# Linear algebra homework problem involving basis and dual basis.

Let $V = R^3$, and define $f_1, f_2, f_3 ∈ V^*$ as follows:
$f_1(x,y,z) = x - 2y$
$f_2(x,y,z) = x + y + z$
$f_3(x,y,z) = y-3z$

Prove that ${f_1,f_2,f,_3}$ is a basis for $V^*$, and then find a basis for $V$ for which it is the dual basis.

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Hints: First show that the $f_i$ are indeed in $V^*$. Then show that they form an independent set. As $V^*$ has dimension 3, this will show they form a basis of $V^*$. To find the dual basis, find $x_i$ in $V$ such that $f_j(x_i)=\cases{1,&$i=j$\cr 0, &$i\ne j$}$. –  David Mitra Feb 15 '12 at 21:34

First, you should verify that the $f_i$ are elements of $V^*$; that is, that they are functions on $\Bbb R^3$ that satisfy:

$\ \ \$1) $f({\bf x}+{\bf y})=f({\bf x})+f({\bf y})$ for all ${\bf x},{\bf y}\in\Bbb R^3$

and

$\ \ \$2) $f(c{\bf x})=cf({\bf x})$ for all $c\in\Bbb R$ and all ${\bf x}\in \Bbb R^3$.

Of course, if you know this has been covered in your class already, you're probably safe just writing something like "as was demonstrated in lecture blah, these are elements of the dual of $V=\Bbb R^3$.

Another fact I assume you have use of is that the dimension of $V^*$ is three (as the dimension of the vector space $\Bbb R^3$ is three).

So, with three linear functionals on $\Bbb R^3$, towards showing that they are a basis of $V^*$, it suffices to show that they are independent. There are many ways towards achieving this end. One way in particular is to show that the matrix $A$ formed by taking as its rows the coefficients of the $f_i$, $$A=\left[\matrix{1&-2&0\cr 1&1&1\cr 0&1&-3}\right],$$ has full rank (that this is so isn't hard to see: $c_1f_1+c_2f_2+c_3f_3={\bf 0}\iff A\bigl[{{\scriptstyle c_1\atop\scriptstyle c_2}\atop c_3}\bigr]=\bf 0$; and the former equation has only the trivial solution if and only if $A$ has full rank).

So, let's row reduce $A$: $$A= \left[\matrix{1&-2&0\cr 1&1&1\cr 0&1&-3}\right] \buildrel{r_2-r_1\rightarrow r_2}\over{\longrightarrow} \left[\matrix{1&-2&0\cr 0& 3& 1\cr 0&1&-3}\right] \buildrel{r_2-3r_3\rightarrow r_3}\over{\longrightarrow} \left[\matrix{1&-2&0\cr 0& 3& 1\cr 0&0&10}\right].$$

At this point we can see that $A$ indeed has full rank. Thus the set $\{f_1,f_2,f_3\}$ is independent and consequently is a basis of $V^*$.

Towards finding the dual basis let's recall what this is: the dual basis of $\{f_1,f_2,f_3\}$ by definition is the basis $\{ {\bf x}_1,{\bf x}_2,{\bf x}_3\}$ of $V$ for which $$f_i({\bf x}_j)=\cases{1,& if i=j\cr 0, & if i\ne j}.$$

So, in particular, the first basis element of dual basis, ${\bf x}_i$, would satisfy $$f_1({\bf x}_1)=1,\ f_2({\bf x}_1)=0, \text{and}, f_3({\bf x}_1)=0.$$ In other words, we have the system: \eqalign{ x-2y&=1\cr x+y+z&=0\cr y-3z&=0 } whose matrix form is $$A {\bf x}_1=\left[\matrix{ 1\cr0\cr 0 }\right]$$ and whose, necessarily unique, solution gives the coordinates of ${\bf x}_1$.

We would have two similar systems to solve in order to find ${\bf x}_2$ and ${\bf x}_3$.

That seems like a lot of work; but wait... suppose we wrote the dual basis as a matrix whose columns were the ${\bf x}_i$. Then we'd have: $$A[ {\bf x}_1\ {\bf x}_2\ {\bf x}_3] =\left[\matrix{1&-2&0\cr 1&1&1\cr 0&1&-3}\right][ {\bf x}_1\ {\bf x}_2\ {\bf x}_3] =\left[\matrix{1&0&0\cr 0&1&0\cr 0&0&1}\right]$$

So $[ {\bf x}_1\ {\bf x}_2\ {\bf x}_3]$ is the inverse of $A$. Rather than solving three systems of equations, we could instead find the inverse of $A$ and then the columns will be our dual basis.

This is what we'll do. Towards that end, we may (and do) adjoin the identity matrix to $A$ and perform a full forward/back row reduction:

\eqalign{ A= \left[\matrix{1&-2&0\cr 1&1&1\cr 0&1&-3} \ \ \Biggl|\ \ \matrix{1& 0&0\cr 0&1&0\cr 0&0&1}\right] &\buildrel{r_2-r_1\rightarrow r_2}\over{\longrightarrow} \left[\matrix{1&-2&0\cr 0& 3& 1\cr 0&1&-3} \ \ \Biggl|\ \ \matrix{1& 0&0\cr -1&1&0\cr 0&0&1}\right]\cr &\buildrel{r_2-3r_3\rightarrow r_3}\over{\longrightarrow} \left[\matrix{1&-2&0\cr 0& 3& 1\cr 0&0&10} \ \ \Biggl|\ \ \matrix{1& 0&0\cr -1& 1&0\cr -1&1&-3}\right]\cr &\buildrel{10r_2-r_3\rightarrow r_2}\over{\longrightarrow} \left[\matrix{1&-2&0\cr 0& 30& 0\cr 0&0&10} \ \ \Biggl|\ \ \matrix{1& 0&0\cr -9&9&3\cr -1&1&-3}\right]\cr &\buildrel{15r_1+r_2\rightarrow r_1}\over{\longrightarrow} \left[\matrix{15&0&0\cr 0& 30& 0\cr 0&0&10} \ \ \Biggl|\ \ \matrix{ 6& 9&3\cr -9&9&3\cr -1&1&-3}\right]\cr &\buildrel{ }\over{\longrightarrow} \left[\matrix{1 &0&0\cr 0& 1& 0\cr 0&0&1 } \ \ \Biggl|\ \ \matrix{ 6/15& 9/15&3/15\cr -9/30&9/30&3/30\cr -1/10&1/10&-3/10}\right].\cr }

So $$A^{-1}=\left[\matrix{2/5& 3/5&1/5\cr -3/10&3/10&1/10\cr -1/10&1/10&-3/10}\right],$$ and the dual basis has as its elements: $${\bf x}_1=\left[\matrix{2/5\cr3/5\cr1/5 }\right],\ {\bf x}_2=\left[\matrix{-3/10\cr3/10\cr1/10 }\right],\ {\bf x}_3=\left[\matrix{-1/10\cr1/10\cr-3/10 }\right].$$

The basis $\{ {\bf x}_1,{\bf x}_2,{\bf x}_3 \}$ is the dual basis of $\{f_1,f_2,f_3\}$.

Note that the relative ordering is important. For instance, we must have $f_3({\bf x}_1)=f_3({\bf x}_2)=0$ and $f_3({\bf x}_3)=1$. As a spot check, let's verify this: \eqalign{ f_3({\bf x}_1)&= (3/5)-3(1/5)=0\cr f_3({\bf x}_2)&= (3/10)-3(1/10)=0\cr f_3({\bf x}_3)&= (1/10)-3(-3/10)=1.\cr }

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$\bf Hint:$ Suppose that $a f_1(x,y,z)+b f_2(x,y,z)+c f_3(x,y,z)=0$. Try different values for $(x,y,z)$, for example, if you substitute $(2,1,-3)$. We obtain $f_1(2,1,-3)=0$, $f_2(2,1,-3)=0$ and $f_3(2,1,-3)=10$, hence $cf_3(2,1,-3)=10c$ which implies $c=0$.

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Proving $\lbrace f_1,f_2,f_3\rbrace$ is a basis for $V^*$ can be done by row reducing the coefficients of $\lbrace f_1,f_2,f_3 \rbrace$ and showing that it has a rank of $3$.

The dual basis of $\lbrace f_1,f_2,f_3 \rbrace$ is found by calculating the inverse of coefficients of $f_i$ which is:

$x_{1}= \ \begin{bmatrix} 4/10 \\ -3/10 \\ -1/10 \end{bmatrix}, x_{2}= \begin{bmatrix} 6/10 \\ 3/10 \\ 1/10 \end{bmatrix}, x_{3}= \begin{bmatrix} 2/10 \\ 1/10 \\ -3/10 \end{bmatrix} ]\$

Checking that $f_{i}(x_{j})=$ $1,$ if $i=j$ and $0$ if $i \ne j$

$f_{1}(x_1)=(4/10)-2(-3/10)=1\\ f_{1}(x_2)=(6/10)-2(3/10)=0\\ f_{1}(x_3)=(2/10)-2(1/10)=0$

$f_{2}(x_1)=(4/10)+(-3/10)+(-1/10)=0\\ f_{2}(x_2)=(6/10)+(3/10)+(1/10)=1\\ f_{2}(x_3)=(2/10)+(1/10)+(-3/10)=0$

$f_{3}(x_1)=(-3/10)-3(-1/10)=0\\ f_{3}(x_2)=(3/10)-3(1/10)=0\\ f_{3}(x_3)=(1/10)-3(-3/10)=1$

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