# Nilpotent Implies Singular

In Finding a Determinant (Part 1) and (Part 2), we saw how to calculate the determinant of a given square matrix. Today, we will work on proving a theorem involving the determinant of matrices.

## Nilpotent Implies Singular

### Theorem: Let $$A$$ be a square matrix. Furthermore, let $$A$$ be nilpotent, that is, $$A^{k}=\mathbf{0}$$ for some natural number $$k$$. Then, $$A$$ is singular, that is, $$|A|=0$$.

Before we get into our proof, we want to make sure that we understand what we get to assume and what we have to show. In this case, we get to assume that if we multiply $$A$$ by itself over and over, we will eventually get out the zero matrix. We should note that does not imply that $$A=\mathbf{0}$$. As an example, suppose that
\begin{align*}
A&=\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}.
\end{align*}
We then get that $$A^{2}=\mathbf{0}$$, so we can have a non-zero matrix which is nilpotent.

However, what the theorem says, is that we cannot have an non-singular matrix that is nilpotent. Recall that a matrix is singular if its determinant is $$0$$ and non-singular otherwise. Therefore, we have seen, that a matrix is invertible if and only if it is non-singular. Hence, we cannot have an invertible matrix which is nilpotent.

But why is this the case? I will follow the process outlined in Direct Proof: a2=b2mod n. by provided a table giving the things we can assume and the things we must show.

Have Given Must Show
$$A$$ is nilpotent. $$A$$ is singular.
$$A^{k}=\mathbf{0}$$. $$|A|=0$$.
$$|A^{k}|=|\mathbf{0}|=0$$.

Now, as long as we recall that $$|A||B|=|AB|$$, we can see that $$0=|A^{k}|=|A|^{k}$$. Furthermore, since the determinant is a real number, we now see that $$|A|$$ multiplied by itself is $$0$$. However, for real numbers, we cannot have the product of two non-zero numbers be zero. Therefore, we have that $$|A|$$ must be $$0$$. We are now ready to put this together into a proof.

## Proof

Let $$A$$ be a square nilpotent matrix. Since $$A$$ is nilpotent, we have that $$A^{K}=\mathbf{0}$$ for some natural number $$k$$. Therefore, $$|A^{K}|=|\mathbf{0}|=0$$. Now, since $$|A|*|B|=|AB|$$, we have that $$0=|A^{k}|=|A|^{k}$$. Furthermore, since $$|A|$$ is a real number, we must have that $$|A|=0$$, since we cannot multiply two non-zero real numbers and get $$0$$. Hence, $$A$$ is a singular matrix.

## Conclusion

As we worked through this, we got to work more with properties of determinants. While you can work with specific examples that ask you to find $$|A|*|B|$$ given $$|A|$$ or $$|B|$$, this gives us a chance to use properties of determinants without having to revert to specific examples. Such theorems are can be extremely helpful in that we now have another way to determine if a matrix is singular.

Hopefully this helped you better understand determinants and get a little extra work with proofs. If this did help, make sure to share this post on Social Media and subscribe to our YouTube channel so that you see the videos that accompany these posts.

## 2 thoughts on “Nilpotent Implies Singular”

1. Anonymous says:

Does singular imply nilpotent?

1. Dr. Justin Albert says:

No, it does not. An example would be the matrix, ((1,0),(0,0)).

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