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

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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 $\bf R^3$