# Change of basis matrix from $\mathbb{R}^3$ to $\mathbb{R}^2$

Let $$f: \mathbb{R}^3 \rightarrow \mathbb{R}^2$$ be a linear application such that: $$\mathcal{M}(f; c.b.\mathbb{R}^3; c.b.\mathbb{R}^2) = \left[ \begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \end{array} \right]$$ and considering the following basis

$$c.b.\mathbb{R}^3 = \{ (1,0,0), (0,1,0), (0,0,1)\}$$

$$c.b.\mathbb{R}^2 = \{ (1,0), (0,1)\}$$

$$\mathcal{B} = \{ (0,1,0), (1,0,1), (1,0,0)\}$$

$$\mathcal{B}' = \{ (1,1), (1,0)\}$$

using the change of basis matrix find:

a) $$\mathcal{M}(f; \mathcal{B}; c.b.\mathbb{R}^2)$$

b) $$\mathcal{M}(f; c.b.\mathbb{R}^3; \mathcal{B}')$$

c) $$\mathcal{M}(f; \mathcal{B}; \mathcal{B}')$$

I've already solved item a) by compting each vector in $$\mathcal{B}$$ with respect to $$c.b.\mathbb{R}^3$$ and then multiplied this vector (on the rhs) by the given matrix. So the solution to a) is the following:

$$(0,1,0) = \alpha_1(1,0,0) + \alpha_2 (0,1,0) + \alpha_3 (0,0,1) \Leftrightarrow \alpha_1 = 0 \land \alpha_2 = 1 \land \alpha_3 = 0$$

$$(1,0,1) = \alpha_1(1,0,0) + \alpha_2 (0,1,0) + \alpha_3 (0,0,1) \Leftrightarrow \alpha_1 = 1 \land \alpha_2 = 0 \land \alpha_3 = 1$$

$$(1,0,0) = \alpha_1(1,0,0) + \alpha_2 (0,1,0) + \alpha_3 (0,0,1) \Leftrightarrow \alpha_1 = 1 \land \alpha_2 = 0 \land \alpha_3 = 0$$

$$\vec{e_1} = \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \end{array} \right] % \left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$$

$$\vec{e_2} = \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \end{array} \right] % \left[ \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$$

$$\vec{e_3} = \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \end{array} \right] % \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$$

$$\mathcal{M}(f; \mathcal{B}; c.b.\mathbb{R}^2) = \left[ \begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right]$$

and I know that the solution for b) is

$$\mathcal{M}(f; c.b.\mathbb{R}^3; \mathcal{B}') = \left[ \begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & -1 \end{array} \right]$$

and solution for c) is

$$\mathcal{M}(f; c.b.\mathbb{R}^3; c.b.\mathbb{R}^2) = \left[ \begin{array}{lll} 1 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$$ but I don't know how to do them. Any tips?

• Show your work for part a). If you understood what you did, the other parts should not be a problem as they work in exactly the same fashion. Commented Jan 18, 2021 at 14:27
• @johnny10 I've eddited the question showing the work, but I can't seem to figure out how to start for b) and c) Commented Jan 18, 2021 at 15:38

There is a systematic way to do this. For such a problem, what you need is simply following the definitions. (It is very instructive to do it at least once to see what is really going on.)

Change the notations to make writing easier.

Let $$\alpha=(e_1,e_2,e_3)$$ be the standard basis of $$\mathbb{R}^3$$ and $$\alpha'=(e_1',e_2')$$ the standard basis of $$\mathbb{R}^2$$.

Then the linear map $$f$$ is such that $$f(e_1)=e_1,\quad f(e_2)=e_1'+e_2',\quad f(e_3)=e_2'.$$ This corresponds to the matrix on the second line of your post.

To find the matrix in (a), what you want is finding the following coefficients $$b_{ij}$$ where $$f(\beta_1)=b_{11}e_1'+b_{12}e_2'\\ f(\beta_2)=b_{21}e_1'+b_{22}e_2'\\ f(\beta_3)=b_{31}e_1'+b_{32}e_2'$$

Here $$\beta_1,\beta_2,\beta_3$$ denote the three vectors of $$\mathcal{B}$$.

Hint:

Simple – denote $$P$$ the change of basis matrix for$$\mathbf R^3$$, $$X$$ a column vector in the standard basis, $$X'$$ its column vector in basis $$\mathcal B$$, and $$Q$$ the change of basis matrix for $$\mathbf R^2$$, $$Y$$ a column vector in the standard basis, $$Y'$$ its column vector in basis $$\mathcal B'$$.

We know that $$\;Y=QY'=\ M X= M PX'$$, whence $$Y'=(Q^{-1}MP)X'$$ which means th matrix you're seeking for is $$Q^{-1}MP$$.