Consider two subspaces $U=\{(a,b,c,d)|b-5c+d=0\}$ and $W=\{(a,b,c,d)|a=d\;,\; b=5c\}$ of $\mathbb{R}^4$. Find a basis and the dimension of $U \cap W$.
So what I have so far is that the intersection must contain vectors which exist in both subspaces so these vectors must meet both $b-5c+d=0$ and $a=d$, $b=5c$ which led me to the conclusion that the vectors in the intersection follow $b(5c)-5c+d(a)=0 \Rightarrow d(a) = 0$ : so vectors in the intersection are of the form $(0,5,1,0)$.
Is this a basis for the intersection, and if so is the dimension just $1$?
Thank you.
Choose basis $\;\{\,(1,1,0,-1)\,,\,\,(1,0,0,0)\,,\,\,(0,5,1,0)\,\;,\;\;\{\,(0,5,1,0)\,,\,\,(1,0,0,1)\,\}\;$ of$\;U,\,W\,$ resp., and now form a matrix with the above vectors as rows and reduce it. Whatever "vanishes" belongs to the intersection:
$$\begin{pmatrix}1&1&0&\!-1\\1&0&0&0\\0&5&1&0\\0&5&1&0\\1&0&0&1\end{pmatrix}\longrightarrow\begin{pmatrix}1&1&0&\!-1\\0&\!-1&0&1\\0&5&1&0\\0&5&1&0\\0&\!-1&0&2\end{pmatrix}\longrightarrow\begin{pmatrix}1&1&0&\!-1\\0&\!-1&0&1\\0&0&1&5\\0&0&1&5\\0&0&0&1\end{pmatrix}$$
we can already see that only the fourth row is going to become all zedros in the next step, thus
$$U\cap W=\text{Span}\,\{\,(0,5,1,0)\,\}$$
which is your answer. Observe the above shows you a general way to attack these problems, as "seeing" the solution straighforward from the definition is, usually, much harder and not always will be possible to do it as you did it.