# Nullity of the linear transformation

Find the nullity of the linear transformation $$\mathbb{T:M_{22} \rightarrow M_{22}}$$ given that it has rank $$2$$.

T can be a matrix transformation, for instance, it can be $$I_2$$. Now it's given that $$I_2$$ has rank 2 so by dimension theorem (rank + nullity = number of columns) it's nullity should be $$2-2 = 0$$. This is what I think is correct.

However, another approach gives a different answer. $$\text{dim(ker T) + rank(T) = dim(M_{22})} \implies \text{nullity} = 2\times 2 - 2 = 2$$

Which approach is correct and what's wrong with the other one?

## 1 Answer

The rank nullity theorem gives that $$M_{22}$$ is of dimension $$4$$, so the answer is $$4-2 = 2$$.

$$I_2$$ is of rank $$2$$ as a linear transformation from $$\Bbb R^2 \to \Bbb R^2$$. But if you take the identity transformation from $$\Bbb M_{22} \to \Bbb M_{22}$$, that has matrix representation $$I_{\mathbf{4}}$$, so has rank $$4$$.

A matrix is used to represent a transformation from $$\Bbb R^n \to \Bbb R^m$$. But when $$\Bbb R^n$$ and $$\Bbb R^m$$ themselves are hidden as spaces of matrices, secretly $$\mathbb M_{22}$$, then we are technically looking at a map from $$\mathbb R^\color{blue}4 \to \mathbb R^{\color{blue}4}$$, not $$\mathbb R^\color{red}2$$.

This confusion should be promptly removed at this stage.

• but for a matrix like I2, isn't rank + nullity = number of columns = 2? Feb 5 '20 at 12:44
• $I_2$ has rank $2$ and nullity $0$ when we look at it as a transformation from $\mathbb R^2$ to $\mathbb R^2$. However, when we look at it as a transformation from $\mathbb M_{22}$ to $\Bbb M_{22}$, it has rank four, because all the matrices like $\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$ etc. ($1$ somewhere, $0$ everywhere else) belong in the image of $I_2$, so we get four linearly independent elements. The point is simple ; $I_2$ is a multiplication operator , but its rank depends on the domain of operation. Feb 5 '20 at 12:46
• I hope you understand what I am saying : essentially, recognizing the dimension of the space is important. If $M$ is a matrix representing the linear transformation $T$, then $M$ is a four dimensional square matrix. So if we think of the matrices as vectors, then multiplication is actually happening by $M$. For example, the identity transformation on $M_{22}$ is the four dimensional identity matrix. So we apply rank nullity on this matrix : if it has rank $2$, then it has nullity $2$. Feb 5 '20 at 12:52