In $\mathbb{R}^5$ there is given vector space $V$. Its dimension is 3. In $\mathbb{R}^{6,5}$ consider the subset $X = \{A \in \mathbb{R}^{6,5} : V \subset \ker A\}$. I have to show that $X$ is a vector space in $\mathbb{R}^{6,5}$ and find its dimension. To show that $X$ is vector space consider $x_1, x_2 \in X$ and $v \in V$. We know that $x_1 v = 0$ and $x_2 v = 0$ so $(\alpha x_1 + \beta x_2) v = \alpha (x_1 v) + \beta (x_2 v) = 0$ so $V \subset \ker (\alpha x_1 + \beta x_2)$. But I don't know how to find $X$'s dimension. Any ideas?

  • $\begingroup$ What's $ker A$? I know what the kernel of a linear transformation is but not of a set... $\endgroup$ Dec 3, 2014 at 19:57
  • $\begingroup$ $A$ is a matrix so it's a linear transformation $\endgroup$
    – alex
    Dec 3, 2014 at 19:59
  • $\begingroup$ so $V$ is a vector space with elements from $R^5$ over what field? $\endgroup$ Dec 3, 2014 at 20:00
  • $\begingroup$ over $\mathbb{R}$ $\endgroup$
    – alex
    Dec 3, 2014 at 20:04
  • 1
    $\begingroup$ Wants to refer to 6 by 5 matrices... $\endgroup$ Dec 3, 2014 at 20:12

3 Answers 3


I understand $V \subseteq \mathbb{R}^5$ is a subspace, $\dim V = 3$

$X = \{A \in \mathbb{R}^{6 \times 5} : V \subseteq \ker A \}$

To show that $X$ is a vector space, it suffices to show it is a subspace of $\mathbb{R}^{6 \times 5}$.

  1. $0 \in X$, clearly because $V \subseteq \mathbb{R}^5 = \ker 0$
  2. For $\alpha_i \in \mathbb{R}$ and $A_i \in X$, if $v \in V$, $A_i v = 0$, and so $ (\sum_i \alpha_i A_i) v = 0$.

So it is a subspace.

Let $B = \{v_1, v_2, v_3 \}$ be a basis of $V$, and extend it to a basis $\{v_1, v_2, v_3, v_4, v_5 \}$ of $\mathbb{R}^5$. As $Av_i = 0$ for $1 \leq i \leq 3$, you only have to say where goes $A v_i$ for $i=4$ and $5$.

So you have 5-3=2 degrees of freedom in the domain and 6 in the codomain, that gives $2 \cdot 6 = 12$. I suspect the dimension is 12. Hope that helps.

I add this: you can think of $A$ as the matrix representation of some linear transformation $f: \mathbb{R}^5 \to \mathbb{R}^6$ with respect to the bases the extended version of $B$ above for $\mathbb{R}^5$ and and the standard basis for $\mathbb{R}^6$, so $f(v_i) = 0$ for $1 \leq i \leq 3$, and you can decide where goes $f(v_i)$ for $i=4,5$.

  • $\begingroup$ For OP: I've made some edits to make it clear what dami wants to say, even with these edits a complete proof should have more detail in it. $\endgroup$ Dec 3, 2014 at 21:02
  • $\begingroup$ What does it mean "you only have to say where goes $A v_i$ for $i = 4$ and $5$" and "you can decide where goes $f(v_i)$ for $i = 4,5$"? $\endgroup$
    – alex
    Dec 3, 2014 at 21:27
  • $\begingroup$ Every linear transformation is uniquely determined by setting $f(v_{i})$(for every $v_{i}$ of a basis in the domain) to a vector in the codomain. $\endgroup$ Dec 3, 2014 at 21:52

Let's take a quick look. I'll try a sketch/give a hint. It seems that so far, so good. Indeed, take $A,B \in X$, $\lambda \in \Bbb R$. To show that $A+\lambda B \in X$, we have to show that $V \subset \ker(A+\lambda B)$, assuming that $V \subset \ker A \ \cap \ker B $. But that's true, since given ${\bf v} \in V$ we have $$(A+\lambda B){\bf v} = A{\bf v}+\lambda B{\bf v} = 0+0 = 0.$$

Now, we have $A \in \Bbb R^{6,5}$, that is, $A: \Bbb R^5 \to \Bbb R^6$, and we know that: $$5 = \dim \ker A + \dim {\rm Im} \ A.$$ Since $\dim V = 3$ and $V \subset \ker A$, we have that $3 = \dim V \leq \dim \ker A.$ This way we can have three types of $A$: $$\begin{cases} \dim \ker A = 3, & \dim {\rm Im} \ A = 2 \\ \dim \ker A = 4, & \dim {\rm Im} \ A = 1 \\ \dim \ker A = 5, & \dim {\rm Im} \ A = 0\end{cases}.$$

Can you go on?

  • $\begingroup$ I got this as well, but there's still some to go till finding X's dimension, i.e. there are many matrices with $Im A=2$ and so on. $\endgroup$ Dec 3, 2014 at 20:37

For the dimension of $X$, if $A\in X$, $A$ has to nullify $V$ and can do anything on the $2$-dimensional orthogonal complement of $V$. So the dimension of $X$ is the same as the dimension of the space of linear transformations from a $2$-dimensional space to a $6$-dimensional space, which is $12$.

EDIT: A commenter is unfamiliar with the direct bijection between $6\times 5$ matrices and linear transformations from $\mathbb{R}^5$ to $\mathbb{R}^6$; we can reinvent like four or five wheels and say the same thing but longer:

Let $V$ have basis $\{v_1,v_2,v_3\}$. Use Gram Schmidt in $\mathbb{R}^5$ so that $\{v_4,v_5\}$ is a basis for $V^{\perp}$. Let $\{e_1,e_2,e_3,e_4,e_5,e_6\}$ be the standard basis for $\mathbb{R}^6$.

Consider the matrices $A_{ij}$ defined as the matrix that takes $v_j$ to $e_i$ and all other $v_k$ to the zero vector in $\mathbb{R}^6$. Explicitly, $A_{ij}=C_{ij}P$, where $C_{ij}$ is a $6\times5$ matrix with all zeros except a $1$ in the $i,j$ position, and $P$ is the change of basis matrix from $\{v_1,v_2,v_3,v_4,v_5\}$ to the standard basis for $\mathbb{R}^5$. Explicitly, $P=\begin{bmatrix}|&|&|&|&|\\v_1&v_2&v_3&v_4&v_5\\|&|&|&|&|\end{bmatrix}^{-1}$. Check: $A_{ij}v_j=C_{ij}Pv_j=C_{ij}\epsilon_j=e_i$ where $\epsilon_j$ are the standard basis vectors in $\mathbb{R}^5$.

There are $30$ of these $A_{ij}$, which are matrices in $M_{6\times 5}$. They are linearly independent because if $\sum_{i,j} c_{ij}A_{ij}=[0]$, then multiplying against each of the $v_j$ one at a time gives $\sum_i c_{ij}e_i=[0]$, and since the $e_i$ are independent, the $c_{ij}$ must equal $0$.

But 18 of them are not in $X$, because they have one of $Av_1,Av_2,Av_3$ nonzero. Only the twelve matrices $A_{i4}$ and $A_{i5}$, which all have $Av_1=Av_2=Av_3=0$ are in $X$.

$M_{6\times5}$ is $30$-dimensional. We have established 12 linearly independent vectors that are in the space $X$. So $\dim V\geq12$. We have established 18 more linearly independent vectors that are not in $X$. So $\dim V\leq30-18-12$. So $\dim V=12$.

  • $\begingroup$ Why is that in second sentence? $\endgroup$
    – alex
    Dec 3, 2014 at 21:46
  • $\begingroup$ $\mathbb{R}^5=V\oplus V^{\perp}$. Any $A$ in $\mathbb{R}^{5,6}$ is uniquely defined by how it acts on $V$ and on $V^{\perp}$. By definition of $X$, if $A\in X$, $A$ acts trivially on $V$ but is free to be any linear transformation when restricted to $V^{\perp}$. $V^{\perp}$ has dimension two, so pick two basis vectors for it. For each of them, for each of six basis vectors for $\mathbb{R}^6$, there is a transformation that sends one of the two to one of the six, and the other one of the two to $0$. These form a basis for $X$ consisting of $12$ transformations. $\endgroup$
    – 2'5 9'2
    Dec 3, 2014 at 21:59
  • $\begingroup$ @alex.jordan The thing is, you are mentioning "A has to nulify"... you probably mean the linear transformation called left multiplication by A has to nulify V. If you'd try to give a detail proof you'd see the argument does not work... at least I could not do it. How do you pick your linear transformation?(the one that comes from A). $\endgroup$ Dec 4, 2014 at 0:27
  • $\begingroup$ @shooting-squirrel If $A$ is in $X$, then by the definition of $X$, $A$ nullifies every vector in $V$. That's what it means to say $V\subset \ker A$. There's no need for any more detail than that. $\endgroup$
    – 2'5 9'2
    Dec 4, 2014 at 0:30
  • $\begingroup$ Dude, A is a matrix not a linear transformation. $\endgroup$ Dec 4, 2014 at 0:31

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