Finding an $\mathbb{R}^3$ basis in which the matrix of a linear trans. is $\left(\begin{smallmatrix}0&1&0\\0&0&1\\0&0&0\\ \end{smallmatrix}\right)$

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!

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

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.

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 $$\bf R^3$$

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.