zachutnat4o

2021-11-17

Let $T:{P}_{3}\left(R\right)\to {M}_{2×2}\left(R\right)$ be a linear transformation, then find the matrix of transformation with respect to standard basis. Also, find the condition number of the matrix of transformation with respect to || ||1.

Phisecome

Step 1
Let $T:{P}_{3}\left(R\right)\to {M}_{2×2}\left(R\right)$ be a linear transformation
To find:
(a) The matrix transformtion with respect to standard basis.
(b) The condition number of the matrix of transformation with
respect to $|||{|}_{1}$.
Step 2
(a) Consider a given transformation:

Now, define a transformation:
$T\left(a+bx+c{x}^{2}+d{x}^{3}\right)=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Tye standard basis of vector space ${P}_{3}\left(R\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{M}_{2time2}\left(R\right)$ are
$B=\left\{a+bx+c{x}^{2}+d{x}^{3}\right\}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Further, we have
$T\left(1\right)=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]=1+\left(0\right)x+\left(0\right){x}^{2}+\left(0\right){x}^{3}$
$T\left(x\right)=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left(0\right)+\left(1\right)x+\left(0\right){x}^{2}+\left(0\right){x}^{3}$
$T\left({x}^{2}\right)=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left(0\right)+\left(0\right)x+\left(1\right){x}^{2}+\left(0\right){x}^{3}$
$T\left({x}^{3}\right)=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left(0\right)+\left(0\right)x+\left(0\right){x}^{2}+\left(1\right){x}^{3}$
Now, the matrix transformation w.r.t to the standard basis B is
${T}_{B}^{{B}^{\prime }}\sim A=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

Donald Proulx

(b)
From a part (a):

Now, the condition number of the matrix A w.r.t norm ${||||{}_{\left\{1\right\}}is||A||}_{1}\cdot {||{A}^{-1}||}_{1}$.
${||A||}_{1}=\sqrt{1+1+1+1}=2\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{||{A}^{-1}||}_{1}=sgrt\left\{1+1+1+1\right\}=2$.
Further solving, we have
Condition number $={||{A}^{-1}||}_{1}\cdot {||{A}^{-1}||}_{1}=\left(2\right)\cdot \left(2\right)=4$.
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