2
$\begingroup$

Let the subspace $S$ from $M_{2\times2}$: $$S=\left\{ \begin{bmatrix} a & b\\ c & d\\ \end{bmatrix} \mid c = a + b\textrm{ and }d=a \right\}.$$ What is the dimension of $S$?

According to the answer key, $\dim (S)=2$. Why is it not $\dim(S)=2\times2=4$? What am I missing?

$\endgroup$
2
  • $\begingroup$ The dimension of $M_{2\times 2}$ is equal to $2\times 2=4$, but $S$ is a smaller subspace of $M_{2\times 2}$, so the dimension is less. $\endgroup$ Aug 10, 2018 at 0:01
  • $\begingroup$ Since $c, d$ can both be described in terms of $a,b$ two parameters define all of the matrices in S. S is a subspace of $M_{2\times 2}, M_{2\times 2}$ is indeed a 4 dimensional vector space. $\endgroup$
    – Doug M
    Aug 10, 2018 at 0:01

3 Answers 3

4
$\begingroup$

Once you have determined $a$ and $b$, you already know what $c$ and $d$ should be. It's imposed on you. Hence, the dimension of the subspace, which can be thought of as the number of independent variables in this case, is $2$. You can also prove this rigorously by showing that

$$B=\left\{\begin{bmatrix}1 & 0\\1 & 1\\\end{bmatrix},\begin{bmatrix}0 & 1\\ 1 & 0\\\end{bmatrix}\right\}$$

is a basis for this subspace, i.e. $S$ is spanned by those two matrices and they are linearly independent.

$\endgroup$
4
  • 1
    $\begingroup$ Just to clarify: if the number of independent variables were $3$, then $dim(S)=3$? $\endgroup$ Aug 10, 2018 at 0:08
  • 1
    $\begingroup$ @MauricioMendes Yes. You can say so. For example, if $d$ didn't need to be equal to $a$ and it could be arbitrary, the dimension of $S$ would've been $3$ then. $\endgroup$ Aug 10, 2018 at 0:09
  • 1
    $\begingroup$ @CameronBuie Yeah. I changed this like 2 seconds before you write this comment. xD My notation was bad because I was using $S$ both for the set itself and its span and I decided to change my notation completely. $\endgroup$ Aug 10, 2018 at 0:16
  • 1
    $\begingroup$ +1: Glad you figured out the formatting. :-) $\endgroup$ Aug 10, 2018 at 0:18
1
$\begingroup$

The subspace $S$ consist of matrices of the form $ \begin{pmatrix} a & b \\ a+b& a \\ \end{pmatrix} $. Every such matrix can be written as a linear combination $$ \begin{pmatrix} a & b \\ a+b& a \\ \end{pmatrix}= a\begin{pmatrix} 1 & 0 \\ 1 & 1 \\ \end{pmatrix}+ b\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}. $$ If you also check that the matrices $\begin{pmatrix} 1 & 0 \\ 1 & 1 \\ \end{pmatrix}$, $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$ are linearly independent, you get that this is a basis for $S$.

$\endgroup$
0
$\begingroup$

$S$ is $2$ dimensional. $M_{2×2}\cong\mathbb F^4$, where $\mathbb F$ is the field you are working over.

A basis for $S$ is $\{\begin {pmatrix} 1&0\\1&1\end{pmatrix},\begin{pmatrix}1&1\\2&1\end{pmatrix}\}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .