(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.
Thanks!
Best Answer
Hints:
(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).