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$\phi \in \alpha(R^3,R^3)$

$B_c:\left(\begin{array}{rrr} 1&2&3 \\ 0&3&4 \\ -1&2&1 \end{array}\right) = [F]_C$

  1. Determine the analytic form of the operator $f(x,y,z)$ $x,y,z \in R^3$

  2. $f^{-1}$ in $B=\{(1,1,0),(1,0,1),(1,0,0)\}$

Can anybody give me the general steps in solving this kind of problems?

1. $B_c=\{e_1,e_2,e_3\}$


I understand it so far.


Why do we apply the function to the bases vectors and not to the x,y,z?


Tell me please, what is the $f(e_i)$ doing, I guess then I'll understand why I should take this combination

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A linear transformation always leaves the coefficients unchanged, and thus we have only to know what it does to the basis vectors, in order to determine the action of the transformation. – awllower Dec 6 '11 at 14:16
up vote 2 down vote accepted

The matrix of the operator $f$ with respect to the basis $B_c$ (is this what your notation $[F]_C$ refers to?) tells you the values of $f$ on this basis. For example, the first column tells you that

$$f(e_1) = 1\cdot e_1 + 0\cdot e_2 + (-1)\cdot e_3 = e_1-e_3.$$

Then if we want to compute the value of $f(x,y,z)$ for an arbitrary vector $(x,y,z)\in\mathbb{R}^3$, it would be useful to decompose this vector in terms of $B_c$ and then exploit the linearity of $f$.

Now, the notation $e_i$ usually refers to the standard basis vector of $\mathbb{R}^n$ whose entries are all zero except for the $i$th entry, which is 1. In the case above, we have $$e_1 = (1,0,0), e_2 = (0,1,0), e_3 = (0,0,1).$$

An arbitrary vector $(x,y,z)$ can be written as a linear combination of this particular basis in a simple way:

$$(x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1)$$

$$= xe_1 + ye_2 + ze_3.$$

So if we want to calculate $f(x,y,z)$, just apply $f$ to both sides:

$$f(x,y,z) = f( xe_1 + ye_2 + ze_3).$$

Now using the fact that $f$ is a linear operator, we can expand the right-hand side:

$$f(x,y,z) = xf(e_1) + yf(e_2) + zf(e_3).$$

From here, you can read off the values $f(e_i)$ from the matrix.

To answer your question about why we don't apply $f$ to the numbers $x,y,z$: the linear operator $f$ is a function from $\mathbb{R}^3$ to $\mathbb{R}^3$. So it only makes sense to evaluate $f(v)$ where $v$ is an element of $\mathbb{R}^3$, i.e. a triple of real numbers, rather than a single real number. However, the fact that $f$ is a linear operator means that it is compatible with multiplication by real numbers, in the sense that

$$f(cv) = cf(v)$$ whenever $c$ is a real number and $v$ is an element of $\mathbb{R}^3$. Linearity also means that $f(v+w) = f(v) + f(w)$ for all $v,w\in\mathbb{R}^3$. Combining these two facts is how the right-hand side of the above equation was expanded.

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