2
$\begingroup$

If I have a projection matrix L in $\mathbb {R^4}$ , I'm just wondering how L would transform vectors in the nullspace of $[L]$ and the column space. I'm also trying to figure out how these pieces of information allow me to find the rank and nullity of $[L]$ without elementary row operations.

For context, here is the question:

$\text{The linear transformation of $L:\mathbb {R^4}\rightarrow \mathbb {R^4}$ projects $\mathbb {R^4}$ orthogonally}$ $\text{onto the subspace $V=\text{span}\{a,b\}$, with}:$

$a=(1,1,1,1)$

$b=(4,2,1,2)$

$\text { (a) How does L transform vectors transform vectors in the null space of [L]?}$

$\text { (b) How does L transform vectors transform vectors in the Column space of [L]?}$

$\text {(c)Explain how the answers to parts (a) and (b) }$ $\text{enable you to find the rank and the nullity of [L] without row reduction.}$

I'm looking at notes here, but I'm having a hard time coming up with some reasoning. I can see that the null space and the column space are orthogonal to each other but I am not really sure how that would explain the transformation or help with part $(c)$ in any way.

If someone could nudge me in the right direction that would be great!

I mean I feel like null space comes into play somehow because I am doing projections and since I want the matrix to be orthogonal, the dot product has to be $0$ so I am trying to see if I can relate that somehow. The fact I am writing the vectors as columns I feel like has to do with something in the column space but I'm not entirely sure about that...

For part $(c)$, I feel like I have to use the rank nullity theorem somehow but I am not sure about this...

$\endgroup$
1
$\begingroup$

Part (a): By definition, the null space of the matrix $[L]$ is the space of all vectors that are sent to zero when multiplied by $[L]$. Equivalently, the null space is the set of all vectors that are sent to zero when the transformation $L$ is applied. $L$ transforms all vectors in its null space to the zero vector, no matter what transformation $L$ happens to be.

Note that in this case, our nullspace will be $V^\perp$, the orthogonal complement to $V$. Can you see why this is the case geometrically?

Part (b): In terms of transformations, the column space $L$ is the range or image of the transformation in question. In other words, the column space is the space of all possible outputs from the transformation. In our case, projecting onto $V$ will always produce a vector from $V$ and conversely, every vector in $V$ is the projection of some vector onto $V$. We conclude, then, that the column space of $[L]$ will be the entirety of the subspace $V$.

Now, what happens if we take a vector from $V$ and apply $L$ (our projection onto $V$)? Well, since the vector is in $V$, it's "already projected"; flattening it onto $V$ doesn't change it. So, for any $x$ in $V$ (which is our column space), we will find that $L(x) = x$.

Part (c): The rank is the dimension of the column space. In this case, our column space is $V$. What's it's dimension? Well, it's the span of two linearly independent vectors, so $V$ is 2-dimensional. So, the rank of $[L]$ is $2$.

We know that the nullity is $V^\perp$. Since $V$ has dimension $2$ in the $4$-dimensional $\Bbb R^4$, $V^\perp$ will have dimension $4 - 2 = 2$. So, the nullity of $[L]$ is $2$.

Alternatively, it was enough to know the rank: the rank-nullity theorem tells us that since the dimension of the overall (starting) space is $4$ and the rank is $2$, the nullity must be $4 - 2 = 2$.

$\endgroup$
  • $\begingroup$ Thanks for this! I just have one question. If I project a vector onto the column space, wouldn't it actually be a linear combination of the column space and not itself? I am not entirely clear on that part. $\endgroup$ – Future Math person Mar 26 '17 at 6:10
  • 1
    $\begingroup$ @FutureMathperson If you take any old vector and project it onto $V$ (which is the column space), then you shouldn't necessarily get the same thing you put in. However, the point I made was that if you start with something that's in $V$, the output (its projection) will be exactly the same as the input. $\endgroup$ – Omnomnomnom Mar 26 '17 at 6:21
  • $\begingroup$ Yep that makes sense! I just wanted to clarify. Thank you so much! $\endgroup$ – Future Math person Mar 26 '17 at 6:22

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.