# Set of real polynomials, organized as a vectorspace over $\mathbb{R}$.

I'm currently studying for a Linear Algebra exam in january, thus I'm going through some older exam questions, and I'm at the following question.

Let $$\mathbb{R}[x]$$ denote the set of all real polynomials, which is organized as a vectorspace over $$\mathbb{R}$$.

• A. Assume that $$U=\{p\in\mathbb{R}_{1}[x]|p(2)=0\}$$. Determine a basis for $$U$$ and its dimension.
• B. Add $$p(x)=x-2$$ to a basis $$B$$ for $$\mathbb{R}_3[x]$$, and determine the coordinate vector $$[q]_B$$ for $$q(x)=3+2x+x^2$$
• C. Determine a subspace $$W\subset \mathbb{R}_3[x]$$ such that $$U\oplus W=\mathbb{R}_3[x]$$. Determine $$\dim W$$.
• D. Find the matrix $$M(T)$$ with respect to $$B$$ when $$T=4D+3I$$, where $$D$$ denote differentiation and $$I$$ denotes the identical map. Determine whether $$T$$ is injective, surjective or bijective.
• E. Let $$L$$ denote the set of all real numbers $$\lambda$$ such that $$S=4D-\lambda I$$ has $$span(1,x^2,x^3)$$ as a subspace which is invariant under $$S$$. Determine $$L$$.

A. $$q(x)=x-2$$ is a basis for $$U$$ and its dimension is 1.

B. I add $$p(x)=x-2$$ to the standard basis of $$\mathbb{R}_3[x]$$, such that $$B=(x-2,1,x,x^2,x^3)$$, this is not a basis of $$\mathbb{R}_3[x]$$ as $$x$$ can be written as a linear comb. of the previous elements (vectors?), thus I reduce it to a basis by removing $$x$$ from the list so $$B=(x-2,1,x^2,x^3)$$. Now I write the matrix with respect to this basis and add $$\begin{bmatrix}3\\2\\1\\0\end{bmatrix}$$ to the matrix, and then I compute the coordinate vector $$[q]_B$$ by gauss elimination.

\begin{align} \begin{bmatrix} -2 & 1 & 0 & 0 & 3\\ 1 & 0 & 0 & 0 & 2\\ 0 & 0 & 1 & 0& 1\\ 0 & 0 & 0 & 1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 0 & 2\\ 0 & 1 & 0 & 0 & 7\\ 0 & 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 1 & 0 \end{bmatrix} \end{align}

Thus $$[q]_B = \begin{bmatrix}2 \\7\\1\\0\end{bmatrix}$$

C. From B. $$\mathbb{R}_3[x]$$ had the basis $$(x-2,1,x^2,x^3)$$, and the basis of $$U=(x-2)$$, so $$W=(1,x^2,x^3)$$, thus $$U\oplus W$$ will span $$\mathbb{R}_3[x]$$. $$\dim W=3$$

I'm stuck at D. and E. I assume that I start by differentiating the basis of $$B$$, then representing $$B'$$ as a matrix which I substitute into $$D$$ and then add the $$3I$$? I'm not sure that I'm capable of solving E. without knowing the methods from D.

I hope that some of it is right, and I hope to get some tips, correction and some help with the last two questions.

Best regards Jens.

Up till now all correct :) For exercise D) you'll need The differentiation Matrix $$D= \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ $$T:=4D+3I=\begin{bmatrix} 3 & 4 & 0 & 0 \\ 0 & 3 & 8 & 0 \\ 0 & 0& 3 & 12 \\ 0 & 0 & 0 & 3 \end{bmatrix}$$ T is bijective because $$\det(T)=3^4\neq 0$$ If you have not learned about determinants jet you can simply use the Gauss-algorithm to Calculate $$A^{-1}$$. Then have to write T with as a Matrix with Respect to the Basis B. For this you will need the Transformation Matricies $$[id]^B_S$$ and $$[id]^S_B$$ (where S denotes the standart-Basis) then $$[T]^B_B=[id]^B_S T [id]^S_B$$
hint: $$[id]^B_S$$ the transformation matrix from S to B is: $$[id]^B_S=\begin{bmatrix} -2 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1& 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ to understand this look at $$[id]^B_S e_j$$.
Also note $$[id]^B_S=([id]^S_B)^{-1}$$
• I think I'm doing something wrong when trying to calculate the $[id]^S_B$ matrix. I get a matrix with a zero in the diagonal such that it isn't invertible. – Jens Kramer Jan 2 '19 at 15:54
• I have misunderstood some things, but correct me if I'm wrong. I just calculated the matrix in the hint, but i thought that it was $[id]^S_B$ not $[id]^B_S$. When you write $e_j$, do you refer to the j'th orthonormal basisvector. I'm not really sure what that should clarify. Last I calculate the inverse of $[id]^B_S$ to be $\begin{bmatrix}0 & 1 & 0 & 0\\ 1 &2 &0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}$. My previous comment said that I didn't think this was invertible because one of the diagonals had a zero, which seems to be wrong. Is it right to say it isn't diagonizable then? – Jens Kramer Jan 2 '19 at 16:58
• the matrix seems right, it is the inverse of the matrix I've written (i'm not quite sure about my notation i might have done it the wrong way round) , $e_j$ is the j-basisvektor $e_1=(1,0,0,0),e_2=(0,1,0,0),e_3=(0,0,1,0),e_4=(0,0,0,1)$ – A. P Jan 2 '19 at 17:07
• You might be right about it, its just I don't use the same notation, it took a minute to figure the id meant the identity :p So multiplying $[id]^B_Se_j$ for each $e_j$ would yield a basis for $B$, right? After finding $[id]^B_S$ and its inverse I determine $[T]^B_B=\begin{bmatrix}3&4&0&0\\0&3&0&0\\8&16&3&0\\0&0&12&3\end{bmatrix}$ To summarize to find the matrix of $T$, $M(T)=[T]^B_B$, I first compute $T$ from the assumptions, then find an invertible matrix $[id]^B_S$ wrt the basis B, finding its inverse, $[id]^S_B$ then I compute $M(T)=[id]^B_ST_S[id]^S_B$? – Jens Kramer Jan 2 '19 at 17:24