Lets say I have a vector space $V$ and it's basis $B=(b1,b2,b3)$. How do I verify that for any $t$, the vectors $$c_{1,t} := (b1), c_{2,t} := (b1+b2), c_{3,t} := (t*b1+b2+b3)$$

are a basis of $V$ as well? Furthermore, if I then want to find the dual basis of $C$, where do I begin?

I tried to start off by proving linear independence of the $c$ vectors. I came up with the following equation: $$a_1*c_1+a_2*c_2+a_3*c_3 = 0$$ If I could verify that $a_1,a_2,a_3$ must all be zero, then I would have at least verified that they are linearly independent. But I cannot seem to solve this, even if I split it up into three seperate equations. I think I have to solve this another way. Any hints?

Edit: I tried to apply the hint I got, so I ended up with the coordinate matrix:

$$\langle B^*,C_t\rangle = \begin{pmatrix}1\ 1\ t\\ 0\ 1\ 1\\ 0\ 0\ 1 \end{pmatrix}$$

I can see now that they are indeed linearly independent. How do I find the dual basis now though?

  • $\begingroup$ Express $C$ as coordinate vectors in basis $B$. Write them in a matrix, you will see it has determinante $\neq 0$ and so they are linear independent, do you know how the dual basis changes when we have a basis change in the vector space? $\endgroup$ – user302982 Nov 28 '16 at 15:30
  • $\begingroup$ Edit: Nevermind, I think you can also see that they are linearly independent by looking at the collumns of the matrix, right? $\endgroup$ – Jack4t3 Nov 28 '16 at 15:40

Hint: To find the dual basis, go back to its definition: If $(\mathbf e_1,\dots,\mathbf e_n)$ is a basis for $V$, then its dual basis is a set of vectors $(\mathbf\varepsilon_1,\dots,\mathbf\varepsilon_n)$ such that $\mathbf\varepsilon_i[\mathbf e_j]=\delta_{ij}$ for $i,j=1..n$. If we represent the $\mathbf\varepsilon_i$ as row vectors and the $\mathbf e$ as column vectors, this can be expressed in matrix form as $$\pmatrix{—&\mathbf\varepsilon_1&—\\\vdots&\vdots&\vdots\\—&\mathbf\varepsilon_n&—}\pmatrix{\mid&\cdots&\mid\\\mathbf e_1&\cdots&\mathbf e_n\\\mid&\cdots&\mid}=I.$$ Does this product suggest anything to you?

  • $\begingroup$ I think I found it now. By matrix transformations I could find the matrix which, multiplied by my basis matrix produces $I$... Thanks! $\endgroup$ – Jack4t3 Nov 29 '16 at 16:52

If you have a basis $e_i$ for a inner product space $(V, < , >)$, then the dual basis in $V^*$ will be the functions that map $v$ to $<v,e_i>$


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