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$f: \mathbb{R}^5 \rightarrow \mathcal{M}_{2}(\mathbb{R}) $

$ (x_1,x_2,x_3,x_4,x_5) \rightarrow \begin{pmatrix} x_1 & x_2 \\ x_1 + x_3 & x_5 \end{pmatrix}$

  1. Determine the basis of $Kerf$
  2. Determine a basis of $Imf$
  3. Determine the matrix $M_1 = M(f', \mathcal{B}, \mathcal{B '})$ of $f$

1 . $x = (x_1,x_2,x_3,x_4,x_5) \in Kerf \iff f(x_1,x_2,x_3,x_4,x_5) = \begin{pmatrix} x_1 & x_2 \\ x_1 + x_3 & x_5 \end{pmatrix} =\begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}$

I got $(x_1,x_2,x_3,x_4,x_5) = (0,0,0,x_4,0)$.

A base of $Kerf$ is $(0,0,0,1,0)$.

  1. $Imf^= Vect <f(e_1), f(e_2), f(e_3), f(e_4), f(e_5) > = <E_{11}, E_{12}, E_{21}, E_{22}>$

So the family $(E_{11}, E_{12}, E_{21}, E_{22})$ is basis.

with $E_{ij} $ is the elementary matrix.

The coordinate $x_4$ is not on the matrix.

Are my answers correct?

I need help with question 3, too. Thank you.

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    $\begingroup$ $f$ should be a map from $\mathbb{R}^5$. $\endgroup$ – tattwamasi amrutam Jul 11 '17 at 21:29
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    $\begingroup$ For starters: you have ignored $x_3$ in the kernel. So it should be $(0,0,x_3,x_4,0) \in K$. This mean kernel has dimension $2$. Likewise the image also needs fixing. Because the image can only have dimension $3$. $\endgroup$ – Anurag A Jul 11 '17 at 21:31
  • $\begingroup$ @AnuragA I just made a mistake typing. I have just corrected it. Thank you for noticing. $\endgroup$ – Zouhair El Yaagoubi Jul 11 '17 at 21:33
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    $\begingroup$ @Jacob. Okay that changes things in what I said above. Now with the edited version your answers are correct. $\endgroup$ – Anurag A Jul 11 '17 at 21:34
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Well $$f(e_1)=\begin{bmatrix} 1 &0 \\ 1 &0\\ \end{bmatrix},f(e_2)=\begin{bmatrix} 0 &1 \\ 0 &0\\ \end{bmatrix},f(e_3)=\begin{bmatrix} 0 &0 \\ 1 &0\\ \end{bmatrix},f(e_5)=\begin{bmatrix} 0 &0 \\ 0 &1\\ \end{bmatrix}$$

The matrix w.r.t the standard bases should look like $$\begin{bmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{bmatrix}$$

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  • $\begingroup$ You mean the matrix with respect to the basis $(f(e_1) , f(e_2), f(e_3), f(e_5))$ which is the basis of $Imf$, which is$ M (f,\mathcal{B}, \mathcal{B'}) $. Right? $\endgroup$ – Zouhair El Yaagoubi Jul 11 '17 at 21:47
  • $\begingroup$ No. I thought that $\mathcal{B'}$ is the standard basis for $M_{2\times2}$ $\endgroup$ – tattwamasi amrutam Jul 11 '17 at 21:49
  • $\begingroup$ Ah, that is what I was asked. Thank you. $\endgroup$ – Zouhair El Yaagoubi Jul 11 '17 at 21:50

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