For what $k$ is $g_k\circ f_k$ invertible?

Question

Let $\ f_k:\Bbb R[x]_{\le1}\to\Bbb R^3, g_k:\Bbb R^3\to\Bbb R[x]_{\le1}$ be linear maps such that $f_k(x-1)=\left( {\begin{array}{*{20}{c}} {{0}} \\ k \\ {1} \end{array}} \right),\ f_k(2-x)=\left( {\begin{array}{*{20}{c}} {{k}} \\ 0 \\ {-1} \end{array}} \right),\ g_k\left( {\begin{array}{*{20}{c}} {{a}} \\ b \\ {c} \end{array}} \right)=(-2a+2kb)x+b-c.$

I'm asked to find the values of $k$ for which $g_k\circ f_k$ is invertible. Now, I think I ought to find its representative matrix $M$ and study its determinant - it shall be a 3x3 matrix. In order to do this, I should find the ones for $g_k$ and $f_k$, say $G$ and $F$ respectively, and then calculate $M=G\cdot F$. Am I right so far?

It seemed to me that $x-1$ and $2-x$ form a basis of $\Bbb R[x]_{\le1}$, so I thought $F=\begin{bmatrix} 0& k \\ k&0 \\ 1&-1 \end{bmatrix}.$

Instead I'm not sure about $g_k$....

Answer 1

HINT: observe that $\Bbb R[x]_{\le 1}$ and $\Bbb R^2$ are isomorphic as vector spaces, and because the functions $f_k$ are linear we have that

$$f_k(x-1)+f_k(2-x)=f_k(1)=(k,k,0)\\ 2f_k(x-1)+f_k(2-x)=f_k(x)=(k,2k,1)$$

Thus, the matrix that represent the action of $f_k$ over some polynomial $sx+r\in\Bbb R[x]_{\le 1}$ is

$$M(f_k)=\begin{bmatrix}k&k\\k&2k\\0&1\end{bmatrix}$$

where I had used the previous isomorphism, that is $\Bbb R^2\ni(r,s)\equiv sx+r\in\Bbb R[x]_{\le 1}$ (and the standard basis of $\Bbb R^2$ and $\Bbb R^3$). And for $g_k$ you have, using also the standard basis of $\Bbb R^3$ and $\Bbb R^2$ that

$$M(g_k)=\begin{bmatrix}0&1&-1\\-2&2k&0\end{bmatrix}$$

That is $M(g_k)(a,b,c)^T=(b-c,-2a+2kb)\equiv (-2a+2kb)x+b-c$.

And you want to find $k$ such that $\det[M(g_k)M(f_k)]\neq 0$. You only need to multiply the two matrices and find it determinant to see what values of $k$ make it different from zero.

Answer 2

If I did my calculations correctly, the representative matrix with respect to the canonical basis of $\mathbb{R}^3$ and the basis $\{x-1,2-x\}$ for $\mathbb{R}[x]_{\leq1}$ should be

$$G=\begin{bmatrix} -4& 4k+1&-1 \\ -2&2k+1 &-1\end{bmatrix}$$ The product $GF$ has determinant $$\det(GF)=4k^2-2k$$ which is non zero for $k\neq 0,1/2$.