$W$ is a vector space of all real $n\times n$ matrices $X$ such that $AX=0$ where $A$ is a real $n\times n$ matrix of rank $r$. Find dimension of $W$.

Question

Let $W$ be the vector space of all real $n\times n$ matrices $X$ such that $AX=0$ where $A$ is a real $n\times n$ matrix of rank $r$. Find the dimension of $W$.

The number of free choices for constructing such an $A$ is clearly $nr$ since $r$ rows of $A$ determine the rest of it.

But, I don't get how to classify the choice of $X$'s that are in $W$.

Answer 1

We have $\text{rank}(A)+ \text{nullity}(A)=n$.

Then $\text{nullity}(A)=n-r$.

Let $X\in W$ so that $AX=0$. Let $X=(C_1,C_2,...,C_n)$ where each $C_i$, $i=1,2,\dots ,n$ is a column vectors of $X$. Since $AX=0$, we have $AC_i=0$ for each $i=1,2,\dots ,n$, that is we have $n$ copies of columns lies in then $\ker(A)$ and $\dim \ker(A)=n-r$

So, $\dim W= n(n-r)$.

Answer 2

Hint: find $\phi : W \to {\rm Ker}(A)^n$ which is isomorphism.

Answer 3

Note that since $A$ is of rank $r$, we can write :

$$ A = P \left(\begin{array}{cc} \mathcal{I}_r & \mathcal{O}_{r, n-r} \\ \mathcal{O}_{n-r, r} & \mathcal{O}_{n-r, n-r} \end{array}\right) Q $$

where $P,Q$ are invertible matrices. $\mathcal{I}_r$ is identity matrix of size $r \times r$ and $\mathcal{O}_{n-r,n-r}$ is the null matrix with the corresponding size etc.

Then $AX=0$ can be rewritten $ P\left(\begin{array}{cc} \mathcal{I}_r & \mathcal{O}_{r, n-r} \\ \mathcal{O}_{n-r, r} & \mathcal{O}_{n-r, n-r} \end{array}\right) QX=0$.

Since $P$ is invertible, it is equivalent to $ \left(\begin{array}{cc} \mathcal{I}_r & \mathcal{O}_{r, n-r} \\ \mathcal{O}_{n-r, r} & \mathcal{O}_{n-r, n-r} \end{array}\right) QX=0$

Now we have a much simpler equation ($A$ is much simpler) and also multilplying by $Q$ wont change the dimension because $X \to QX$ is bijective since $Q$ is invertible. In otherwords, $\dim W = \dim QW$ ($QW$ is the vector space of all matrices of $W$ but multiplied by $Q$).

For simplicity, let $M=QX$. We can write $M= \left(\begin{array}{cc} \mathcal{A}_r & \mathcal{B}_{r,n-r} \\ \mathcal{C}_{n-r,r} & \mathcal{D}_{n-r,n-r} \end{array}\right)$ as a block matrix. the sizes of the block are put as index to make it clearer.

Then $\left(\begin{array}{cc} \mathcal{I}_r & \mathcal{O}_{r,n-r} \\ \mathcal{O}_{n-r,r} & \mathcal{O}_{n-r,n-r} \end{array}\right)\left(\begin{array}{cc} \mathcal{A}_r & \mathcal{B}_{r,n-r} \\ \mathcal{C}_{n-r,r} & \mathcal{D}_{n-r,n-r} \end{array}\right)=\left(\begin{array}{cc} \mathcal{A}_r & \mathcal{B}_{r,n-r} \\ \mathcal{O}_{n-r,r} & \mathcal{O}_{n-r,n-r} \end{array}\right) $

Hence it is immediate that solving $\left(\begin{array}{cc} \mathcal{A}_r & \mathcal{B}_{r,n-r} \\ \mathcal{O}_{n-r,r} & \mathcal{O}_{n-r,n-r} \end{array}\right) = 0$ is equivalent to $\mathcal{A}_r=\mathcal{O}_{r}$ and $\mathcal{B}_{r,n-r}=\mathcal{O}_{r,n-r}$.

In other words, we can choose $\mathcal{C}_{n-r,r}$ and $\mathcal{D}_{n-r,n-r}$ freely ! Which boils down to choosing $n(n-r)$ coefficients (the concatenated matrix of $\mathcal{C}_{n-r,r}$ and $\mathcal{D}_{n-r,n-r}$ has $n$ columns and $(n-r)$ rows).

At first glance it seems very cumbersome and tedious, but at least to me it feels very natural to simplify the equation and then just find all the zeros that will make the dimension appears naturally with the free coefficients. Then it's just writing some block matrices to make it rigorous.