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Let $ W_1 = \textrm{span}\left\{\begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}2\\1\\1\end{pmatrix}\right\}$, and $ W_2 = \textrm{span}\left\{\begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}3\\0\\-1\end{pmatrix}\right\}$. Find a basis for $W_1 \cap W_2$

I first thought of solving the augmented matrix:

$ \begin{pmatrix}1 && 2 && 1 && 3\\2 && 1 && 0 && 0\\3 && 1 && 1 && -1\end{pmatrix}$

But this matrix can have 3 pivots and so it's column space dimension can be at most 3 (which doesn't make sense since the basis I'm looking for must have dimension 2.

So, what is the correct way to solve these exercise.

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3 Answers 3

up vote 1 down vote accepted

Note that both the spaces have dimension $2$, this means that if their intersection had dimension $2$ they were equal. However it is immediate that no vector in $W_2$ can be written as $(x,y,z)$ with $y\neq 0$, and therefore $W_1\neq W_2$.

From this follows that the intersection has to have dimension $1$ or $0$. But we can also rule out the case where the intersection is zero, because then the direct sum of these spaces would be a $4$ dimensional subspace of $\Bbb R^3$. Impossible! So the intersection must have dimension $1$.

Since we already know that vectors in $W_2$ have their $y$ coordinate $0$, it is sufficient to see if we can find such vector in $W_1$, and then we can try to see whether or not this vector is in $W_2$ as well. The simplest and most obvious way would be: $$\begin{pmatrix}1\\2\\3\end{pmatrix}-2\begin{pmatrix}2\\1\\1\end{pmatrix}=\begin{pmatrix}1-4\\2-2\\3-2\end{pmatrix}=\begin{pmatrix}-3\\0\\1\end{pmatrix}=-\begin{pmatrix}3\\0\\-1\end{pmatrix}$$

Clearly this vector is in $W_2$ and therefore in the intersection. Since the intersection is of no more than one dimension, this is indeed a basis for $W_1\cap W_2$.


Some remarks: Note that I did not present a "general method" of solving this problem. I pointed out several heuristics which helped me solve this very quickly.

The reward for understanding the approach above, except the approach itself, is to understand that sometimes the general method is tedious and riddled with indices, whereas when one stops to ponder and observe the objects, one can draw immediate conclusions which cut the work in half.

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If I just could give you +10000! Thank you very much for the clear explanation. I was breaking my back to understand. –  Susana Mar 9 '13 at 2:46
    
You're welcome, @Susana. I also added another heuristic to the answer, which you may find useful. –  Asaf Karagila Mar 9 '13 at 2:47

Let's try to make it easier: if you take a minute to read carefully what is $\,W_2\,$ , you'll see that

$$\begin{pmatrix}x\\y\\z\end{pmatrix}\in W_2\iff y=0$$

Now, since an element of $\,w_1\in W_1\,$ has the general form

$$w_1=a\begin{pmatrix}1\\2\\3\end{pmatrix}+b\begin{pmatrix}2\\1\\1\end{pmatrix}=\begin{pmatrix}a+2b\\2a+b\\3a+b\end{pmatrix}$$

we see that $\,w_1\in W_2\iff 2a+b=0\,$

Since $\,\dim(W_1\cap W_2)=1\,$ (why?) , we get that the above condition completely determines the elements of the intersection, and thus

$$W_1\cap W_2=\left\{\,\begin{pmatrix}a+2b\\2a+b\\3a+b\end{pmatrix}\;;\;2a+b=0\,\right\}=\left\{\,\begin{pmatrix}-3a\\0\\a\end{pmatrix}\;;\;a\in\Bbb R\,\right\}=\operatorname{Span}\left\{\,\begin{pmatrix}-3\\0\\1\end{pmatrix}\,\right\}$$

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The two sub spaces are not the same, because $W_2$ has no extent in the second axis while $W_1$ does. The intersection is then a line in the $xz$ plane, which $W_2$ spans. If you can find a vector in $W_1$ in that plane that is your basis

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