Mathematics Asked on December 29, 2021
We have tha matrix begin{equation*}A:=begin{pmatrix}1 & 5 & 8 & 2 \ 2 & 4 & 6 & 0 \ 3 & 3 & 8 & 2 \ 4 & 2 & 6 & 0 \ 5 & 1 & 8 & 2end{pmatrix}in mathbb{R}^{5times 4}end{equation*}
Determine a basis $B$ of $mathbb{R}^4$ and a basis $C$ of $mathbb{R}^5$ such that $M_C^B(A)$ contains at the left upper corner the unit matrix and everywhere else zeroes.
Could you give me a hint for that?
So do we have to write each column of $A$ as a linear combination of $C$ and the corresponding coefficients have to satisfy the desired form that $M$ will have?
So do we have to write each column of $A$ as a linear combination of $C$ and the corresponding coefficients have to satisfy the desired form that $M$ will have?
If I understand you correctly, then yes. More explicitly, if $B={b_1,dots,b_4}$ and $C={c_1,dots, c_5}$ are bases of of $mathbb R^4$ and $mathbb R^5$, respectively, and $M^B_C(A)$ looks like
$$M^B_C(A)=begin{pmatrix}alpha_1&beta_1&gamma_1&delta_1\ alpha_2&beta_2&gamma_2&delta_2\ alpha_3&beta_3&gamma_3&delta_3\ alpha_4&beta_4&gamma_4&delta_4\ alpha_5&beta_5&gamma_5&delta_5end{pmatrix},$$
then that means that $b_1$ is mapped to $alpha_1 c_1+dots+alpha_5c_5$, while $b_2$ is mapped to $beta_1c_1+dots+beta_5c_5$, and so on. Now you want it to look like
$$M_C^B(A)=begin{pmatrix}1&0&0&0\ 0&1&0&0\ 0&0&1&0\ 0&0&0&1\ 0&0&0&0end{pmatrix}.$$
So $b_1$ should be mapped to $c_1$, $b_2$ to $c_2$, and so on. The easiest way to find such bases would be to just take the standard base as $B$. Then look at the vector to which $b_i$ is mapped and choose that as $c_i$. Then choose some additional vector as $c_5$ which completes the base.
Answered by Vercassivelaunos on December 29, 2021
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