Matrix Representation: Change of Basis

Let \(\alpha: \mathcal{P}_2(\mathbb{R}) \longrightarrow M_{2\times 2}(\mathbb{R})\) be defined by

\(\alpha(f(x))=\left(\begin{array}{cc}f^{'}(0)& 2f(1)\\0& f^{''}(3)\end{array}\right)\)

First, let’s show that \(\alpha\) is a linear transformation. Let \(f(x),g(x) \in \mathcal{P}_2(\mathbb{R})\) and \(a,b\in \mathbb{R}\). Then by definition, we have

\(\alpha(af(x)+bg(x))=\left(\begin{array}{cc}af'(0)+bg'(0)& 2af(1)+2bg(1)\\0& af''(3)+bg''(3)\end{array}\right)\)

\(\hspace{1.6in}\)=\(\left(\begin{array}{cc}af'(0)& 2af(1)\\0& af''(3)\end{array}\right)\)+\(\left(\begin{array}{cc}bg'(0)& 2bg(1)\\0& bg''(3)\end{array}\right)\)

\(\hspace{1.6in}\)=\(a\left(\begin{array}{cc}f'(0)& 2f(1)\\0& f''(3)\end{array}\right)\)+\(b\left(\begin{array}{cc}g'(0)& 2g(1)\\0& g''(3)\end{array}\right)\)

\(\hspace{1.6in}=a\alpha(f(x))+b\alpha(g(x))\)

So that \(\alpha\) is a linear transformation.

Second, we find the kernel space \(ker(\alpha)\), then use the Dimension Theorem (formula) to decide the rank of \(\alpha\)

The kernel of \(\alpha\) is defined as

\(ker(\alpha)=\{v\in V|\alpha(v)=0_{M_{2\times2}(\mathbb{R})}\}\)

\(\alpha(f(x))=\left(\begin{array}{cc}f^{'}(0)& 2f(1)\\0& f^{''}(3)\end{array}\right)=[0]\)

\(\implies f'(0)=0, 2f(1)=0, f''(3)=0\)

If \(f(x)=a+bx+cx^2\) then we have,

\(\begin{array}{c}f'(0)\implies b=0\\2f(1)=0\implies 2(a+b+c)=0\\f''(3)=0\implies 2c=0\end{array}\)

\(\implies a=b=c=0 \implies ker(\alpha)=\{0_{\mathcal{P}_2(\mathbb{R})}\}\)

Then \(nullity(\alpha)=\dim ker(\alpha)=0\) and if we use the dimension formula then, \(rank(\alpha)=\dim \mathcal{P}_2(\mathbb{R})-nullity(\alpha)=3-0=3\)

Third, we will find the representation matrix \(\phi_{BD}(\alpha)\), where \(B=\{1+x,1-x,x^2\}\) is an ordered basis for \(\mathcal{P}_2(\mathbb{R})\)

and

\(D=\begin{Bmatrix}\begin{bmatrix}1 & 0\\0 &0\end{bmatrix},\begin{bmatrix}0 & 1\\0 &0\end{bmatrix},\begin{bmatrix}0 & 0\\1 &0\end{bmatrix},\begin{bmatrix}0 & 0\\0 &1\end{bmatrix}\end{Bmatrix}\)

is an ordered basis for \(\mathbf{M}_{2\times 2}(\mathbb{R})\)

Now if \(f(x)=1+x\) then

\(\alpha(f(x))=\left(\begin{array}{cc}f^{'}(0)& 2f(1)\\0& f^{''}(3)\end{array}\right)\)

\(\hspace{1in}=\left(\begin{array}{cc}1& 4\\0& 0\end{array}\right)\)

\(\hspace{1in}=\left(\begin{array}{c}1\\4\\0\\0\end{array}\right)\)

Now if \(f(x)=1-x\) then

\(\alpha(f(x))=\left(\begin{array}{cc}f^{'}(0)& 2f(1)\\0& f^{''}(3)\end{array}\right)\)

\(\hspace{1in}=\left(\begin{array}{cc}-1& 0\\0& 0\end{array}\right)\)

\(\hspace{1in}=\left(\begin{array}{c}-1\\0\\0\\0\end{array}\right)\)

Now if \(f(x)=x^2\) then

\(\alpha(f(x))=\left(\begin{array}{cc}f^{'}(0)& 2f(1)\\0& f^{''}(3)\end{array}\right)\)

\(\hspace{1in}=\left(\begin{array}{cc}0& 2\\0& 2\end{array}\right)\)

\(\hspace{1in}=\left(\begin{array}{c}-2\\2\\0\\2\end{array}\right)\)

because,

\(\left(\begin{array}{cc}0& 2\\0& 2\end{array}\right)\)\(=-2\left(\begin{array}{cc}1& 0\\0& 0\end{array}\right)\)\(+2\left(\begin{array}{cc}0& 1\\0& 0\end{array}\right)\)\(+0\left(\begin{array}{cc}0& 0\\1& 0\end{array}\right)\)\(+2\left(\begin{array}{cc}1& 0\\0& 1\end{array}\right)\)

Therefore, \(\phi_{BD}(\alpha)=\)\(\left(\begin{array}{ccc}1& -1& -2\\4& 0& 2\\0& 0& 0\\0& 0& 2\end{array}\right)\)

Share on

Share

Tweet

You may also like