# The rank of a multiplication map $L\colon M_{2 \times 3} \to M_{3 \times 3}$

Let $L\colon M_{2 \times 3} \to M_{3 \times 3}$ be the linear transformation defined by $L(A) = \left[\begin{array}{rr}2 & -1 \\ 1 & 2 \\ 3 & 1\end{array}\right]\,A$. Find the dimension of the range of $L$.

Answer: $6$

How is the answer $6$? Isn't it $2$?

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I assumed that "M23" was the space of $2 \times 3$ matrices. It would not hurt to make this explicit, nor would it hurt to learn $\TeX$! – Dylan Moreland Jun 20 '12 at 23:41
Welcome to math.SE! In order to get better answers, you could explain a bit what you've tried, and what you know. For example, why do you think its rank is $2$? – talmid Jun 21 '12 at 0:01

As is often the case with linear algebra, how easy this is depends on which theorems you know.

The vector spaces $M_{2 \times 3}$ and $M_{3 \times 3}$ have dimensions $6$ and $9$, respectively. It seems like you know the rank-nullity theorem, and hence that $L$ having rank $6$ is equivalent to $L$ having trivial kernel. How could you verify this? Note that, for example, $\left(\begin{array}{rr} 2 & -1 \\ 1 & 2 \\ 3 & 1 \end{array}\right) \left(\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) = \left(\begin{array}{rrr} 2 & 0 & 0 \\ 1 & 0 & 0 \\ 3 & 0 & 0 \end{array}\right).$

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If we identify $M_{2 \times 3}$ with $\Bbb{R}^6$ and similarly with the other vector space, we see that $L$ is a linear transformation from $\Bbb{R}^6$ to $\Bbb{R}^9$. Now I claim that the kernel of $L$ is trivial. Indeed, suppose there is a matrix

$$A = \left[\begin{array}{ccc} a & b & c & \\ d & e & f \end{array}\right]$$

such that $L(A) = 0$. Then we get the following system of equations:

$$\begin{eqnarray*} 2a - d &=& 0 \\ 2b - e &=& 0 \\ 2c - f &=& 0 \\ a + 2d &=& 0 \\ b + 2e &=& 0 \\ c + 2f&=& 0 \\ 3a + d &=& 0\\ 3b + e &=& 0 \\ 3c + f &=& 0. \end{eqnarray*}$$

In other words we are trying to find the null space of the matrix

$$\left[\begin{array}{cccccc} 2 & 0 & 0 & -1 & 0 & 0 \\ 0& 2 & 0 & 0 & -1 & 0 \\ 0& 0& 2 & 0 & 0 & -1 \\ 1 & 0 & 0 & 2 & 0 & 0 \\ 0& 1 & 0 & 0 & 2 & 0 \\ 0& 0& 1 & 0 & 0 & 2\\ 3 & 0 & 0 & 1 & 0 & 0 \\ 0& 3 & 0 & 0 & 1 & 0 \\ 0& 0& 3 & 0 & 0 & 1 \end{array}\right].$$

Upon row reduction, this matrix in reduced row echelon form is

$$\left[\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0& 1 & 0 & 0 & 0 & 0 \\ 0& 0& 1 & 0 & 0 & \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0& 0 & 0 & 0 & 1 & 0 \\ 0& 0& 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right].$$

You can see that there are no free variables, so that the dimension of the kernel if zero. By rank nullity, we have

$$\begin{eqnarray*} \dim \textrm{ran} L &=& \dim \Bbb{R}^6 - \dim \textrm{ker} L \\ &=& 6 - 0 \\ &=& 0 \end{eqnarray*}$$

from which it follows that the range of $L$ has dimension 6.

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