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I have the linear space $V = \{ P \in \mathbb{R}_4[x] : xP'''(x) + P''(x) = P'(-1) + P(0) = 0 \}$ and have to find its dimension. In terms of tools I'm only allowed to use there's the theorem that for any linear $F : A \to B$ where $A$ and $B$ are linear spaces, it's true that $\dim (\ker (F)) + \dim (\mathrm{Im}(F)) = \dim (A)$, so my idea is to define some linear $F$ such that $F: \mathbb{R}_4[x] \to \mathbb{R}_4[x]$, but I have no idea for $F(P)$. $V$ looks like the kernel of some $F$, so it should work, but I don't know how to continue. Could you give me any hints?

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You can consider two linear maps: \begin{align} F\colon \mathbb{R}_4[x] & \to \mathbb{R}_4[x], & F(P)&=xP'''(x)+P''(x) \\ G\colon \mathbb{R}_4[x] & \to \mathbb{R}, & G(P)&=P'(-1)+P(0) \end{align} Thus $V=\ker F\cap \ker G$.

The matrix of $F$ with respect to the basis $\{1,x,x^2,x^3,x^4\}$ is \begin{bmatrix} 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 12 & 0 \\ 0 & 0 & 0 & 0 & 36 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} and the matrix of $G$ is \begin{bmatrix} 1 & 1 & -2 & 3 & -4 \end{bmatrix} Can you finish?

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  • $\begingroup$ At the moment there wasn't much talk about bigger matrices than 3x3 on my lecture. However, I think (using theorem I wrote) that $\dim (\ker (F)) = 2$ and $\dim (\ker (G)) = 4$ and now I could use theorem for finite dimensions $\dim (A \cap B) = \dim (A) + \dim (B) - \dim (A+B)$, but what would be the dimension of $A+B$? $\endgroup$ – chandx Mar 23 at 13:15
  • $\begingroup$ @chandx It's easier to realize that the kernel of $F$ consists of the polynomials having degree at most one; then apply the second condition and you'll have the answer. $\endgroup$ – egreg Mar 23 at 13:57
  • $\begingroup$ Right, so I see that $\ker (F) = \{ dx + e : d,e \in \mathbb{R} \}$ and second condition tells me that $-4a+3b-2c+d+e = 0$ so in other words $d+e = 0$, thus $e = -d$ and $V = \{ ax - a : a \in \mathbb{R} \}$, so its base is just $\{ x-1 \}$, thus $\dim (V) = 1$, is it right? $\endgroup$ – chandx Mar 23 at 14:22
  • $\begingroup$ @chandx Yes, that's right. $\endgroup$ – egreg Mar 23 at 14:22
  • $\begingroup$ @greg Thank you for help! $\endgroup$ – chandx Mar 23 at 14:24

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