Prove that there are no bases in $\mathbb R^3$ in which $f$ has the matrix $B$

Question

Suppose the linear map $f : \mathbb R^3 \to \mathbb R^3$ has a matrix $\begin{bmatrix} 1&0&2\\ 7&2&9\\ 1&0&5 \end{bmatrix}$ in some bases. Prove that there are no bases in $\mathbb R^3$ in which $f$ has the matrix $\begin{bmatrix} 1&0&2\\ 1&-1&-1\\ 1&1&5 \end{bmatrix}$.

My solution:
First matrix has order $3$, second matrix has order $2$ and in particular has a determinant $0$. If such a basis existed, then for some invertible change of basis matrix $P$ we would have $A = P^{−1}BP$, where $A,B$ are the matrices from the problem. However, the determinants are different, so equality cannot hold, so they are not matrices of the same linear map.

I wrote this answer in my homework, but got the answer that "I only considered similar matrices, which is the wrong approach". I don't understand where I assumed that these matrices are similar and what the correct answer should look like.

Answer 1

Suppose and $B_1=(v_1,v_2,v_3)$ and $C_1=(w_1,w_2,w_3)$ are ordered basis of $\mathbb{R}^3$ in which $f$ has matrix representation $$A=\begin{bmatrix} 1&0&2\\ 7&2&9\\ 1&0&5 \end{bmatrix}$$ That is \begin{align} f(v_1)&=w_1+7w_2+w_3\\ f(v_2)&=2w_2\\ f(v_3)&=w_1+9w_2+5w_3 \end{align} Suppose $B_2=(u_1,u_2,u_3)$ and $C_2=(z_1,z_2,z_3)$ are other ordered bases of $\mathbb{R}^3$ and that (a) $M$ transforms expressions in base $B_2$ into base $B_1$ and (b) $N$ transforms expressions in base $C_1$ into base $C_2$. Then, the matrix $NAM$ is the representation of $f$ in the ordered bases $B_2$-$C_2$. The matrices $M$ and $N$ are invertible (why?) and so, $\operatorname{det}(NAM)=\operatorname{det}(N)\operatorname{det}(A)\operatorname{det}(M)\neq0$ since $\operatorname{det}(A)=6\neq0$.

On the other hand, the matrix $$ L=\begin{bmatrix} 1&0&2\\ 1&-1&-1\\ 1&1&5 \end{bmatrix} $$ has $\operatorname{det}(L)=0$. Hence $L\neq NAM$ for any invertible matrices $M$, $N$.

Answer 2

If $B$ and $B'$ are bases of $\Bbb R^3$, then what you did shows that the matrix $[f]_B^B$ of $f$ with respect to the basis $B$ cannot possible be equal to $[f]_{B'}^{B'}$. But the idea is to prove that there are no bases $B_1$; $B_2$, $B_3$, and $B_4$ of $\Bbb R^3$ such that $[f]_{B_1}^{B_2}=[f]_{B_3}^{B_4}$.

You can do that by showing that the second matrix times $(2,3,-1)$ is equal to $(0,0,0)$, but there is no non-null vector $v\in\Bbb R^3$ such that the first matrix times $v$ is the null vector.