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I have to determine the $Matrix F$ which describes f in comparison to the base A(a1,a2,a3,a4), and the matrix B - which describes regarding the basis E=(e1,e2,e3,e4) for $R^4$ That is
$F =a[f]a$
$B=e[f]e$

From a previous calculation, i know that:

$f(x)=(x \cdot a1)a2+(x \cdot a2)a3 +(x \cdot a3)a4$
Where $a1,a2...$ is vectors forming an orthogonal basis $R^4$

But I dont get the meaning of F=a[f]a

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  • $\begingroup$ Too confusing. Do you have two basis for $\mathbb{R}^4$ and you want to write the matrix of the map $f$? $\endgroup$
    – Sigur
    Dec 27, 2012 at 16:43
  • $\begingroup$ Sorry for the confusion. Yeah I have two basis for $R^4$ E(e1,e2,e3,e4) and A(a1,a2,a3,a4). $\endgroup$
    – user54425
    Dec 27, 2012 at 16:47
  • $\begingroup$ To determine the matrix you should apply the map on the base, write the image as linear combinations of the base and transpose to obtains the matrix. $\endgroup$
    – Sigur
    Dec 27, 2012 at 16:49
  • $\begingroup$ Could u elaborate how to map on the base? Thanks btw $\endgroup$
    – user54425
    Dec 27, 2012 at 16:52

1 Answer 1

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I don't understand you notation, but I believe that you want $F:=[f]_A^A$, the matrix of the map $f:(\mathbb{R}^4,A)\to (\mathbb{R}^4,A)$ where $A=\{a_1,\ldots, a_4\}$ is a base for $\mathbb{R}^4$.

Also similar for the other base.

Well, if this is the case, you have to compute $f(a_i)$ (using the definition of $f$) and write it as linear combination of vectors on $A$, that is, $$f(a_i)=\sum_{j=1}^4\alpha_{ij}a_j, \quad 1\leq i\leq 4.$$ Then you write the matrix $$F:=[f]_A^A=\begin{pmatrix}\alpha_{11} & \cdots & \alpha_{41} \\ \vdots & \ddots & \vdots \\ \alpha_{14} & \cdots & \alpha_{44}\end{pmatrix}.$$

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  • $\begingroup$ Thanks man I will try and then respond :) $\endgroup$
    – user54425
    Dec 27, 2012 at 17:04
  • $\begingroup$ Not sure i quite got it i45.tinypic.com/m7ru5d.png Is that anything near? I plotted the f(x) in Maple as M(m). $\endgroup$
    – user54425
    Dec 27, 2012 at 17:58
  • $\begingroup$ By definition, $f(a_1)=(a_1\cdot a_1)a_2+ (a_1\cdot a_2)a_3+ (a_1\cdot a_3)a_4$. So its coordinates are $f(a_1)=(0,a_1\cdot a_1, a_1\cdot a_2,a_1\cdot a_3)$. If the base $A$ is orthonormal, then you get $(0,1,0,0)$. $\endgroup$
    – Sigur
    Dec 27, 2012 at 18:05
  • $\begingroup$ Ah u simply get f(a1),f(a2),f(a3),f(a4) and put them in the rows of F? $\endgroup$
    – user54425
    Dec 27, 2012 at 20:21
  • $\begingroup$ i50.tinypic.com/11b4vuh.png $\endgroup$
    – user54425
    Dec 27, 2012 at 20:38

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