Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have $B= \{b_1, \dots,b_n \}$ be basis for vector space $V$ over reals. Then if $A = \{a_1, \dots,a_n \}$ be basis for dual space $V^*$ (dual space is defined as set of all linear function mapping $V$ to reals). $A$ is not defined to be basis of $B$. Matrix $M$ is defined as: \begin{array}{cccc} a_1(b_1) & a_2(b_1) & \ldots & a_n(b_1) \\ a_1(b_2) & a_2(b_2) & \ldots & a_n(b_2) \\ \vdots & \vdots & \ddots & \vdots \\ a_1(b_n) & a_2(b_n) & \ldots & a_n(b_n) \\ \end{array} I have to show (a) $M$ is invertible and (b) for some $f \in V^*$, for vectors $x,y$ where $x = [f]_A $ and $ y = (f(b_1), \dots, f(b_n))^T$, that $y = Mx$.

For part (a) I want to show that columns of $M$ are linearly independent. However, I tried doing a proof by contradiction. I assume that say the last column is a linearly combination of the first $n-1$ columns. I was unable to obtain a contradiction in this way.

For part (b), I have no idea how to approach this. I noticed that if $M = I$, then we simply have that $A$ is the dual basis of $B$. For $ x= (x_1,...x_n)^T$, then $Mx = \begin{pmatrix} \sum_i^n x_ia_i(b_1) \\ \vdots \\ \sum_i^n x_ia_i(b_n) \end{pmatrix}$ How do I show that for some $j$ that $f(b_j) = \sum_i^n x_ia_i(b_j)$?

$a_i$ is a function mapping $V$ to reals.

share|improve this question
2  
Welcome to Math.SE! If this is homework or revision, please say so with a tag :) –  Shaun Nov 18 '13 at 18:29
    
What do you mean by $a_j(b_i)$? –  Student Nov 18 '13 at 18:31

2 Answers 2

Suppose the matrix is singular, so that there exists a column linearly dependent in the preceeding columns, say

$$\begin{pmatrix}a_i(b_1)\\a_i(b_2)\\\ldots\\a_i(b_n)\end{pmatrix}=\sum_{k=1}^{i-1}c_k\begin{pmatrix}a_k(b_1)\\a_k(b_2)\\\ldots\\a_k(b_n)\end{pmatrix}\iff$$

$$\forall\,1\le m\le n\;\;,\;\;a_i(b_m)=\sum_{k=1}^{i-1}c_ka_k(b_m)=\sum_{k=1}^{i-1}a_k(c_kb_m)$$

But since a linear functional (and, in fact, any linear map) is completely and uniquely determined by its values on some basis of the domain, we get that $\;a_i\;$ is a linear combination of $\;a_1,...,a_{i-1}\;$ , which of course contradicts the fact that $\;a_1,...,a_n\;$ is a basis of $\;V^*\;$

share|improve this answer

The solution to part (a) is, of course, pretty straightforward, as DonAntonio's answer illustrates, but to put it in my own words, perhaps with a slightly different spin: the matrix $M$ is indeed singular if and only if the columns are linearly dependent. Thus $M$ singular implies there exist $c_i \in \Bbb R$, with some $c_i \ne 0$, such that

$\sum_i c_i a_i(b_j) = 0, \, 1 \le j \le n, \tag{1}$

but (1) merely states that the linear functional $a = \sum_i c_i a_i \in V^*$ vanishes for each element of the basis $B$ of $V$, hence we must have $a = 0$ indentically. But if

$a = \sum_i c_i a_i = 0, \tag{2}$

it follows that $c_i = 0$ for $1 \le i \le n$, since the $a_i$ form a basis of $V^*$, hence a linearly independent set. But this contradicts the original choice of the $c_i$ in (1); hence $M$ must be nonsingular.

As for part (b), for any $f \in V^*$, taking $x = f_[A]$, which I interpret as meaning the components of $x$ are the numbers $f_i$ where

$f = \sum_if_ia_i \tag{3}$

is the expansion of $f$ in the basis $A = \{a_1, \dots,a_n \}$ of $V^*$; that is,

$x = (x_1, x_2, . . . , x_n)^T = (f_1, f_2, . . . , f_n)^T. \tag{4}$

In the light of (4) we have, since

$M_{ij} = a_j(b_i), \tag{5}$

that

$(Mx)_i = \sum_jM_{ij}x_j = \sum_j a_j(b_i)x_j = \sum_j a_j(b_i)f_j$ $= \sum_j f_ja_j(b_i) = (\sum_j f_ja_j)(b_i) = f(b_i) = y_i, \tag{6}$

whence

$y = Mx. \tag{7}$

QED!!!

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.