To show that a nilpotent matrix is not invertible, we proceed as follows:

Definition of Nilpotent Matrix:

A square matrix $A$ is nilpotent if there exists a positive integer $k$ such that: $A^k = 0$ where $0$ is the zero matrix.

Invertibility Condition:

A matrix $A$ is invertible if there exists another matrix $B$ such that: $AB = BA = I$ where $I$ is the identity matrix. For this to hold, the determinant of $A$, denoted as $\det(A)$, must be nonzero.

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Proof:

Step 1: Assume $A$ is nilpotent

By definition, there exists some $k \in \mathbb{N}$ such that: $A^k = 0$ If $A$ were invertible, there would exist a matrix $B$ such that $AB = BA = I$.

Step 2: Contradiction from $A^k = 0$

Using the assumption $AB = I$, we multiply both sides of $AB = I$ by $A^{k-1}$: $A^{k-1}(AB) = A^{k-1}I$ $(A^k)B = A^{k-1}$ Since $A^k = 0$, this simplifies to: $0 \cdot B = A^{k-1}$ $0 = A^{k-1}$ This implies $A^{k-1}$ is also the zero matrix.

Step 3: Determinant of $A$

The determinant of a nilpotent matrix $A$ is $\det(A) = 0$ because $A^k = 0$ implies that $A$ cannot have full rank. A matrix with $\det(A) = 0$ is not invertible.

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Conclusion:

The assumption that $A$ is nilpotent leads to the conclusion that $A$ cannot be invertible because it satisfies $A^k = 0$ for some $k$, which forces $\det(A) = 0$. Hence, a nilpotent matrix is not invertible.

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Do you want further clarification or a numerical example to solidify the concept?

Related Questions:

1. What is an example of a nilpotent matrix and its powers?
2. How is the determinant related to the eigenvalues of a matrix?
3. Can a matrix be nilpotent and diagonalizable? Why or why not?
4. How does a nilpotent matrix arise in real-world applications or mathematical modeling?
5. What are the implications of nilpotency in linear algebra?

Tip: To check whether a matrix is nilpotent, compute successive powers of the matrix until you reach the zero matrix (or prove it won't happen).