# Matrix $M_1 = M(f', \mathcal{B}, \mathcal{B '})$ of $f$

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

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

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.

• $f$ should be a map from $\mathbb{R}^5$. – tattwamasi amrutam Jul 11 '17 at 21:29
• 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 $3$. – Anurag A Jul 11 '17 at 21:31
• @AnuragA I just made a mistake typing. I have just corrected it. Thank you for noticing. – Zouhair El Yaagoubi Jul 11 '17 at 21:33
• @Jacob. Okay that changes things in what I said above. Now with the edited version your answers are correct. – Anurag A Jul 11 '17 at 21:34

## 1 Answer

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

• 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? – Zouhair El Yaagoubi Jul 11 '17 at 21:47
• No. I thought that $\mathcal{B'}$ is the standard basis for $M_{2\times2}$ – tattwamasi amrutam Jul 11 '17 at 21:49
• Ah, that is what I was asked. Thank you. – Zouhair El Yaagoubi Jul 11 '17 at 21:50