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Let $A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ be a matrix in $M_2(\mathbb{R})$. Then $U$ and $V$ are defined by:

$U=\{M\in M_2(\mathbb{R})|MA=AM\}$

$V=\{M\in M_2(\mathbb{R})|MA^T=A^TM\}$

I want to show that U and V are subspaces of $M_2(\mathbb{R})$ but I am unsure how to do it. I know that in order for a subset to be a subspace, it has to be closed under addition, scalar multiplication and must contain the zero vector. I know how to prove scalar multiplication and the zero vector, but I am unsure about addition.

$\begin{bmatrix} \mu & \lambda \\ 0 & \mu \end{bmatrix}$ some $\mu,\lambda\in\mathbb{R}$. How does this work under addition?

Also, I am not sure that I understand the concept of dimension correctly, because I thought it was just the product of the number of columns and rows in a matrix, but then I stumbled upon this formula:

$dim(U+V)=dim(U)+dim(V)-dim(U\cap{V})$ where $U, V$ are some sets of matrices. So when calculating dimension of a set of matrices, do we perform operations with the dimensions of the individual matrices? (This is probably a confusing question, I probably don't understand the whole concept of dimension properly).

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  • $\begingroup$ You seem to claim that all matrices of the form $\begin{bmatrix}\mu&0\\0&\lambda\end{bmatrix}$ are in $V$. Re-check that. $\endgroup$ – Andreas Blass Nov 27 '20 at 21:33
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For adittion, if $M_1, M_2\in U$ you get $(M_1+M_2)A=M_1A+M_2A=AM_1+AM_2$, because $M_1$ and $M_2$ are in $U$. Then $U$ is a subspace of $M_2(\mathbb{R})$. The process is similar for $V$.

For the dimensions, you need to find a base for every subspace. If $M\in U$ and $M$ is like

$$M=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]$$

Then you get the following relation

$$\left[\begin{matrix}a&a+b\\c&c+d\end{matrix}\right]=\left[\begin{matrix}a+c&b+d\\c&d\end{matrix}\right]$$

And with this $c=0$, $a=d$ and $b\in\mathbb{R}$. So your matrix $M$ is like

$$M=\left[\begin{matrix}a&b\\0&a\end{matrix}\right]=a\cdot\left[\begin{matrix}1&0\\0&1\end{matrix}\right]+b\cdot\left[\begin{matrix}0&1\\0&0\end{matrix}\right]$$

You can see that the base of $U$ has two elements, and with this $dim(U)=2$. Sorry for my english.

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