=0(1,2)+3(1,1)\text{.} \end{equation*}

   We conclude that

   \begin{equation*} [\boldv]_{B}=(0,3)\text{.} \end{equation*}

   More generally, to compute \([\boldv]_B\) for an arbitrary \(\boldv=(a,b)\in \R^2\text{,}\) we must find the pair \((c_1,c_2)\) satisfting \((a,b)=c_1(1,2)+c_2(1,1)\text{,}\) or equivalelenty

   \begin{equation*} \begin{linsys}{2} c_1\amp +\amp c_2 \amp =\amp a\\ 2c_1\amp +\amp c_2\amp =\amp b \end{linsys}\text{.} \end{equation*}

   The usual techniques yield the unique solution \((c_1,c_2)=(-a+b,2a-b)\text{,}\) and thus

   \begin{equation*} [\boldv]_B=(-a+b, 2a-b) \end{equation*}

   for \(\boldv=(a,b)\text{.}\)

2. Let \(V=P_2\text{,}\) \(B=(x^2+x+1, x^2-x, x^2-1)\text{,}\) and \(p(x)=x^2\text{.}\) To compute \([p(x)]_B\) we must find the unique triple \((c_1,c_2,c_3)\) satisfying

   \begin{equation*} x^2=c_1(x^2+x+1)+c_2(x^2-x)+c_3(x^2-1)\text{.} \end{equation*}

   The equivalent linear system once we combine like terms and equate coefficients is

   \begin{equation*} \begin{linsys}{3} c_1\amp +\amp c_2\amp +\amp c_3\amp =\amp 1\\ c_1\amp -\amp c_2\amp \amp \amp =\amp 0\\ c_1\amp \amp \amp -\amp c_3\amp =\amp 0\\ \end{linsys}\text{.} \end{equation*}

   The unique solution to this system is \((c_1,c_2,c_3)=(1/3, 1/3, 1/3)\text{.}\) We conclude

   \begin{equation*} [x^2]_B=\frac{1}{3}(1, 1, 1)\text{.} \end{equation*}

   The same reasoning shows that more generally, given polynomial \(p(x)=ax^2+bx+c\text{,}\) we have

   \begin{equation*} [p(x)]_B=\frac{1}{3}(a+b+c, a-2b+c, a+b-2)\text{.} \end{equation*}

Let \(V=\R^2\) and \(B=((1,1),(-1,2))\text{.}\) Find a general formula for \([(a,b)]_B\text{.}\) Note: \(B\) is orthogonal with respect to the weighted dot product

The set

is a subset of the space \(W=\{ A\in M_{22}\colon \tr A=0\}\text{.}\) Let \(W'=\Span S\text{.}\) Contract \(S\) to a basis of \(W'\) and determine whether \(W'=W\text{.}\)