Matrices and linear transformations, find f based on it's matrix in a basis

Question

Let $f:\mathbb{R}^3\to\mathbb{R}^3$, knowing that it's matrix based on the basis $B=\{(1,1,0),(0,1,0),(0,1,1)\}$ is: $M(f)_B=\begin{pmatrix}1&1&2\\-1&2&1\\1&3&4\end{pmatrix}$ then find $f(x,y,z).$

First, I really want to understand this, in special, what the matrix of the function based on a basis actually is and what it does to the function? What's it's importance? also what it's properties? Why does it help us and can we rewrite the function based on that? And why is that matrix important to the question? (finding $f$)

My attempt:

We know that the coordinates of an arbitrary vector $v=(x,y,z)$ based on a basis are (let's name the vectors from $B$ in order $b_1,b_2,b_3$): $$v=\lambda_1b_1+\lambda_2b_2+\lambda_3b_3$$ Where $\lambda_{1,2,3}$ are the actual coordinates. To calculate the function matrix, I know that we have:

$$f(b_1) = \lambda_{1,1}b_1'+\lambda_{1,2}b_2'+\lambda_{1,3}b_3'$$ $$...$$

and so on, where $b_1',b_2',b_3'$ are some other vectors from some other basis from the domain, right? Now... How I know that the matrix I was given is based on the canonic basis if it's not mentioned? Since the matrix of a function must be corelated to 2 basis, right?

Answer 1

Take $(x,y,z)\in\mathbb{R}^3$ and compute the coordinates of $(x,y,z)$ in the basis $B$. That is, find $\alpha,\beta,\gamma\in\mathbb R$ such that $(x,y,z)=\alpha(1,1,0),\beta(0,1,0)+\gamma(0,1,1)$. Then you will know that\begin{multline}f(x,y,z)=\alpha\bigl((1,1,0)-(0,1,0)+(0,1,1)\bigr)+\\+\beta\bigl((1,1,0)+2(0,1,0)+3(0,1,1)\bigr)+\gamma\bigl(2(1,1,0)+(0,1,0)+4(0,1,1)\bigr),\end{multline}because asserting that $M$ is the matrix of $f$ with respect to the basis $B$ (twice) means that:

• $f(1,1,0)=(1,1,0)-(0,1,0)+(0,1,1)$;
• $f(0,1,0)=(1,1,0)+2(0,1,0)+3(0,1,1)$;
• $f(0,1,1)=2(1,1,0)+(0,1,0)+4(0,1,1)$.

Answer 2

I guess you are looking for the expression in terms of the cannonical basis $B_c=\{(1,0,0),(0,1,0),(0,0,1)\}$.

Calculate the matrix $M$ of the change of basis, or equivalently, the matrix of the linear transformation $Id:({\mathbb R}^3,B_c)\to ({\mathbb R}^3,B)$ (that is, the identity map but considering the canonical basis on the beginning and your besis in the end.

Now the matrix of $f$ with respect to the canonical basis will be $$M(f)_{B_c}=M\cdot M(f)_B\cdot M^{-1}$$ and you just have to compute $$f(x,y,z)=M(f)_{B_c}\cdot (x,y,z)^t.$$

Answer 3

It's just a discussion:

I think all of your firstly asked questions have answers here. And for if the basis is not mentioned conventionally we take the canonical basis. Here is an Example.