Invertible Matrix Theorem and Rank-Nullity Theorem

Let’s talk about a way to connect two theorems that come up in Linear Algebra: the Invertible Matrix Theorem (IVT) and the Rank-Nullity Theorem.

The Invertible Matrix Theorem has a lot of equivalent statements of it, but let’s just talk about two of them. Here’s the first one.

$\bf{Definition}$: An $n\times n$ matrix $A$ is invertible if and only if $A\vec{x} = 0$ has only the solution $\vec{x} = 0$.

Another way of saying this is that the null space is zero-dimensional. Usually, when a set is written as the span of one vector, it’s one dimensional. But in this case, since no scalar multiple of the zero vector actually scales it at all, the null space is zero-dimensional. We’ll use this fact when we talk about the Rank-Nullity Theorem.

Here’s another statement of the Invertible Matrix Theorem.

$\bf{Definition}$: An $n\times n$ matrix $A$ is invertible if and only if its columns map onto $\mathbb{R}^n$

We are going to use these two characterizations of the Invertible Matrix Theorem to connect it to the Rank-Nullity Theorem. First, let’s define the rank and nullity of a matrix.

$\bf{Definition}$: The rank of a matrix is the dimension of the column space of the matrix.

Since the column space lives in $\mathbb{R}^m$, the rank of a matrix is less than or equal to $m$.

$\bf{Example}$: The matrix $A = \begin{pmatrix}1 & 0 \\ 0 &0 \\ 0&0 \end{pmatrix}$ has rank $1$, since any vector in the column space is of the form $\begin{pmatrix}k \\ 0\\ 0 \end{pmatrix}$.

Now let’s define the nullity of a matrix.

$\bf{Definition}:$ The nullity of a matrix is the dimension of the null space.

Since the null space lives in $\mathbb{R}^n$, the nullity of a matrix is less than or equal to $n$.

Now we’re ready to state the Rank-Nullity Theorem.

$\textbf{Rank-Nullity Theorem:}$ Let $A$ be an $m\times n$ matrix. Then $rank(A) + nullity(A) = n$.

To connect this to the Invertible Matrix Theorem, we just need to remember that an invertible matrix has full rank ($rank = n$), since it maps onto $\mathbb{R}^n$.

If we substitute into the equation describing the rank-nullity theorem, we see that $n+nullity(A) = n$. This shows that if a matrix has full rank, then its null space is zero-dimensional, which connects the two statements of the IVT above.

In the next post, we’ll show talk about the connection between the Rank-Nullity Theorem and the First Isomorphism Theorem.

As a preview of the connection, the Rank-Nullity Theorem relates the dimension of the null space of a matrix to the dimension of the column space of a matrix for linear transformations between Euclidean spaces. The First Isomorphism Theorem relates the dimension of the kernel (similar to the null space) of a transformation between groups to the dimension of the image of a transformation between groups.

Stay tuned!

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