How to find matrix of linear map sample of vectors and their projection?

Question

I would like to find matrix of linear transformation $f$, which projects vectors like this:

$$f((x, y, z)^T)= (a, b, c, d, e)^T$$

How Can I find this matrix ? I am confused that vectors have different amount of elements.

Can you provide please minimal explanation / proof why your solution works ?

Thank you for help

Answer 1

As @Lukas suggested the matrix can be found using the bases, or direct computation. Let's assume that the matrix that we are looking for is M then, M is 5x3 (since it transforms a 3x1 vector to 5x1) and let assume that

$$ M=\begin{bmatrix} c_1 & c_2 & c_3 \\ \end{bmatrix} $$

where $c_i$ is the columns of M and each of them is 5x1

$$ M \begin{pmatrix} 1 \\ 0 \\0 \\ \end{pmatrix} = \begin{bmatrix} c_1 & c_2 & c_3 \\ \end{bmatrix} \begin{pmatrix} 1 \\ 0 \\0 \\ \end{pmatrix} = c_1 = \begin{pmatrix} 3 \\ 5 \\7 \\ 9 \\ 6\end{pmatrix} $$

$$ M \begin{pmatrix} 1 \\ 2 \\0 \\ \end{pmatrix} = \begin{bmatrix} c_1 & c_2 & c_3 \\ \end{bmatrix} \begin{pmatrix} 1 \\ 2 \\0 \\ \end{pmatrix} = c_1 + 2 c_2 =\begin{pmatrix} 2 \\ 3 \\4 \\ 5 \\ 1\end{pmatrix} $$

Hence,

$$ c_2 = 1/2\left( \begin{pmatrix} 2 \\ 3 \\4 \\ 5 \\ 1\end{pmatrix} - \begin{pmatrix} 3 \\ 5 \\7 \\ 9 \\ 6\end{pmatrix}\right) = \begin{pmatrix} -.5 \\ -1 \\-1.5 \\ -2 \\ -2.5\end{pmatrix} $$

$$ M \begin{pmatrix} 1 \\ 2 \\3 \\ \end{pmatrix} = \begin{bmatrix} c_1 & c_2 & c_3 \\ \end{bmatrix} \begin{pmatrix} 1 \\ 2 \\3 \\ \end{pmatrix} =c_1 + 2c_2 + 3c_3 = \begin{pmatrix} 1\\2 \\ 3 \\4 \\ 5 \end{pmatrix} $$

Hence,

$$ c_3 = 1/3\left( \begin{pmatrix} 1 \\ 2 \\ 3 \\4 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\4 \\ 5 \\ 1\end{pmatrix}\right) = \begin{pmatrix} -1/3 \\ -1/3 \\-1/3 \\ -1/3 \\ 4/3\end{pmatrix} $$

Answer 2

the fact that the vectors have different amounts of elements is not a problem. The linear map simply maps from $K^3$ to $K^5$ where $K$ is any field (probably $K= \mathbb R$ or $K = \mathbb Q$ is meant in this case). Now we know that the columns of the matrix of the transformation consist of the values of $f(e_1), f(e_2), f(e_3)$ where $e_i$ is the $i$-th canonical basis vector (or in general: Let $f: V \rightarrow W$ be a linear transformation of finite dimensional vector spaces. The $i$-th column of the matrix representing the transformation consists of the coordinates of $f(b_i)$ in the chosen basis of $W$ where $b_i$ is the $i$-th basis vector of the chosen basis of $V$).

In this case we use the canonical bases of $K^3$ and $K^5$ and we need to know $f(\begin{pmatrix} 1 \\ 0 \\0 \\ \end{pmatrix})$, $f(\begin{pmatrix} 0 \\ 1 \\0 \\ \end{pmatrix})$ and $f(\begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix})$ which we can figure out because $f$ is linear and we know the given three values. We just write the three values as columns of the matrix and we are done.

Hope this helps you.