0
$\begingroup$

Consider $5$ functions $f_1, …, f_5$ defined on the interval $[0,5]$, and $\textbf{assume they have the following property:}$ given any numbers $b_1, …, b_5$ there always exist coefficients $c_1, ..., c_5$ such that
$\,\,\,\,\,\,\,\,\,\,$ $\displaystyle \sum_{j=1}^5 c_jf_j(i) = b_i$
Then, $\textbf{show}$ that under these circumstances that once the number $b_1, …, b_5$ are fixed the coefficients $c_1, ..., c_5$ that solve the above problem are uniquely determined, this will in particular mean there is a linear function that determines the $c_j's$ from the $b_i's$, which means there is a matrix $M_{ij}$ such that
$\,\,\,\,\,\,\,\,\,\,$ $\displaystyle c_j = \sum_{i=1}^5 M_{ij}b_i$
Without knowing anything else about $M$, say what is $\left(M^{-1}\right)_{ij}$.


I am not sure if I am setting this up correctly or taking the proper approach. First, I wrote out the first equation as a system of equations and got:
$c_1f_1(1)+c_2f_2(1)+c_3f_3(1)+c_4f_4(1)+c_5f_5(1)=b_1$
$c_1f_1(2)+c_2f_2(2)+c_3f_3(2)+c_4f_4(2)+c_5f_5(2)=b_2$
$\vdots$
$c_1f_1(5)+c_2f_2(5)+c_3f_3(5)+c_4f_4(5)+c_5f_5(5)=b_5$.

So, I set up a matrix out of these systems:
$\begin{pmatrix} c_1f_1(1) & c_2f_2(1) & ... & c_5f_5(1)\\ \vdots \\ c_1f_1(5) & ... & ... & c_5f_5(5) \end{pmatrix} = \begin{pmatrix} b_1 \\ \vdots \\ b_5 \end{pmatrix}$ $\implies$ $\begin{pmatrix} f_1(1) & f_2(1) & ... & f_5(1)\\ \vdots \\ f_1(5) & ... & ... & f_5(5) \end{pmatrix}\begin{pmatrix} c_1 \\ \vdots \\ c_5 \end{pmatrix} = \begin{pmatrix} b_1 \\ \vdots \\ b_5 \end{pmatrix}$

Let $M = \begin{pmatrix} f_1(1) & f_2(1) & ... & f_5(1)\\ \vdots \\ f_1(5) & ... & ... & f_5(5) \end{pmatrix}$, $\overline{\textbf{c}} = \begin{pmatrix} c_1 \\ \vdots \\ c_5 \end{pmatrix}$, and $\overline{\textbf{b}} = \begin{pmatrix} b_1 \\ \vdots \\ b_5 \end{pmatrix}$.

So, we have $M \,\overline{\textbf{c}} = \overline{\textbf{b}}$.

**Here I wasn't too sure what to do next. I wanted to show that $M$ is invertible, but the only way I know is that the determinant is not zero. I didn't know how to do that with this $5 \times 5$ matrix. I wanted to show this becuase I figured it would help me show that $\overline{\textbf{c}} = M^{-1} \overline{\textbf{b}}$ which would mean that $\left(M^{-1}\right)_{ij} = M^T$. Or am I mistaken about the inverse being the transpose? As far as uniqueness goes for $\overline{\textbf{c}}$, I thought if we assume there is a $\overline{\textbf{c}}^{'}$, then we could obtain $\overline{\textbf{b}}$ which int turn means $\overline{\textbf{c}}^{'} = \overline{\textbf{c}}$.

$\endgroup$
6
  • $\begingroup$ By the way: you can more conveniently type in bold text by putting everything between double-asterisks **like this**. $\endgroup$ Feb 1 at 16:21
  • $\begingroup$ Other formatting things: type $[\vdots]$ with \vdots and $[\ddots]$ with \ddots. $\endgroup$ Feb 1 at 16:23
  • $\begingroup$ @BenGrossmann What is the purpose of the diagonal dots? Is it to show that the matrix is filled up? For instance, vertical dots show that the matrix extends similarly down the columns. So what is the purpose of diagonal dots? $\endgroup$
    – Doodle Bob
    Feb 1 at 16:31
  • $\begingroup$ @D For example, I would have written $$ \pmatrix{c_1f_1(1) & c_2f_2(1) & \cdots & c_5f_5(1)\\ c_1f_1(2) & c_2f_2(2) & \cdots & c_5 f_5(2)\\ \vdots &\vdots & \ddots & \vdots\\ c_1f_1(5) & c_2f_2(5) & \cdots & c_5f_5(5) }. $$ Yes, you could leave that middle blank with $$ \pmatrix{c_1f_1(1) & c_2f_2(1) & \cdots & c_5f_5(1)\\ c_1f_1(2) & c_2f_2(2) & \cdots & c_5 f_5(2)\\ \vdots &\vdots & & \vdots\\ c_1f_1(5) & c_2f_2(5) & \cdots & c_5f_5(5) }, $$ but I find the first version to be nicer. $\endgroup$ Feb 1 at 16:33
  • $\begingroup$ @BenGrossmann Thank you, that helps me with formatting. I am new to writing matrices in latex. $\endgroup$
    – Doodle Bob
    Feb 1 at 16:36
2
$\begingroup$

Note that the matrix that you have called $M$ does the reverse of what is being requested: given the vector $\mathbf c$, you have found a matrix $A$ for which $A\mathbf c = \mathbf b$. What is required is a matrix "that determines the $c_j$'s from the $b_i$'s". In other words, $M$ should satisfy $\mathbf c = M \mathbf b$.

However, your work so far makes this easy. Note that $$ A \mathbf c = \mathbf b \implies \mathbf c = A^{-1} \mathbf b. $$ In other words, the desired matrix $M$ is the inverse of the matrix you found (which I have called $A$). Note that this requires the inverse of $M$ to exist, but we can conclude that this must hold because the $c_j$'s are uniquely determined from the $b_i$'s.

With that, the last part of the question is easy: the inverse of $M$ is simply $(A^{-1})^{-1} = A$. In other words, we have $$ (M^{-1})_{ij} = f_j(i). $$

$\endgroup$
6
  • $\begingroup$ So you begin with $M=A^{-1}$ thus obtaining $M^{-1}=(A^{-1})^{-1})=A$. But, what is $A$? $\endgroup$
    – Doodle Bob
    Feb 1 at 23:48
  • $\begingroup$ @DoodleBob As I said, $A$ is the matrix that you called $M$. It is the matrix whose $i,j$ entry is $f_j(i)$. $\endgroup$ Feb 1 at 23:53
  • $\begingroup$ So do you end up with $M=M^{-1}$? $\endgroup$
    – Doodle Bob
    Feb 2 at 0:13
  • $\begingroup$ @DoodleBob No, I end up with $M = A^{-1}$ $\endgroup$ Feb 2 at 0:14
  • 1
    $\begingroup$ @DoodleBob No, I said that $A$ is the matrix that you (mistakenly) called $M$. I have chosen to call this matrix $A$ so that I could use the letter $M$ to refer to the matrix that is called $M$ in the question. $\endgroup$ Feb 2 at 1:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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