# Find a common direct complement for $K$ and $L$ in $M_n$

Given the matrices $$A=\begin{pmatrix} 1 & 0 \\ -1 & -1\end{pmatrix}, \ B=\begin{pmatrix} 1 & 2 \\ 0 & -1\end{pmatrix}$$ and subspaces of $M_2$:

$$K=\{X\in M_2: tr(XA)=0\}, L=\{X\in M_2: tr(XB)=0\}$$

Find bases for $K$ and $L$ and determine one common direct complement for $K$ and $L$ in $M_n$.

My attempt:

$$X\in K\Leftrightarrow tr(XA)=0,\ X=\begin{pmatrix} x_1 & x_2 \\ x_3 & x_4\end{pmatrix}\Leftrightarrow \begin{pmatrix} x_1 & x_2 \\ x_3 & x_4\end{pmatrix}\cdot\begin{pmatrix} 1 & 0 \\ -1 & -1\end{pmatrix}=\begin{pmatrix} x_1-x_2 & -x_2\\ x_3-x_4 & -x_4\end{pmatrix}$$

$$x_1-x_2-x_4=0\Rightarrow x_1=x_2+x_4\Rightarrow X=\begin{pmatrix} x_2+x_4 & x_2 \\x_3 & x_4\end{pmatrix}\Rightarrow \\ X=\begin{pmatrix}x_2+x_4 & x_2 \\ x_3 & x_4\end{pmatrix}=x_2\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}+x_3\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}+x_4\begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}$$

So one basis for $K$ is $\{\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix},\begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}\}$

Similarly I found one basis for $L: \{\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}, \begin{pmatrix} 0 & 1\\0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0\\1 & 2\end{pmatrix}\}$

How do I now find a common direct complement? In general, we get a direct complement as a span of those vectors that complement the base to the base of the whole vector space. So, would some direct complement of $K$ be $\{\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\0 & 1\end{pmatrix}$? And for $L$ $\{\begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix},\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix},\begin{pmatrix} 0 & 0\\ 0 & 1\end{pmatrix}\}$? And then I intersect those two to find a common one?

• $K$ and $L$ are three-dimensional. Their direct complements are one-dimensional. So, if $M$ is any matrix that does not belong to both $K$ and $L$, the span of $M$ would serve as a common direct complement of both subspaces. – user1551 Jan 24 '17 at 9:02
• @user1551 Why does it have to be a matrix that does not belong to both $K$ and $L$? – lmc Jan 25 '17 at 14:04
• @user1551 And how do you know that direct complement is one-dimensional? – lmc Jan 25 '17 at 14:05
• (1) By definition, the only matrix that lies inside the intersection of $K$ and its complementary subspace is zero. So, if $M$ is a nonzero matrix in the complement, it must not belong to $K$. The same goes for $L$. (2) $K$ is 3D and $M_2(\mathbb R)$ is 4D, so any complement of $K$ is 1D. The same goes for $L$. – user1551 Jan 25 '17 at 14:38

First, a general observation. Let $V$ be an $n$-dimensional vector space and let $\varphi \colon V \rightarrow \mathbb{F}$ be a non-zero linear functional. By the rank-nullity theorem, we know that $\dim \ker(\varphi) = n - 1$. Let us choose any vector $w \in V$ such that $\varphi(w) \neq 0$. Then

$$V = \ker(\varphi) \oplus \operatorname{span} \{ w \}.$$ To see this is true, not that if $v \in \ker(\varphi) \cap \operatorname{span} \{ w \}$ then $\varphi(v) = 0$ and $v = aw$ for some $a \in \mathbb{F}$. This implies that $\varphi(aw) = a\varphi(w) = 0$ but then $a = 0$ and $v = 0$. Since the dimensions add up, we have a direct sum decomposition.

Now if $\varphi_1, \varphi_2$ are two non-zero functionals and we want to find a direct sum complement for both $\ker(\varphi_1),\ker(\varphi_2)$, by the observation above we only need to choose $w \in V$ such that $\varphi_1(w) \neq 0$ and $\varphi_2(w) \neq 0$.

In your case, $V = M_2(\mathbb{F})$, $\varphi_1(X) = \operatorname{tr}(XA)$ and $\varphi_2(X) = \operatorname{tr}(XB)$. Hence, to find a direct sum complement for both $K = \ker(\varphi_1)$ and $L = \ker(\varphi_2)$, we need to find $X \in M_2(\mathbb{F})$ such that $\varphi_1(X) \neq 0$ and $\varphi_2(X) \neq 0$.

Let

$$X = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.$$

Then for any $Y \in M_2(\mathbb{F})$, the matrix $XY$ will be a matrix whose first row is the first row of $Y$ and whose second row is zero. Hence,

$$\operatorname{tr}(XA) = \operatorname{tr}(XB) = 1$$

so $\operatorname{span}(X)$ complements both $K$ and $L$ to $M_2(\mathbb{F})$.

• It seems that someone wanted a more detailed answer so I added many details. – levap Jan 25 '17 at 19:28