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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) I`m 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 don`t know how to give an appropriate base for this set.

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  • $\begingroup$ The Zero vector cannot belong to any basis $\endgroup$ – David P Sep 22 '19 at 17:32
  • $\begingroup$ True, thanks. So the intersection set is not correct? $\endgroup$ – Sebas Martinez Santos Sep 22 '19 at 17:34
  • $\begingroup$ I did not check that, but a basis of the zero subspace is the empty set. $\endgroup$ – David P Sep 22 '19 at 17:35
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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$$B_2=\left\{\begin{bmatrix}1&0&0\\0&0&0\\0&0&1\end{bmatrix},\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix},\begin{bmatrix}0&0&0\\0&0&1\\1&0&0\end{bmatrix}\right\}.$$

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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?

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