# Different transformation matrices corresponding to different basis pairs for the same linear map

Consider a linear map $A: V \to W$. Using $B_1$ as the basis in $V$, and $F_1$ as the basis in $W$, the corresponding transformation matrix is $$A_1 = \begin{pmatrix} 1&2&3\\ 2&4&x \end{pmatrix}.$$

The same linear map, using $B_2$ as the basis in $V$, and $F_2$ as the basis in $W$, is $$A_2 = \begin{pmatrix} 1&1&1\\ 4&y&z \end{pmatrix}.$$

1) Assume that $x = 5$. Can we determine $y$ and $z$?

2) Does there exist a coordinate triple that always represents different vectors of $V$ for $A_1$ and $A_2$, regardless of the $x, y, z$ values? If yes, what is it? If no, why doesn't such a triple exist?

The only thing I know for sure at this point is that both $A_1$ and $A_2$ must have the same rank. Right now I'm of the opinion that we can only go so far as to say that, given $x$ = 5, we know that one out of $y$ and $z$ must be 4 (rank of $A_2$ must be 2), but we can't definitely say which of the two will be 4, and can't determine the value of the other variable with this level of information. No idea how to tackle part b, though.

EDIT: To clarify the second part, consider a coordinate triple, $C = (c_1,c_2,c_3)$ which represents a vector $v_1 \in V$ in terms of $B_1$ and a vector $v_2 \in V$ in terms of $B_2$. Is there any set of values for which $C$ is guaranteed to represent these vectors such that $v_1 \neq v_2$, regardless of the $x, y, z$ values?

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For your first question, you are essentially asking that if $PA_1Q=A_2$ for some (invertible) change-of-basis matrices $P$ and $Q$, can we determine $y$ and $z$? The answer is negative. For example, take $P=I$ (i.e. $F_1=F_2$). We have \begin{align*} \begin{pmatrix}1&2&3\\2&4&5\end{pmatrix} \begin{pmatrix}7&1&-1\\0&0&1\\-2&0&0\end{pmatrix} &=\begin{pmatrix}1&1&1\\4&2&2\end{pmatrix},\\ \begin{pmatrix}1&2&3\\2&4&5\end{pmatrix} \begin{pmatrix}7&-5&0\\0&0&-1\\-2&2&1\end{pmatrix} &=\begin{pmatrix}1&1&1\\4&0&1\end{pmatrix}. \end{align*} So, different choices of $B_2$ will result in different appearances of $A_2$.
But since $A_1$ and $A_2$ are similar matrices, what I added about the rank should hold, right? – wemblem Feb 20 '13 at 23:13
@forthecookies (1) $V$ and $W$ have different dimensions, so $B_i\not=F_i$ and $A_1,A_2$ are not similar matrices in the conventional sense. (Two square matrices $X$ and $Y$ are called similar if $X=PYP^{-1}$ for some invertible matrix $P$.) (2) No, as long as $(y,z)\not=(4,4)$, the rank of $A_2$ would be 2. – user1551 Feb 20 '13 at 23:20