1
$\begingroup$

I have this exercise and I want to know if my answer is correct. The exercise is:

Consider the linear space $\mathbb{R}^{2\times2}$ of $2\times2$ matrices with real entries. Consider $W$ contained in this space:

$$W = \{[a_{i,j}] \in \mathbb{R}^{2\times2}\mid a_{1,1} + a_{2,2} = 0\}$$

Calculate a basis of $\mathbb{R}^{2\times2}$ that contains a basis of $W$

So my doubt is in the question. Do they simply want a basis for $W$? Because that's easy

$\left\{\begin{pmatrix} 1 & 0 \\0 & -1\end{pmatrix}, \begin{pmatrix} 0 & 1 \\0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}\right\}$

But my doubt is if this is really what they want because they refer a basis of $\mathbb{R}^{2\times2}$ and for that we should have an extra fourth matrix, right?

Can someone clarify this for me?

$\endgroup$
  • $\begingroup$ No basis of $\mathbb{R}^{2\times2}$ can contain $W$, but it can contain a basis of $W$. $\endgroup$ – egreg Apr 24 '16 at 15:40
1
$\begingroup$

You seem to be missing some words:

Calculate a basis of $\mathbb{R}^{2\times2}$ that contains a basis of $W$

is the correct wording.

You have computed correctly a basis for $W$; now a general result is that

if $\{v_1,\dots,v_m\}$ is a linearly independent set in the vector space $V$ and $v\in V$ with $v\notin\operatorname{Span}\{v_1,\dots,v_m\}$, then the set $\{v_1,\dots,v_m,v\}$ is linearly independent.

So you can just find a matrix not in $W$, which is easy, and add it to the set you found. The resulting set has four elements and is linearly independent, so it is a basis for $\mathbb{R}^{2\times2}$ (because this space has dimension $4$).


More generally, the exchange lemma says that if you have a linearly independent set $\{v_1,\dots,v_m\}$ and a basis $\{w_1,\dots,w_n\}$ of a vector space, you can replace $m$ vectors in the basis with $v_1,\dots,v_m$ so that the resulting set is again a basis.

$\endgroup$
0
$\begingroup$

Use $\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$ to complete a set of 4 linearly independent matrices in $\Bbb R^{2\times2}$.

$\endgroup$

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.