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I would appreciate some help with a linear algebra practice question, I'm studying for my final and I am stuck, this is a screenshot of the question:

Screenshot of question

Are my answers correct?

a)

$P_{2}$:

$ \begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}$

$P_{1}$:

$ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}$

b)

Rank: 2 (as I have two pivots)

Nullity: 0 (as dim - rank= 2 - 2= 0)

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  • $\begingroup$ No, the kernel of your map contains $1$. In general if $A:V\to W$ we have $\dim\ker A+\dim{\rm im} A=\dim V$. In your case $\dim V=3$, and $\dim{\rm im}\; A=2$. $\endgroup$ – Pedro Tamaroff Dec 12 '13 at 21:37
  • $\begingroup$ Meaning your nullity cannot be 0. $\endgroup$ – Vladhagen Dec 12 '13 at 21:38
  • $\begingroup$ You have to find one matrix $T$ so why do you write two matrices? The rank nullity theorem: $\dim P_2=3=\dim\ker T+\mathrm{rank}(T)=1+2$ $\endgroup$ – user63181 Dec 12 '13 at 21:40
  • $\begingroup$ a) The answer must be a $2\times 3$ matrix: you are going from $3$ dimensions to $2$ dimensions. $\endgroup$ – Julien Dec 12 '13 at 21:40
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Let $T_A$ be the matrix: $\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$

This matrix will remove the zero degree coefficient from any degree 2 or smaller polynomial (written as a vector), and will permute the other two coefficients.

As Sami Ben Romdhane indicates, your rank is 2 and your nullity is 1.

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  • $\begingroup$ How did you know that the bottom row will no longer be there? $\endgroup$ – Abushawish Dec 12 '13 at 22:15
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    $\begingroup$ We are multiplying a vector with three entries by the matrix. We need to obtain a vector with two entries. The dimension of your codomain space is 2. So we cannot have three rows. We only want two rows so that our dimensions are correct. $\endgroup$ – Vladhagen Dec 12 '13 at 23:06
  • $\begingroup$ Perfect, thanks! $\endgroup$ – Abushawish Dec 14 '13 at 2:03

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