TIFR-2018-Maths-A: 15

Question

True/False Question :

The set of nilpotent matrices in $M_{3}\left ( \mathbb{R} \right )$ spans $M_{3}\left ( \mathbb{R} \right )$ considered as an $\mathbb{R}$-vector space (a matrix $A$ is said to be nilpotent if there exists $n \in \mathbb{N}$ such that $A^{n}=0$).

Answer 1

No (False), because nilpotent matrices always have trace = 0, and linear combinations of them also always have trace $0$.
That means you can never reach matrices with nonzero trace (like the identity matrix).

Comments

Things I learned:

In a nilpotent matrix, trace is zero ( Diagonal elements need not be zero)

Here the matrix space has 9 Dimensions, but the nilpotent matrix has the condition that, the trace should be zero "a_11 + a_22 = - a_33", so one variable is restricted. Hence, nilpotent matrix can span only 8 Dimensions.

Is my reasoning correct?

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Not Exactly Correct. Read this once again :

For a \(3 \times 3\) matrix to be nilpotent, its characteristic polynomial must be
\[
p(\lambda) = \lambda^3.
\]
This imposes three conditions on the matrix entries (from the coefficients of the characteristic polynomial):

\[
\text{Trace} = 0 \quad (a_{11} + a_{22} + a_{33} = 0),
\]
\[
\text{Sum of principal minors} = 0,
\]
\[
\det(A) = 0.
\]

The set of nilpotent matrices is not a simple vector space, but a complex algebraic variety.
The dimension of the set of \(n \times n\) nilpotent matrices is given by the formula
\[
n(n-1).
\]

Therefore, for \(n=3\), the dimension of the variety of nilpotent matrices is
\[
3(3-1) = 3 \times 2 = 6.
\]

 

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Thank you very much for detailed explanation 👌