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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...

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1 Answer 1

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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 $4 - 2 = 2$.

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  • $\begingroup$ 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. $\endgroup$ Commented Mar 26, 2017 at 6:10
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    $\begingroup$ @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. $\endgroup$ Commented Mar 26, 2017 at 6:21
  • $\begingroup$ Yep that makes sense! I just wanted to clarify. Thank you so much! $\endgroup$ Commented Mar 26, 2017 at 6:22

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