0
$\begingroup$

Let $S = \left\{A\in M_2(\mathbb{R}):A\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}1&1\\0&1\end{bmatrix}A\right\}$ be a linear space over $\mathbb{R}$. Find a basis for $S$.

Here's what I did:

Let $ A=\begin{bmatrix}a_1&a_2\\a_3&a_4\end{bmatrix}\in S$ , $a_1,a_2,a_4,a_4 \in \mathbb{R}$ . I want $A\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}1&1\\0&1\end{bmatrix}A$, $ A = \begin{bmatrix}a_1&a_2\\0&a_1\end{bmatrix}$. That means that $S=\left\{\begin{bmatrix}a_1&a_2=ka_1\\0&a_1\end{bmatrix} :a_1,k\in\mathbb{R}\right\}$

So, a basis for $S$ is : $\left\langle\begin{bmatrix}1&k\\0&1\end{bmatrix}\right\rangle$? What am i missing?

$\endgroup$
  • $\begingroup$ A parameter for the leading diagonal. $\endgroup$ – Donald Splutterwit Oct 1 '17 at 9:21
2
$\begingroup$

Let $A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Using the fact that $A\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$A, we get the following: \begin{align} a&=d\\ c&=0. \end{align}

And so, the set $S$ contains matrices of the form $\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}$.

A basis for this set is $\left\{\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\right\}$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

\begin{eqnarray*} \begin{bmatrix} a_1 & a_2 \\ a_3 & a_4 \\ \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} a_1 & a_1+a_2 \\ a_3 & a_3+a_4 \\ \end{bmatrix} \end{eqnarray*} \begin{eqnarray*} \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} a_1 & a_2 \\ a_3 & a_4 \\ \end{bmatrix} =\begin{bmatrix} a_1+a_3 & a_2+a_4 \\ a_3 & a_4 \\ \end{bmatrix} \end{eqnarray*} Consider the leading diagonal $a_3=0$ , now the $(1,2)$ entry will be satisfied. The $(2,1)$ gives $a_1=a_4=j$. So \begin{eqnarray*} \begin{bmatrix} j & k \\ 0 & j \\ \end{bmatrix} \end{eqnarray*} is a parameterisation of the space.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.