Let $W_1,W_2$ be the subspaces of $M_2(\Bbb R)$ such that,

$$W_1=\{\begin{pmatrix} x && y \\ z && 0 \end{pmatrix}, x,y,z\in \Bbb R \}, ~~W_2=\{\begin{pmatrix} x && y \\ 0 && x \end{pmatrix}, x,y\in \Bbb R \}$$

then Prove/Disprove that $W_1+W_2=M_2(\Bbb R)$.

Solution: $\dim(W_1)=3 \,,\dim(W_2)=2$

dim$(W_1\cap W_2)=1$

Since $$\dim (W_1+W_2)=\dim W_1+\dim W_2-\dim(W_1\cap W_2)$$ we have $\dim (W_1+W_2)=4$ and hence $W_1+W_2=M_2(\Bbb R)$.

Am I right? If I am wrong then please help me to solve this problem . Thank you.

  • $\begingroup$ Suppose $A$ is a $2 \times 2$ matrix. Can you find $B_1 \in W_1, B_2 \in W_2$ such that $B_1 + B_2 = A$? $\endgroup$ – AJY Sep 6 '16 at 16:30
  • $\begingroup$ It seems right to me. $\endgroup$ – Luca Bressan Sep 6 '16 at 16:32
  • $\begingroup$ The step proving $\mathrm{dim}(W_1 \cap W_2) = 1$ is omitted, although it is not difficult to fill in ($x$ and $z$ must be $0$, so only $y$ is left to vary). However, it is also not difficult to give a direct proof, as suggested by AJY, rather than appealing to the linear inclusion-exclusion principle. $\endgroup$ – Robert Furber Sep 6 '16 at 16:35

That way works, but if I were the grader I'd want you to show why $\dim(W_1 \cap W_2) = 1$.

You can also just do this directly: if $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ is some arbitrary matrix, can you select $x_1,y_1,z,x_2,y_2$ so that $\begin{bmatrix}x_1 & y_1 \\ z & 0\end{bmatrix} + \begin{bmatrix}x_2 & y_2 \\ 0 & x_2\end{bmatrix}$ is equal to this matrix?


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