Brock Byrd
2022-07-16
Answered

Let $T:{M}_{2x2}\to {\mathbb{R}}^{3}$ have matrix $[T{]}_{B,A}=\left[\begin{array}{cccc}1& 2& 0& 1\\ 0& 1& 1& 0\\ 1& 1& -1& -1\end{array}\right]$ relative to $A=\{\left[\begin{array}{cc}2& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 5& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 6\end{array}\right]\}$ and $\beta =\{(1,1,1),(1,2,3),(1,4,9)\}$. Find the matrix of T relative to the bases ${A}^{\prime}=\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 4\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 2& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 7\end{array}\right]\}$ and ${\beta}^{\prime}=\{(1,1,1),(1,0,0),(1,1,0)\}$

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Kaya Kemp

Answered 2022-07-17
Author has **18** answers

First of all, notice that ${M}_{2x2}\cong {R}^{4}$ with $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\to (a,b,c,d)$.

Your first step is the right way to go, but I would be careful with the notation. Let $\alpha $ be the standard basis auf ${\mathbb{R}}^{3}$, then

$T(\left[\begin{array}{cc}2& 0\\ 0& 0\end{array}\right])=(1,0,1{)}_{B}=(2,5,10{)}_{\alpha}=(10,-3,-5{)}_{{B}^{\prime}}$

$T(\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right])=(2,1,1{)}_{B}=(4,8,9{)}_{\alpha}=(9,-4,-1{)}_{{B}^{\prime}}$

$T(\left[\begin{array}{cc}0& 0\\ 5& 0\end{array}\right])=(0,1,-1{)}_{B}=(2,5,12{)}_{\alpha}=(12,-3,-7{)}_{{B}^{\prime}}$

$T(\left[\begin{array}{cc}0& 0\\ 0& 6\end{array}\right])=(1,0,-1{)}_{B}=(0,-3,-8{)}_{\alpha}=(-8,3,5{)}_{{B}^{\prime}}$

Thus, we have

$[T{]}_{{B}^{\prime},A}=\left[\begin{array}{cccc}10& 9& 12& -8\\ -3& -4& -3& 3\\ -5& -1& -7& 5\end{array}\right]$

Now we are almost done. The new basis ${A}^{\prime}$ is good to handle because every new basis vector is a scalar multiple of a vector of $A$. For example the first basis vector of ${A}^{\prime}$ is half of its counterpart of $A$ (So $(2,0,0,0{)}_{{A}^{\prime}}=(1,0,0,0{)}_{A}$). So if you map $(1,0,0,0{)}_{{A}^{\prime}}$ with our wanted matrix, it will have the same result as if we map $(\frac{1}{2},0,0,0{)}_{A}$ into $[T{]}_{{B}^{\prime},A}$ (because they both map into the vector space generated by ${B}^{\prime}$) Thus, the first column of $[T{]}_{{B}^{\prime},{A}^{\prime}}$. You can get the other columns with the same method. The solution is:

$[T{]}_{{B}^{\prime},{A}^{\prime}}==\left[\begin{array}{cccc}5& 9\frac{4}{3}& 12\frac{2}{5}& -8\frac{7}{6}\\ -\frac{3}{2}& -4\frac{4}{3}& -3\frac{2}{5}& 3\frac{7}{6}\\ \frac{5}{2}& -1\frac{4}{3}& -7\frac{2}{5}& 5\frac{7}{6}\end{array}\right]$

Your first step is the right way to go, but I would be careful with the notation. Let $\alpha $ be the standard basis auf ${\mathbb{R}}^{3}$, then

$T(\left[\begin{array}{cc}2& 0\\ 0& 0\end{array}\right])=(1,0,1{)}_{B}=(2,5,10{)}_{\alpha}=(10,-3,-5{)}_{{B}^{\prime}}$

$T(\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right])=(2,1,1{)}_{B}=(4,8,9{)}_{\alpha}=(9,-4,-1{)}_{{B}^{\prime}}$

$T(\left[\begin{array}{cc}0& 0\\ 5& 0\end{array}\right])=(0,1,-1{)}_{B}=(2,5,12{)}_{\alpha}=(12,-3,-7{)}_{{B}^{\prime}}$

$T(\left[\begin{array}{cc}0& 0\\ 0& 6\end{array}\right])=(1,0,-1{)}_{B}=(0,-3,-8{)}_{\alpha}=(-8,3,5{)}_{{B}^{\prime}}$

Thus, we have

$[T{]}_{{B}^{\prime},A}=\left[\begin{array}{cccc}10& 9& 12& -8\\ -3& -4& -3& 3\\ -5& -1& -7& 5\end{array}\right]$

Now we are almost done. The new basis ${A}^{\prime}$ is good to handle because every new basis vector is a scalar multiple of a vector of $A$. For example the first basis vector of ${A}^{\prime}$ is half of its counterpart of $A$ (So $(2,0,0,0{)}_{{A}^{\prime}}=(1,0,0,0{)}_{A}$). So if you map $(1,0,0,0{)}_{{A}^{\prime}}$ with our wanted matrix, it will have the same result as if we map $(\frac{1}{2},0,0,0{)}_{A}$ into $[T{]}_{{B}^{\prime},A}$ (because they both map into the vector space generated by ${B}^{\prime}$) Thus, the first column of $[T{]}_{{B}^{\prime},{A}^{\prime}}$. You can get the other columns with the same method. The solution is:

$[T{]}_{{B}^{\prime},{A}^{\prime}}==\left[\begin{array}{cccc}5& 9\frac{4}{3}& 12\frac{2}{5}& -8\frac{7}{6}\\ -\frac{3}{2}& -4\frac{4}{3}& -3\frac{2}{5}& 3\frac{7}{6}\\ \frac{5}{2}& -1\frac{4}{3}& -7\frac{2}{5}& 5\frac{7}{6}\end{array}\right]$

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Find the transformation matrix

Let $B=\{2x,3x+{x}^{2},-1\},{B}^{\prime}=\{1,1+x,1+x+{x}^{2}\}$

Need to find the transformation matrix from $B$ to ${B}^{\prime}$.

I know that:

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$(a{x}^{2}+bx+c{)}_{{B}^{\prime}}=(a-b,b-c,c)$

How to proceed using this info in order to find the transformation matrix?

Let $B=\{2x,3x+{x}^{2},-1\},{B}^{\prime}=\{1,1+x,1+x+{x}^{2}\}$

Need to find the transformation matrix from $B$ to ${B}^{\prime}$.

I know that:

$(a{x}^{2}+bx+c{)}_{B}=(\frac{b-3c}{2},c,-a)$

$(a{x}^{2}+bx+c{)}_{{B}^{\prime}}=(a-b,b-c,c)$

How to proceed using this info in order to find the transformation matrix?