I have this linear transformation: $f(x,y,z) = (x+y+z, 2x+2y+2z, 3x+3y+3z)$ From $\mathbb{R}^3 \rightarrow \mathbb{R}^3$
I need to find the associated matrix with respect to the standard basis.
How can I solve this?
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3 Answers
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You are moving from $\mathbb{R}^3 \rightarrow \mathbb{R}^3$ and you know that a linear transformation is fully determined by the values it assumes on the vectors of a basis. So let's take the canonical basis, $B=\{e_1,e_2,e_3\}$ and let's see how the map works on them.
$f(1,0,0)=(1,2,3), \qquad f(0,1,0)=(1,2,3), \qquad f(0,0,1)=(1,2,3)$
So the linear transformation has this representation as a matrix:
$$\begin{pmatrix} 1 & 1 & 1 \\ 2&2&2\\3&3&3\end{pmatrix}$$
You can also observe that $\text{dim}\ker(f)=2$ because the rank of the matrix is $1$. In fact, $\ker(f)=Span\left( \begin{pmatrix}-1\\1\\0\end{pmatrix},\begin{pmatrix}-1\\0\\1\end{pmatrix}\right)$
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By definition, the $i$-th column of the matrix is the image of the $i$-th basis vector.
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This is addressed in a previous question that I gave an answr to: Check that operator is linear transformation and find matrix in the same basis
To go from the linear transformation $T$ to the matrix $A$ with respect to the standard basis, simply find $T(e_1),T(e_2),$ and $T(e_3)$. Those make up the columns of the matrix.
