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Clarification: this is a review problem, not a homework problem or anything—I'm not getting graded on this. That being said, I cannot seem to figure out how to do the last part.

Consider the matrix $A = u v^T$, where $u, v \in \mathbb R^{m}$.

(a) What is the rank of $A$? Find a basis for the range of $A$.

This part is easy: we know that Range($uv^T$) is just span({$u$}).

(b) List all eigenvalue of $A$. What are their geometric and algebraic multiplicities?

Since the columns of $A$ are just linear combinations of $u$, it should be easy to say that 0 is an eigenvalue with geometric and algebraic multiplicity $m - 1$ since the eigenspace associated with $\lambda = 0$ is just Null($A$). The other eigenvalue I feel can only be found through observation: $uv^T*u$ = $u<u, v>$ = $<u, v> u$, so the eigenvalue is $<u,v>$ with geometric and algebraic multiplicities 1. If there's any other way to find this, please let me know.

(c) Find the eigenvector for the nonzero eigenvalue of $A$.

From above: $u$

(d) Find an orthogonal projector onto the range of $A$.

This is pretty obvious again just by looking at the definition of a projector and because Range($A$) = span({$u$}): $\frac{1}{\|u\|^2}uu^T$.

(e) Find an orthogonal projector onto the nullspace of $A$.

No idea. I'm having a sort of disconnect here and really can't figure it out.

Community
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CoolRobloxKid12
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  • This has been answered before here: https://math.stackexchange.com/questions/1686223/finding-orthogonal-projectors-onto-the-range-null-space-of-the-matrix-a – tia Jun 06 '19 at 22:36
  • Wow, clearly I did not search hard enough. From my quick read of that answer, however, I don't think the answerer is answering the question at hand. – CoolRobloxKid12 Jun 06 '19 at 23:44
  • Hint: The null space is the orthogonal complement of the row space. – amd Jun 07 '19 at 00:40
  • Also see https://math.stackexchange.com/questions/904926/determinant-of-a-rank-1-update-of-a-scalar-matrix-or-characteristic-polynomia. – StubbornAtom Jun 08 '19 at 13:44
  • Some of the answers may need special cases for the case where $u=0$ or $v=0$. – Daniel Schepler Jun 08 '19 at 20:18

1 Answers1

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Hint: For $V=U\oplus W$ and respective projectors $P_{U}$ and $P_{W}$

$$I=P_{U}+P_{W}$$

eranreches
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  • So since Range$(A^T)^{\perp}$ = Null($A$) and the projector onto Range($A^T$) is $\frac{1}{|v|^2}vv^T$, the answer is $I - \frac{1}{|v|^2}vv^T$? Still trying to wrap my head around this, but I'm slowly getting it. Additionally, do you think there's a better way of doing parts (b) and (c) without just like noticing that $u$ is the eigenvector? Thanks for the help! – CoolRobloxKid12 Jun 06 '19 at 23:43
  • Why $A^{T}$? You can simply divide the space into what is $u$ and what is perpendicular to $u$ (null space of $A$). Thus the projector to the null space is $I$ minus the projector into $u$, which is $I-\frac{uu^{T}}{\left|u\right|^{2}}$ as you write. – eranreches Jun 06 '19 at 23:48
  • I might be wrong on this, but isn't Range$(A^T)^\perp$ = Null$(A)$ what the fundamental theorem of linear algebra states? I'm pretty sure that Range($A$) is not perpendicular to Null($A$). – CoolRobloxKid12 Jun 07 '19 at 06:26
  • @CoolRobloxKid12 You're right, my bad. Note that the projector you wrote in the first comment is into ${\rm Range}\left(A^{T}\right)^{\perp}$, since you used the decomposition $\mathbb{R}^{m\times m}={\rm Range}\left(A^{T}\right)^{\perp}\oplus {\rm Range}\left(A^{T}\right)$. – eranreches Jun 07 '19 at 06:38