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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 $L$.

My answers are,

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.

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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 $[id]^B_S=([id]^S_B)^{-1}$

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  • $\begingroup$ 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. $\endgroup$ – Jens Kramer Jan 2 '19 at 15:54
  • $\begingroup$ 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? $\endgroup$ – Jens Kramer Jan 2 '19 at 16:58
  • $\begingroup$ 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 $e_1=(1,0,0,0),e_2=(0,1,0,0),e_3=(0,0,1,0),e_4=(0,0,0,1)$ $\endgroup$ – A. P Jan 2 '19 at 17:07
  • $\begingroup$ 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 $M(T)=[id]^B_ST_S[id]^S_B$? $\endgroup$ – Jens Kramer Jan 2 '19 at 17:24
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    $\begingroup$ 👍Wolfram alpha gives me the same answer, $\endgroup$ – A. P Jan 2 '19 at 18:11

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