1
$\begingroup$

I am working on problems from my textbook. However, I am lost as to how to show this.

A. Let $M_{2\times 2}$ be the vector space of all $2\times 2$ matrices. Show that the set of non-singular $2\times 2$ matrices is NOT a subspace of $M_{2\times 2}$.

B. Let $M_{2\times 2}$ be the vector space of all $2\times 2$ matrices. Show that the set of singular $2\times 2$ matrices is NOT a subspace of$M_{2\times 2}$.

C. Describe the smallest subspace of $M_{2\times 2}$ that contains matrices $$\begin{pmatrix}2 & 1 \\ 0 & 0\end{pmatrix},\ \ \begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix}\ \ \text{and}\ \ \ \begin{pmatrix}0 & -1 \\ 0 & 0\end{pmatrix}$$

Find the dimension of the subspace.

I think I can prove that addition for A and B is not closed, thus disproving the potential for subspace. Though, I am not sure about C.

$\endgroup$
5
$\begingroup$

For a, is the zero matrix in the set?

For b, show that addition is not closed (can you think of two matrices which are non-invertible but add to the identity?)

For c, notice that any subspace containing the three matrices necessarily contains all linear combinations of the three matrices. Conversely, what can we say about the span of the three matrices?

$\endgroup$
0
$\begingroup$

For item A, I'd like you to consider the following matrices: $A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$ and $B = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}$. Both $A,B \in M_{2\times2}$. These two matrices are both non-singular since $|A|=|B|= 1\ne 0$. but their sum, $A + B = 0_{2\times2}$ is singular. Thus, because of violation of closure, the set of non-singular $2\times2$ matrices is not a subspace of $M_{2\times2}$.

For item B, now consider, $C = \begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$ and $D = \begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}$. For both matrices, we have a row of zeroes, and consequently, since $|C|=|D| = 0$, both matrices are singular. But the sum, $C + D = I_{2\times2}$ which is non-singular. Again, in violation of closure, the set of singular $2\times2$ matrices is not a subspace of $M_{2\times2}$.

Now for item C, notice that they are all of the form, $E = \begin{bmatrix}a & b\\0 & d\end{bmatrix}$. Thus, we can show that $E = a\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix} + b\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix} + d\begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$. You can show that $E_1 = \begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}$, $E_2 = \begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$, $E_3 = \begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$ forms a linearly independent set and will span $E$, and the set $\{E_1, E_2, E_3\}$ forms a basis for $E$. Thus, the dimension of $E$ is 3.

$\endgroup$

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.