Let $U$ and $V$ be the null spaces of $ A=\begin{bmatrix} 1&1&0&0\\0& 0&1&1 \end{bmatrix} $ and $ B=\begin{bmatrix} 1&2&3&2\\0&1&2&1 \end{bmatrix} $. Then what will be the dimension of $U+V.$

I calculated the null space of $U$ and $V$ as follows: \begin{align*} U=\{x\in \mathbb{R}^4: Ax=0 \}\implies U=\text{span}\{(1,-1,0,0), \ (0,0,1,-1) \}\\ V=\{x\in \mathbb{R}^4: Bx=0 \}\implies V=\text{span}\{(1,-2,1,0), \ (0,-1,0,1) \}\\ \end{align*} For calculating the dimension of $U+V$, we have to see what is the dimension of $U\cap V$. For this I consider the matrix $$ \begin{bmatrix} 1&-1&0&0\\0&0&1&-1\\1&-2&1&0\\0&-1&0&1 \end{bmatrix}\sim \begin{bmatrix} 1&-1&0&0\\0&-1&1&0\\0&0&1&-1\\0&0&0&0 \end{bmatrix} ,$$ which tells that the dimension of $U\cap V=1.$ So, $\dim(U+V)=3$.

Now I want to ask is there any shorter method to do this? At least for the dimension of $U\cap V$.


In your particular question, a quick observation can be as follows: since $U \cap V \subseteq U (\text{or } V)$ so $\text{dim}(U \cap V) \leq 2$. It cannot be $2$ because then $U=V$ which is clearly not the case. It cannot be $0$ either because $(1,-1,1,-1) \in U \cap V$ (note that this vector is the sum of the basis vectors of $U$ and also the difference of the basis vectors of $V$). Thus $\text{dim}(U \cap V) =1$.

Then use $$\text{dim}(U + V)=\text{dim}(U)+\text{dim}(V)-\text{dim}(U \cap V).$$

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To find $U+V$ without finding $U$ or $V$:

Note that $U \cap V$ is essentially $\{x \in \Bbb R^4: Ax=0 \land Bx=0\}$.

Rewrite $Ax=0 \land Bx=0$ as $\begin{bmatrix}A\\B\end{bmatrix}x=0$, one can see that $\dim(U \cap V) = \dim\left(\ker\begin{bmatrix}A\\B\end{bmatrix}\right)$, i.e. $\dim\left(\ker\begin{bmatrix}1&1&0&0\\0&0&1&1\\1&2&3&2\\0&1&2&1\end{bmatrix}\right)$.

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  • $\begingroup$ But this only gives $\dim(U\cap V)$. How to conclude that $\dim(U+V)=3$. $\endgroup$ – Sachchidanand Prasad Sep 9 '17 at 18:57
  • $\begingroup$ @SachchidanandPrasad You also found $\dim(U+V)$ from $\dim(U \cap V)$ in your question. Why are you asking me of this now? $\endgroup$ – Kenny Lau Sep 9 '17 at 18:59
  • $\begingroup$ No, you have mentioned that "To find $U+V$ without finding $U$ or $V$". $\endgroup$ – Sachchidanand Prasad Sep 9 '17 at 19:01
  • $\begingroup$ That is irrelevant to your comment on my question. $\endgroup$ – Kenny Lau Sep 9 '17 at 19:03
  • $\begingroup$ No, I am not getting how will we find $U+V$ without finding $U$ and $V$. That is what I asked to you. $\endgroup$ – Sachchidanand Prasad Sep 9 '17 at 19:04

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