Linear mapping of a room, Show that F is bijective and find matrix F^-1

**Riazy** · October 3rd 2012, 11:21 PM

Hello guys I have attached the problem.. could someone help me solve it?

I have made "an attempt" on the problem

**FernandoRevilla** · October 4th 2012, 02:45 AM

The matrix of $A$ of $F$ with respect to $B=\{e_1,e_2,e_3\}$ , is the matrix satisfying $Y=AX$ where $X$ and $Y$ are respectively the coordinates of $x\in\mathbb{R}^3$ anf $F(x)$ with respect to $B$ . By a well known theorem:

$\begin{Bmatrix}F(e_1)=a_1e_1+a_2e_2+a_3e_3\\F(e_2) =b_1e_1+b_2e_2+b_3e_3\\F(e_3)=c_1e_1+c_2e_2+c_3e_3 \end{matrix}\Rightarrow A=\begin{bmatrix}{a_1}&{b_1}&{c_1}\\{a_2}&{b_2}&{c _2}\\{a_3}&{b_3}&{c_3}\end{bmatrix}$

Besides, $F$ is bijective if and only if $\det (A)\neq 0$ , and if $F$ is bijective, the matrix of $F^{-1}$ with respect to $B$ is $A^{-1}$ .

**johnsomeone** · October 4th 2012, 04:15 AM

Stupid me - I didn't read the problem correctly. My comments beneath the "%%%%%..." were thinking that you were supposed to find $F$ , but you're actually asked to find $F^{-1}$ .

1. Your approach was correct, but you didn't order your columns correctly.

(When a basis is *labelled* $\{e_1, e_2, e_3 \}$ for the purpose of writing linear maps as matricies, it will always be assumed that that's the order being used when referencing a position in the matrix.)

$F^{-1}(e_1) = e_1 - e_2 + e_3, F^{-1}(e_2) = e_1 + e_2 + e_3, F^{-1}(e_3) = e_1 + e_2$

Thus $F^{-1} = \left( \begin{matrix} | & | & | \\ F^{-1}(e_1) & F^{-1}(e_2) & F^{-1}(e_3) \\ | & | & | \end{matrix} \right) = \left( \begin{matrix} 1 & 1 & 1 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{matrix} \right)$ .

2. Since $\{e_1, e_2, e_3 \} \subset Image(F) \subset \mathbb{R}^3$ , it follows that $F$ is surjective (onto).

Thus, from dim(Null(F)) + dim(Image(F)) = 3, it follows that Null(F) = {0}. Therefore F is injective (one to one).

Therefore F is bijective.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Given $F \in L(\mathbb{R}^3, \mathbb{R}^3), \beta = \{e_1, e_2, e_3 \}$ , a linear map of real vector spaces and a basis,

and that $F(e_1 + e_2) = e_3, F(e_1 + e_2 + e_3) = e_2, F(e_1 - e_2 + e_3) = e_1$ ,

then show that $F$ is a bijection and find its matrix representation with respect to $\beta$ .

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Regarding your work, what are the columns of the matrix you made? What is $e_1 + e_2$ ? It's $F^{-1}(e_3)$ . Likewise, you 2nd column is $F^{-1}(e_2)$ and your 3rd is $F^{-1}(e_1)$ . So that's not going to be the matrix representation of $F$ .

Also, I don't understand why you thought volume was relevant.

It appears that you didn't really know what to do. See FernandoRevilla's post above. The first step is to get those equations for $F(e_1), F(e_2)$ , and $F(e_3)$ in terms of $e_1, e_2$ , and $e_3$ . Then, as his post says, you can write down the matrix and, by computing its determinant, show that it's bijective.

--------------------------------

The way you acheive that 1st step is to use that $F$ is linear and the equations,

$F(e_1 + e_2) = e_3, F(e_1 + e_2 + e_3) = e_2, F(e_1 - e_2 + e_3) = e_1$ ,

to solve for $F(e_1), F(e_2)$ , and $F(e_3)$ in terms of $e_1, e_2$ , and $e_3$ .

To you show how you do that, I'll do one:

$F(e_1 + e_2 + e_3) = e_2, F(e_1 - e_2 + e_3) = e_1$ , so $F(e_1 + e_2 + e_3) - F(e_1 - e_2 + e_3) = e_2 - e_1$ .

Since F is linear, you can simplify the left hand side by:

$F(e_1 + e_2 + e_3) - F(e_1 - e_2 + e_3) = F\left( (e_1 + e_2 + e_3) - (e_1 - e_2 + e_3) \right)$

$= F(2e_2) = 2F(e_2)$ .

Thus $F(e_1 + e_2 + e_3) - F(e_1 - e_2 + e_3) = 2F(e_2) = e_2 - e_1$ , so $F(e_2) = - \frac{1}{2}e_1 + \frac{1}{2}e_2$ .

$F(e_2) = - \frac{1}{2}e_1 + \frac{1}{2}e_2$ solves for $F(e_2)$ in terms of $e_1, e_2$ , and $e_3$ . Now do it for the others.

---------------------------------

One thing that might help is to label the *vectors* $F(e_1), F(e_2),$ and $F(e_3)$ by $u_1, u_2,$ and $u_3$ .

Then, using linearity, the equations $F(e_1 + e_2) = e_3, F(e_1 + e_2 + e_3) = e_2, F(e_1 - e_2 + e_3) = e_1$ become:

$u_1 + u_2 = e_3, u_1 + u_2 + u_3 = e_2, u_1 - u_2 + u_3 = e_1$ , where $u_1 = F(e_1), u_2 = F(e_2), u_3 = F(e_3)$ .

You then solve that for $u_1, u_2$ and $u_3$ just as you would a regular system of simultaneous equations, *except* remembering that all those things are vectors, not numbers. You can't multiply two vectors or divide by a vector (though you can multiply and divide vectors by real numbers!). However, to solve that system of equations, you won't need to do any of those forbidden acts. Thus, what you write on your paper when you solve it will look identical to how you'd solve a typical system of equations with 3 equations and 3 unknowns.

Anyway, you solve that system for $u_1, u_2$ and $u_3$ , which are $F(e_1), F(e_2),$ and $F(e_3)$ , and then you've finished the 1st step, and can proceed to solve the problem as described in FernandoRevilla's post above.

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