Let $V$ be a subspace of $\mathbb R^{13}$ of dimension $6$ and $W$ be a subspace of $\Bbb R^{31}$ of dimension $29$. What's the dimension of the space of all linear maps from $\Bbb R^{13}$ to $\mathbb R^{31}$ whose kernel contains $V$ and whose image is contained in $W$.

Here, $ \dim(V)=6$ and $\dim(W)=29$. Also , $V\subset \ker(T)$ and $\text{Im}(T) \subset W$.

So, $\dim(\text{Im}(T))\le 29$ and $\dim(\ker(T))\ge 6$.

Now, $\dim(\ker(T))+\dim(\text{Im}(T))=13=\begin{cases}6+7\\7+6\\8+5\\9+4\\10+3\\11+2\\12+1\end{cases}$

So there are $7$ cases. So dimension of the space of all linear maps from $\Bbb R^{13}$ to $\mathbb R^{31}$ is $7$.

Is it correct or there are something wrong?


3 Answers 3


A better way of looking at it: think of the vector spaces as groups, and try to understand how the isomorphism theorems would work.

Any map from $\mathbb R^{13} \to \mathbb R^{31}$ with $V$ contained in the kernel, is a map from $\dfrac {\mathbb R^{13}}V \to \mathbb R^{31}$.

Similarly, if the image is contained in $W$, then this is just a map from $\dfrac {\mathbb R^{13}}V \to W$.

Now, since $\dfrac{R^{13}}{V} \cong \mathbb R^7$, we are basically looking at maps from $\mathbb R^7$ to $\mathbb R^{29}$, without any restrictions.

Now, we know this set is isomorphic to the set of all matrices with dimension $29 \times 7$. The basis for such a set of matrices is just the following: an empty matrix except for one non-zero entry, varying over all possible entries.

However, there are $29\times 7 = 203$ entries, hence it follows that the dimension of the set of all linear maps with the given constraints is actually $203$.


To make things clear (correct answers having been already provided) and in order to understand what is needed, it's easier to think in terms of the corresponding matrices.

Let's construct the matrix of a linear map $f:\mathbb{R}^{13} \rightarrow \mathbb{R}^{31}$ that fits your description.

Let $(u_{1},..., u_{13})$ be the canonical base of $\mathbb{R}^{13}$ and consider without loss of generality that $V = span(u_{1},...,u_{6})$.

Similarly, let $(v_{1},..., v_{31})$ be the canonical base of $\mathbb{R}^{31}$ and consider that $W = span(v_{1},...,v_{29})$.

The matrix representation of $f$ has 13 columns and 31 rows, the $i^{th}$ column corresponding to the image of $u_{i}$ i.e., and sorry for stating the obvious, if $f(u_{1}) = \sum_{j = 1}^{31}a_{j}.v_{j}$ then the first column will be

$$\begin{pmatrix} a_{1}\\ .\\ .\\ .\\ a_{31} \end{pmatrix}$$

Now you need $V \subseteq Ker(f)$ which means that the first 6 columns (corresponding to the image of $(u_{1},...,u_{6})$) must be zero. You also need $Im(f) \subseteq W$ which means that the last two rows must be zero. The Matrix $A$ looks like this:

$$ A= \left[ \begin{array}{c|c} 0 & B \\ \hline 0 & 0 \end{array} \right] $$

Where the bottom zeros are 2 rows of zeros and $B$ has ($29$ rows $\times$ $7$ colmuns).

The dimension of the space of the corresponding linear maps is simply the size of $B$ which is $203$ (a possible base is formed by matrices with only one element of $B$ being equal to $1$).


No, it is wrong. For example, I can come up with $29$ independent maps: suppose that $w_1, w_2, \ldots w_{29}$ span $W$, and $v_1, \ldots v_6$ span $v$ which extends by $v_7, \ldots v_{13}$ to a basis of $\mathbb{R}^{13}$. Then we take $v_1\rightarrow0$, $v_2\rightarrow0, \ldots v_{12}\rightarrow 0$, and $v_{13}\rightarrow w_i$ for $i=1, 2, \ldots 29$.

Can you extend this to give $7\cdot 29$ independent linear maps? Is this all of them?


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