Find basis given linear map and transformation matrix

Question

Consider the linear map $f\colon\mathbb{R}^3\to\mathbb{R}^3$ and let $E=\{e_1,e_2,e_3\}$ be the standard basis of $\mathbb{R}^3$. We are given $f(e_1)=2e_1+e_2+e_3, f(e_2)=-4e_1+2e_2, f(e_3)=-4e_2-2e_3$. The task is to find basis $A=\{a_1,a_2,a_3\}$ and $B=\{b_1,b_2,b_3\}$ of $\mathbb{R}^3$ such that the transformation matrix from basis $A$ to basis $B$ is $$ M_B^A=\begin{pmatrix}1 & 2 & 3\\4 & 5 & 6\\ 7 & 8 & 9\end{pmatrix}. $$

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My idea was to use $$ M_B^A=T_B^E\cdot M_E^E\cdot T_E^A, $$ where $M_E^E$ is the transformation matrix I can deduce from above, namely, $$ M_E^E=\begin{pmatrix}2 & -4 & 0\\1 & 2 & -4\\1 & 0 & -2\end{pmatrix}. $$ And the columns of $T_E^A$, the change of basis matrix from basis $A$ to basis $E$, should just consist of the vectors of $A$ since $E$ is the standard basis, i.e., $$ T_E^A=\begin{pmatrix}a_1 & a_2 & a_3\end{pmatrix}\in\mathbb{R}^{3\times 3}. $$ Does this help to determine $A$ and $B$?

It seems to me that I first need to determine $T_B^E$ but I am missing information to do so.

Do you have an idea? Maybe my ansatz is not the right idea. However, I have no alternative idea.

Answer 1

Suppose that for each $i\in\{1,2,3\}$ we have $a_i=a_{i1}e_1+a_{i2}e_2+a_{i3}e_3$ and $b_i=b_{i1}e_1+b_{i2}e_2+b_{i3}e_3$. Let $M_A=\|a_{ij}\|$ and $M_B=\|b_{ij}\|$. Let $F=\|f(a_i)_j\|$, where the components of the vectors $f(a_i)$ are taken in the standard basis. Following Digitallis' comment, we obtain that $f(a_i)=a_{i1} f(e_1)+a_{i2}f(e_2)+a_{i3}f(e_3)$ for each $i\in\{1,2,3\}$. Thus $F=M_A (M^E_E)^t$. On the other hand, $F=(M^A_B)^t M_B$.

So we have to find some nonsingular real $3\times 3$ matrices $M_A$ and $M_B$, satisfying the equality $$M_A (M^E_E)^t = (M^A_B)^t M_B.$$ This can be done, for instance, as follows. The matrix $(M^E_E)^t$ has rank $2$, so it can be transformed by a sequence of basic transformations to the diagonal matrix $\Delta=\operatorname{diag}(1,1,0)$. The sequence of basic transformations provides us nonsingular real $3\times 3$ matrices $X$ and $Y$ such that $X(M^E_E)^tY=\Delta$. Similarly we can find nonsingular real $3\times 3$ matrices $U$ and $V$ such that $U(M^A_B)^tV=\Delta$. Then $X(M^E_E)^tY=U(M^A_B)^tV$, so it suffices to put $M_A=U^{-1}X$ and $M_B=VY^{-1}$.

Answer 2

One approach to handle this question is to note that every matrix has a unique reduced row echelon form (rref): For every matrix $A\in M_{m,n}(\mathbb{R})$, if $\hat{A}\in M_{m,n}(\mathbb{R})$ is the rref of $A$, there exists an elementary matrix $S\in M_m(\mathbb{R})$ such that $\hat{A}=SA$.

Here $S$ reflect the elementary row operations needed to recover $A$ from $\hat{A}$. To obtain such $S$, one may consider the augmented matrix $(A\mid I_3)$. Observe that $$S(A\mid I_3)=(SA\mid SI_3)=(\hat{A}\mid S).$$ Then it suffices to compute the rref of $A$ along with the augmented matrix $(A\mid I_3)$.

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First, we may find the rref of $M_B^A$:

\begin{align*} (M_B^A\mid I_3)&=\begin{pmatrix} 1 & 2 & 3 & 1 & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1 \end{pmatrix}\xrightarrow{r_3\gets r_3-r_2}\begin{pmatrix} 1 & 2 & 3 & 1 & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 3 & 3 & 3 & 0 & -1 & 1 \end{pmatrix} \\ &\xrightarrow{r_2\gets r_2-r_1}\begin{pmatrix} 1 & 2 & 3 & 1 & 0 & 0 \\ 3 & 3 & 3 & -1 & 1 & 0 \\ 3 & 3 & 3 & 0 & -1 & 1 \end{pmatrix}\xrightarrow{r_3\gets r_3-r_2}\begin{pmatrix} 1 & 2 & 3 & 1 & 0 & 0 \\ 3 & 3 & 3 & -1 & 1 & 0 \\ 0 & 0 & 0 & 1 & -2 & 1 \end{pmatrix} \\ &\xrightarrow{r_2\gets r_2-3r_1}\begin{pmatrix} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & -3 & -6 & -4 & 1 & 0 \\ 0 & 0 & 0 & 1 & -2 & 1 \end{pmatrix}\xrightarrow{r_2\gets r_2/(-3)}\begin{pmatrix} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & 4/3 & -1/3 & 0 \\ 0 & 0 & 0 & 1 & -2 & 1 \end{pmatrix} \\ &\xrightarrow{r_1\gets r_1-2r_2}\begin{pmatrix} 1 & 0 & -1 & -5/3 & 2/3 & 0 \\ 0 & 1 & 2 & 4/3 & -1/3 & 0 \\ 0 & 0 & 0 & 1 & -2 & 1 \end{pmatrix}. \end{align*} This shows that $$X_1:=\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{pmatrix}=\boxed{\begin{pmatrix} -5/3 & 2/3 & 0 \\ 4/3 & -1/3 & 0 \\ 1 & -2 & 1 \end{pmatrix}}M_B^A=:\boxed{S_1}M_B^A.$$

By similar arguments, we also have $$X_2:=\begin{pmatrix} 1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}=\boxed{\begin{pmatrix} 1/4 & 1/2 & 0 \\ -1/8 & 1/4 & 0 \\ -1/4 & -1/2 & 1 \end{pmatrix}}M_E^E=:\boxed{S_2}M_E^E.$$

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Now we may make some further observations: Here $X_1=S_1M_B^A$ where $S_1,S_2$ are non-singular, so we have $M_B^A=S_1^{-1}X_1$. Because $X_1$ and $X_2$ are in rref, if we can find another non-singular matrix $T$ such that $X_1=X_2T$, then $$M_B^A=S_1^{-1}X_1=S_1^{-1}X_2T=S_1^{-1}S_2M_E^ET.$$

To find such $T$, one can consider the dual version of our previous argument by performing elementary column operations to $\begin{pmatrix} X_1 \\ I_3 \end{pmatrix}$ and $\begin{pmatrix} X_2 \\ I_3 \end{pmatrix}$, and multiply the corresponding elementary matrices in some fashion.

In this question, however, some easy observations are sufficient as the structures of $X_1$ and $X_2$ are rather simple: $$X_1=\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{pmatrix}=\begin{pmatrix} 1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}\boxed{\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}}=X_2\boxed{T}$$

Let me explain how I found such $T$:

• Because the first two columns of $X_1$ and $X_2$ are identical, we can then put the first two columns of $T$ to be simply $(1,0,0)^T$ and $(0,1,0)^T$.

• Now since $T$ is supposed to be non-singular, I can put the last column to be of the form $(x,y,1)^T$ (here $1$ of course can be replaced by other real numbers). Then by solving linear systems, I got $x=1$ and $y=3$.

As a result, such $T$ is not necessarily unique, so it is possible for you to find different solutions.

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Anyway, now we may let

$$T_B^E:=S_1^{-1}S_2=\begin{pmatrix} -5/3 & 2/3 & 0 \\ 4/3 & -1/3 & 0 \\ 1 & -2 & 1 \end{pmatrix}^{-1}\begin{pmatrix} 1/4 & 1/2 & 0 \\ -1/8 & 1/4 & 0 \\ -1/4 & -1/2 & 1 \end{pmatrix}=\boxed{\begin{pmatrix} 0 & 1 & 0 \\ 3/8 & 13/4 & 0 \\ 1/2 & 5 & 1 \end{pmatrix}}$$

and

$$T_E^A:=T=\boxed{\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix}}.$$

As noted above, we now have

$$M_B^A=S_1^{-1}S_2M_E^ET=T_B^EM_E^ET_E^A.$$

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P.S. Here are some theories: Let $A\in M_{m,n}(\mathbb{R})$ be of rank $r$. Then there exist non-singular matrices $P\in M_m(\mathbb{R})$ and $Q\in M_n(\mathbb{R})$ such that $$A=P\begin{pmatrix} I_r & 0_{r\times(n-r)} \\ 0_{(m-r)\times(n-r)} & 0_{(m-r)\times(n-r)} \end{pmatrix}Q.$$ Here $P$ can be obtained by performing elementary row operations to $A$ to find rref, and $Q$ can be obtained by continuing with elementary column operations.

If $B\in M_{m,n}(\mathbb{R})$ represents the same linear transformation as $A$ with respect to different bases, then the rank of $B$ is also equal to $r$, and certainly, we can find $P\in M_m(\mathbb{R})$ and $Q\in M_n(\mathbb{R})$ such that \begin{align*} B&=P'\begin{pmatrix} I_r & 0_{r\times(n-r)} \\ 0_{(m-r)\times(n-r)} & 0_{(m-r)\times(n-r)} \end{pmatrix}Q' \\ &=P'(P^{-1}AQ^{-1})Q'=\boxed{(P'P^{-1})}A\boxed{(Q^{-1}Q')}. \end{align*}