2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.