True or False: Let $n ≥ 4$. There is an $n\times n$ matrix A such that $A^4 = 0$ and $\text{rank}(A+I_n) < 4$.

Question

I don't know how to judge this and I am a beginner at linear algebra so I need an answer which explains all necessary terms.

Welcome to Math.SE. You are likely to get a response if you show what you have done and where you are stuck. — Shailesh, Nov 14 '20 at 00:02
Here is the intuition behind it, nilpotent matrix is similar to strictly upper triangular matrix (e.g. https://math.stackexchange.com/q/1003525/399263). When you add $I_n$ there will be all $1$ on the diagonal and this matrix cannot have rank $<4$. — zwim, Nov 14 '20 at 00:58

score 5 · Accepted Answer · answered Nov 14 '20 at 00:00

5

If $\text{rank}(A + I_n) < 4$, then by the rank-nullity theorem it's nullity is at least $1$, i.e. there exists some non-trivial vector $v \neq 0$ such that $(A + I_n) v = 0 \implies A v = -v$. In other words, $v$ would be an eigenvector of $A$ with eigenvalue $-1$. But then $$A^4 v = A(A(A(Av))) = v \neq 0$$ But why would this lead to a contradiction? (Hint: We haven't yet used the fact that $A^4 = 0$.)

answered Nov 14 '20 at 00:00

paulinho

6,553

Does the contradiction mean that "A^4 can't be zero" (since "A^4 v" is not equal to 0 and v is not 0) but the description said that "A^4 is equal to zero"? – Rae Jiang Nov 14 '20 at 02:17
Yes, that’s correct! – paulinho Nov 14 '20 at 03:07

True or False: Let $n ≥ 4$. There is an $n\times n$ matrix A such that $A^4 = 0$ and $\text{rank}(A+I_n) < 4$.

1 Answers1