(a) Prove that a square matrix $T$ of rank one has $\text{tr}(T)=0$ if and only if $T^2=0$.

(b) Consider a matrix $A$ of the form $A=aI+T$, where $a\ne0$, $I$ is the identity matrix, and $T$ has rank one and zero trace. Find the inverse and the determinant of $A$.

(c) Find the inverse of $A$ as above when $T$ has rank one but nonzero trace $\text{tr}(T)=b$.

For which value of $b$ is $A$ not invertible?

I'm still stuck on part (a), but campus buildings are closing soon, so I'll be working from home but would love to get some hints / comments on this question. I'll have limited access to this site - on my phone.

For part (a), I've been trying to look at the SVD of matrix $A$, since one can read off the rank very easily - by looking at the number of non-zero singular values of $A$. Then I am trying some block matrix multiplication to see whether $T^2 = 0$, from assuming that $\text{tr}(T) =0$. So far, no luck. Do you think I should stick with this SVD approach, or is it better to play around with the definition and properties of nilpotent operators?

Any other hints for the other parts of the question would be greatly appreciated.


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    $\begingroup$ hint: if $A$ is rank one, then $A = ab^T$ for some nonzero vectors $a, b.$ we went over this last year! $\endgroup$ – abel Jan 4 '15 at 5:09
  • $\begingroup$ Yes, we did, @abel :) Not sure how this decomposition of A, along with trace =0, can help me conclude that T^2 = 0, though. I've tried for awhile last night and will try again now. Can I get another hint perhaps? Thanks... $\endgroup$ – User001 Jan 4 '15 at 23:21
  • $\begingroup$ let us stick to $n = 2.$ pick any two nonzero vectors $a$ and $b$. form the rank one matrix $ab^T.$ come back and tell me what you get for the trace of that matrix in terms of the vectors you chose. $\endgroup$ – abel Jan 4 '15 at 23:24
  • $\begingroup$ ...I got trace = 1, if I let u= (1,0), and $v^t=(1,0)^T$. If I scale by $\sigma_1$, then the trace is $\sigma_1$ ... $\endgroup$ – User001 Jan 4 '15 at 23:32
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    $\begingroup$ can you also compute the dot products of the pairs you choose and the trace of the rank one matrix. pick generic vectors not special one like you have been doing. $\endgroup$ – abel Jan 4 '15 at 23:45


(a) If $T$ is rank one, $T$ should be of the form $T=\sigma_1 uv^T$ where $\sigma_1$ is the highest singular value and $u$,$v$ are left and right singular vectors respectively. Convince yourself that this is true using singular value decomposition. You might need the cyclic property of trace as well.

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(b) Use the fact that $A=aI+\sigma_1 uv^T$. Observe what happens when $T$ is rank-one and zero-trace. Now say $B=\alpha I+\beta uv^T$ for some unknown constants $\alpha$ and $\beta$. Try to find if there are any $\alpha$ and $\beta$ such that $AB=I$ and $BA=I$. If you are still struggling, take a look at sherman-morrison formula.

(c) use the same strategy as in (b).

  • $\begingroup$ By trial-and-error (see above), I was able to see that, with specific choice of vectors that form a rank one matrix with zero trace, the square of the matrix is equal to zero, @dineshdileep. Last night, I spent quite a bit of time on the SVD of T, and I do accept that it is of the form $\sigma_1 uv^T$. How could I proceed from this angle(compared to my work above)?How could I use trace = 0 now, from the SVD of T? My guess would be to look at the similarity transforms, since trace(T) is invariant under such transforms. What do you think? Or, go a different route to concluding that $T^2$=0? $\endgroup$ – User001 Jan 5 '15 at 0:50
  • $\begingroup$ And I read up on the cyclic property and linear property of trace, too - but not sure how I could use either one of these properties...thanks, @dineshdileep $\endgroup$ – User001 Jan 5 '15 at 0:53
  • $\begingroup$ I actually got it now, multiplying T^2 out and moving some parts around that are scalars after matrix multiplication :) Thanks again @dineshdileep ... $\endgroup$ – User001 Jan 5 '15 at 1:16
  • $\begingroup$ Thanks so much for your edited answer for part(a) - that's exactly what I did, @dineshdileep :). I'm working on part(b) now, and am using the Sherman-Morrison formula. But I feel that this is very...specialized and doesn't really reinforce any of the core linear algebra that an old exam should be testing. What do you think? Is there a more "fundamental" way to solve this problem, i.e. , find the inverse (and then the determinant) of an invertible matrix that has been perturbed or "corrected" by a rank-one marix? Thanks... $\endgroup$ – User001 Jan 5 '15 at 2:57
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    $\begingroup$ did you try finding $\alpha$ and $\beta$? $\endgroup$ – dineshdileep Jan 5 '15 at 10:34

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