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I have a basis B of $\mathbb{R^3}$ $\\{(1,2,3),(1,3,4),(1,3,5)\\}$ I have this matrix of a linear map F from basis B to canonical basis C of $\mathbb{R^3}$ $$ \begin{pmatrix} 6 & 9 & 12 \\ 2 & 3 & 5 \\ 0 & 0 & 0 \\ \end{pmatrix} $$ I calculated the same matrix from C to C $$ \begin{pmatrix} -3 & 0 & 3 \\ -2 & -1 & 2 \\ 0 & 0 & 0 \\ \end{pmatrix} $$ I have to calculate kernel and image of F. Calculating them from the second matrix I obtained: $\text{Ker(F)}= \{x(1,0,1)\} $ and $\text{Im(F)}= \{x(-3,-2,0)+y(0,-1,0)\} $

My question is: can I calculate them directly from the first matrix?

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Yes. The kernel are those vectors $v$ whose coordinates $x$; $y$, and $z$ with respect to the basis $B$ are such that$$\begin{bmatrix}6&9&12\\2&3&5\\0&0&0\end{bmatrix}.\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}.$$This is equivalent to$$\left\{\begin{array}{l}6x+9y+12z=0\\2x+3y+5z=0\\0=0.\end{array}\right.$$That is, the kernel is$$\left\{-\frac{3y}2\begin{bmatrix}1\\2\\3\end{bmatrix}+y\begin{bmatrix}1\\3\\4\end{bmatrix}\,\middle|\,y\in\mathbb R\right\}.$$

The image of $F$ is$$\left\langle\begin{bmatrix}6\\2\\0\end{bmatrix},\begin{bmatrix}9\\3\\0\end{bmatrix},\begin{bmatrix}12\\5\\0\end{bmatrix}\right\rangle.$$

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  • $\begingroup$ Did you notice that the second column is dipendent from the first? $\endgroup$ – rik99 Jul 17 at 22:24
  • $\begingroup$ No. I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Jul 17 at 22:28

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