# find the matrix $\ A' \$ with respect to the basis $\ B'=\{(-2,1), \ (-1,1) \} \$

If the linear map $\ T:R^2 \to R^2 \$ has the matrix of transformation $$A=\begin{pmatrix} -3 & 1 \\ 3 & -1 \end{pmatrix}$$

with respect to the basis $\ B=\{(1,-1), \ (-1,5) \} \$ , then find the matrix $\ A' \$ with respect to the basis $\ B'=\{(-2,1), \ (-1,1) \} \$ ?

Let $\ A' \$ be the matrix corresponding the basis $\ B'=\{(-2,1), \ (-1,1) \} \$.

Then,

$A'=PAP^{-1} \$ , where $\ P=\begin{pmatrix} -2 & -1 \\ 1 & 1 \end{pmatrix}$

Thus,

$A'=\begin{pmatrix} -2 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} -3 & 1 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} -1 & -1 \\ 1 & 2 \end{pmatrix}=\begin{pmatrix} 3 & -1 \\ 0 & 0 \end{pmatrix}$

Am I right ?

• I don't think you're right. Your $P$ would convert a matrix with the standard basis into the basis B' but A is not written in the standard basis – N8tron Jun 7 '18 at 6:51
• so what would be $\ P \$ ? Help me – user484305 Jun 7 '18 at 6:59
• Look at your linear algebra book in change of coordinates section. I think you need a slightly later part or section than you are currently looking at. There will likely be an augmented matrix with all elements from each basis that you'll need to put in rref – N8tron Jun 7 '18 at 7:17

Let's look at this a little more generally general. Suppose we have $\mathcal B=\left\{b_1,b_2,\ldots,b_n\right\}$ and $\mathcal B'=\left\{b_1',b_2',\ldots,b_n'\right\}$ bases for a vector space $V$ a subspace of $\mathbb{R}^m$ and $T:V \to V$ linear with matrix with respect to $\mathcal{B}$ of $$A= [T]_\mathcal{B}$$

You want a change of basis matrix $P$ such that $$PAP^{-1}=[T]_\mathcal{B'}$$

Notationally if you take $[v]_\mathcal{B}$ to mean vector $v$ written in the coordinates of basis $\mathcal{B}$ that is it can be written uniquely as $$\textbf{v}=x_1\textbf{b}_1+x_2\textbf{b}_2+ \cdots +x_n\textbf{b}_n$$ then $[\textbf{v}]_\mathcal{B}=\begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n\end{bmatrix}$

So matrix $P$ must have the property that $P[v]_\mathcal{B}=[v]_{\mathcal{B}'}$ for all vectors $v$ and it will follow that $P^{-1}[v]_{\mathcal{B}'}=[v]_{\mathcal{B}}$ . So in particular this is true for the vectors $b_1,b_2,\ldots,b_n$ that is

$$\begin{split} p_1=Pe_1=&P[b_1]_\mathcal{B}=[b_1]_{\mathcal{B}'}\\ p_2=Pe_2=&P[b_1]_\mathcal{B}=[b_2]_{\mathcal{B}'}\\ &\vdots \\ p_n=Pe_n=&P[b_1]_\mathcal{B}=[b_n]_{\mathcal{B}'}\\ \end{split}$$

Where $p_1, \ldots, p_n$ are the column vectors of $P=[p_1,\ldots,p_n]$ Note in the proof we are going through the standard basis but we don't need it anymore we have a direct method of finding the columns of $P$ namely $p_k=[b_k]_{\mathcal{B}'}$

So for each $\textbf{b}_k$ we can write

$$\textbf{b}_k=x_1\textbf{b}_1'+x_2\textbf{b}_2'+ \cdots +x_n\textbf{b}_n'$$

Which can be rewritten as the matrix equation $$\textbf{b}_k=B'\textbf{x}$$ where $B'=[\textbf{b}_1',\textbf{b}_2', \ldots \textbf{b}_n']$ and $\textbf{x}=\textbf{p}_k$

Now you need to solve for $x$ for all $k=1,\ldots,n$

$$\begin{split} \textbf{b}_1&=B'\textbf{p}_1\\ \textbf{b}_2&=B'\textbf{p}_2\\ &\vdots& \\ \textbf{b}_n&=B'\textbf{p}_n\\ \end{split}$$

You can convert this into a matrix equation $B=B'P$ where $B=[\textbf{b}_1,\textbf{b}_2, \ldots \textbf{b}_n]$. You could just find $B'^{-1}$ and multiply by both sides to find $P=B'^{-1}B$ that's gimusi's soltion. however you can instead consider the following augmented matrix $$[B' | B]$$

when you row reduce it to rref you get $$[I|B'^{-1}B]=[I|P]$$

In your case the matrix you need to reduce is

$$\left[\begin{array}{rr|rr} -2 & -1 & 1 & -1\\ 1 & 1 & -1 & 5 \end{array}\right]$$

HINT

We can use an intermediate change by the canonical basis, notably let consider

• the change of basis from $B$ to the canonical that is $v_C=P_Bv_B$

$$P_B=\begin{pmatrix} 1 & -1 \\ -1 & 5 \end{pmatrix}$$

• the change of basis from $B’$ to the canonical that is $v_C=P_{B’}v_{B’}$

$$P_{B'}=\begin{pmatrix} -2 & -1 \\ 1 & 1 \end{pmatrix}$$

therefore in the canonical basis we have

$$y_B=Ax_B\implies P_B^{-1}y_C=AP_B^{-1}x_C\implies y_C=P_BAP_B^{-1}x_C=A_Cx_C$$

from here let consider the change in the new basis $B’$ that is

$$y_C=A_Cx_C\implies P_{B’}y_{B’}=A_CP_{B’}x_{B’} \implies y_{B’}=P_{B’}^{-1}A_CP_{B’}x_{B’}=A'x_{B’}=$$

therefore

$$A'=P_{B’}^{-1}\,P_B\,A\,P_B^{-1}\,P_{B’}$$

• You really don't need to use the canonical basis. It is one way to do it, but there is a way to compute the change of basis matrix between any two basis without going through the standard one. The notation is a little worse, but the computation is not so bad – N8tron Jun 7 '18 at 7:37
• @N8tron Ok I would say, we can use the canonical basis, it is a clear way to obtain the result. For the notation you are right but I've followed the notation used in the OP. – gimusi Jun 7 '18 at 7:58
• Also it's worth pointing out that though this method is okay for $2 \times 2$ matrices it scales really poorly with to $n \times n$ finding each inverse will take roughly the same computational power as finding the single change of coordinates matrix, not to mention the extra matrix multiplications – N8tron Jun 7 '18 at 11:11
• @N8tron Anyway I think it can be useful to know at least from the theoretical point of view in order to set up the correct equality to obtain $A'$ from $A$ and then performing the calculation by computational methods. – gimusi Jun 7 '18 at 11:59
• @N8tron That's Always nice have different point of view for any OP. Thanks – gimusi Jun 7 '18 at 12:14