0
$\begingroup$

Let $f:\mathbb{R}^3\to\mathbb{R}^3$ be a linear transformation with the following associated matrix in $\mathbb{R}^3$ canonical basis: \begin{pmatrix} 0 & 2 & 1\\ 0 & 0 & 3\\ 0 & 0 & 0\\ \end{pmatrix} I have to find an $\mathbb{R}^3$ basis such that the matrix of $f$ in this basis is: \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{pmatrix} Could you give me some hints? Thanks!

$\endgroup$
  • $\begingroup$ French language? "Application" here is usually "transformation" in English. $\endgroup$ – Matt Samuel Dec 24 '18 at 12:43
  • $\begingroup$ Your are right. Sorry. $\endgroup$ – Gibbs Dec 24 '18 at 12:45
  • 2
    $\begingroup$ No need to be sorry. Was clear what you meant, just thought you should know in case you didn't already. $\endgroup$ – Matt Samuel Dec 24 '18 at 12:46
3
$\begingroup$

I hope canonical basis means the standard basis $\left\lbrace e_1, e_2, e_3 \right\rbrace$. From the matrix, we can say that

$$f \left( e_1 \right) = \left( 0, 0, 0 \right)$$

$$f \left( e_2 \right) = \left( 2, 0, 0 \right)$$

$$f \left( e_3 \right) = \left( 1, 3, 0 \right)$$

Now, let the other basis be $\left\lbrace v_1, v_2, v_3 \right\rbrace$ for which the matrix of $f$ is given by $\left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right]$. Let $v_1 = \alpha_{11} e_1 + \alpha_{12}e_2 + \alpha_{13}e_3$, $v_2 = \alpha_{21} e_1 + \alpha_{22} e_2 + \alpha_{23}e_3$ and $v_3 = \alpha_{31} e_1 + \alpha_{32} e_2 + \alpha_{33} e_3$. We know from the matrix that

$$f \left( v_1 \right) = \left( 0, 0, 0 \right)$$

$$f \left( v_2 \right) = \left( 1, 0, 0 \right)$$

$$f \left( v_3 \right) = \left( 0, 1, 0 \right)$$

Since $f$ is liner, we get $$2 \alpha_{12} + \alpha_{13} = 0 \\ 3 \alpha_{13} = 0$$

This gives that $\alpha_{12} = 0$ and $\alpha_{11}$ is a free variable.

Similarly, we have

$$2 \alpha_{22} + \alpha_{23} = 1 \\ 3 \alpha_{23} = 0$$

This gives $\alpha_{22} = \dfrac{1}{2}$ and $\alpha_{21}$ is a free variable.

Finally, we also have

$$2 \alpha_{32} + \alpha_{33} = 0 \\ 3 \alpha_{33} = 1$$

Thus, we get $\alpha_{32} = - \dfrac{1}{6}$, $\alpha_{33} = \dfrac{1}{3}$ and $\alpha_{31}$ is a free variable.

While you can choose any real number for the free variable, since our intention to find a basis, we fix those free variables as $1$.

Thus, a basis for which $f$ has the given matrix is $\left\lbrace \left( 1, 0, 0 \right), \left( 1, \dfrac{1}{2}, 0 \right), \left( 1, - \dfrac{1}{6}, \dfrac{1}{3} \right) \right\rbrace$. You can also keep the free variables as $0$ and you will get yet another basis.

$\endgroup$
1
$\begingroup$

Note that in such a basis, $f_1,f_2,f_3$, we have $M f_3=f_2$ and $Mf_2=f_1$, so that the choice of $f_3$ determines the basis. Take for instance $f_3=e_3$, then $f_2=Mf_3= e_1+3e_2$, and $f_1=Mf_2=6 e_2$, for $e_1,e_2,e_3$ the original basis of $\bf R^3$

$\endgroup$
0
$\begingroup$

The image of the basis vectors under a linear transformation are the columns of the matrix in that basis. Do you know any (non-zero) vector which is mapped to $[0,0,0]^T$? That's $v_1$. Do you know any vector which is mapped to $v_1$? That's $v_2$. Do you know any vector which is mapped to $v_2$? That's $v_3$. Now $v_1,v_2,v_3$ is the basis you're looking for.

$\endgroup$

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.