0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ but for a matrix like I2, isn't rank + nullity = number of columns = 2? $\endgroup$
    – Archer
    Feb 5 '20 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. $\endgroup$ Feb 5 '20 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 $2$. $\endgroup$ Feb 5 '20 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.