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.

  • $\begingroup$ What do you mean by $a_j(b_i)$? $\endgroup$ – superAnnoyingUser Nov 18 '13 at 18:31

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


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

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 $\;V^*\;$


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

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


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


$y = Mx. \tag{7}$


Hope this helps. Cheerio,

and as always,

Fiat Lux!!!


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