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 1


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.

  • $\begingroup$ but for a matrix like I2, isn't rank + nullity = number of columns = 2? $\endgroup$
    – Archer
    Commented Feb 5, 2020 at 12:44
  • $\begingroup$ $I_2$ 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. $\endgroup$ Commented Feb 5, 2020 at 12:46
  • $\begingroup$ 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 $2$. $\endgroup$ Commented Feb 5, 2020 at 12:52

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