# a) Find a base for $W_1 \cap W_2$ and b) Find a base for $W_1 + W_2$

Let

$$W_1 = \left\{ \begin{pmatrix} a & b & c \\ -b & a & b \\ c & b & a \\ \end{pmatrix} :\, a, b, c \in \Bbb R \right\}$$ $$W_2 = \left\{ \begin{pmatrix} a & 0 & 0 \\ b & 0 & c \\ c & b & a \\ \end{pmatrix} :\, a, b, c \in \Bbb R \right\}$$

a) Find a basis for $$W_1 \cap W_2$$

b) Find a basis for $$W_1 + W_2$$

For a) Im not sure if the set for $$W_1 \cap W_2 =$$ $$\{ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\}$$ so the base is only the zero matrix.

For b) I belive the set is: $$W_1 + W_2 =$$ $$\{ \begin{pmatrix} 2a & b & c \\ 0 & a & b+c \\ 2c & 2b & 2a \\ \end{pmatrix}$$:$$a, b, c \in R\}$$ But dont know how to give an appropriate base for this set.

• The Zero vector cannot belong to any basis – David P Sep 22 '19 at 17:32
• True, thanks. So the intersection set is not correct? – Sebas Martinez Santos Sep 22 '19 at 17:34
• I did not check that, but a basis of the zero subspace is the empty set. – David P Sep 22 '19 at 17:35

a) Since $$W_1\cap W_2=\{0\}$$, the only basis of $$W_1\cap W_2$$ is the empty set.
b) Since $$W_1\cap W_2=\{0\}$$, a way of finding a basis $$B$$ of $$W_1+W_2$$ is to take a basis $$B_1$$ of $$W_1$$ and a basis $$B_2$$ of $$W_2$$ and then to take $$B=B_1\cup B_2$$. It is natural to take$$B_1=\left\{\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix},\begin{bmatrix}0&1&0\\-1&0&1\\0&1&0\end{bmatrix},\begin{bmatrix}0&0&1\\0&0&0\\1&0&0\end{bmatrix}\right\}$$and$$B_2=\left\{\begin{bmatrix}1&0&0\\0&0&0\\0&0&1\end{bmatrix},\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix},\begin{bmatrix}0&0&0\\0&0&1\\1&0&0\end{bmatrix}\right\}.$$
For b), rewrite what you obtained as $$a \begin{pmatrix} 2& 0& 0\\ 0 & 1 & 0 \\ 0 & 0& 2 \\ \end{pmatrix}+b \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \\ 0 & 2 & 0 \\ \end{pmatrix}+c \begin{pmatrix} 0 & 0 &1\\ 0 & 0 & 1 \\ 2 & 0 & 0 \\ \end{pmatrix}$$ Can you say whether the three above matrices are linearly independent?