Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that any subspace of vector space $V$ is a null space over some linear transformation $V \rightarrow V$.

So far I have:
Let $W$ be the subspace of $V$, let $(e_1, e_2, \ldots, e_r)$ be the basis of $W$, where $r \leq \dim(V)$.
It seems I need to find a linear transformation $T: V \rightarrow V$, so that:.
$T(s) = 0$ if $s \in W$
$T(s) \neq 0$ if $s \notin W$.
So $T(x) = 0$, if $x$ is a linear combination of $(e_1, e_2, \ldots, e_r)$ and $T(x) \neq 0$ if it's not.
How do I construct the matrix of this linear transform?

share|cite|improve this question
I wouldn't say "the" basis: there are lots of possible choices! Now, do you know that you can extend this basis to a basis $\{e_1, \ldots, e_r, f_{r + 1}, \ldots, f_{\dim V}\}$ for $V$? – Dylan Moreland May 6 '12 at 1:37
How do you feel about projections? – mixedmath May 6 '12 at 3:17
up vote 1 down vote accepted

You are doing well: first find a basis $(e_1,\ldots,e_r)$ for $W$.

Then, remember that every linearly independent subset of $V$ can be extended to a basis. Since $(e_1,\ldots,e_r)$ is linearly independent, we can extend it to a basis: $(e_1,\ldots,e_r,f_1,\ldots,f_{n-r})$, where $n=\dim(V)$, $r=\dim(W)$, $r\leq n$.

Then remember that a linear transformation may be specified by saying what it does to a basis: proceeding as you did before, send each and every one of $e_1,\ldots,e_r$ to $0$.

How do we ensure nothing else is mapped to $0$? How about mapping each $f_i$ to itself?

Suppose $a_1e_1+\cdots+a_re_r + b_1f_1+\cdots+b_{n-r}f_{n-r}$ is mapped to $0$. That means that $$\begin{align*} 0 &= T(a_1e_1+\cdots+a_re_r+b_1f_1+\cdots+b_{n-r}f_{n-r})\\ &= a_1T(e_1)+\cdots+a_rT(e_r) + b_1T(f_1) + \cdots + b_{n-r}T(f_{n-r})\\ &= a_10 + \cdots +a_r0 + b_1f_1 + \cdots + b_{n-r}f_{n-r}\\ &= b_1f_1+\cdots+b_{n-r}f_{n-r}. \end{align*}$$ What can we conclude about $b_1,\ldots,b_{n-r}$?

Now, it is easy to find the matrix of $T$ with respect to the basis $(e_1,\ldots,e_r,f_{1},\ldots,f_{n-r})$. What is it? Express the image of each basis vector in terms of the basis vectors, and those are the columns of the matrix of $T$.

share|cite|improve this answer
For $0 = b_1f_1 + ... + b_{n-r}f_{n-r}$ to be true $b_1, ..., b_{n-r}$ have to be 0, since $f_1, ..., f_{n-r}$ are linearly independent. This means that $T(x) = 0$ only if $x \in W$. But before that I don't understand why $T(e_i) = 0$ $1<=i<=r$. Why does it have to be so? – user1376993 May 6 '12 at 13:51
Actually I got that part, I'm looking for a transformation where $T(e_i) = 0$ and $T(f_j) = f_j$ – user1376993 May 6 '12 at 15:00
Ok, so let $B$ be the change of basis matrix from basis $(e_1,\ldots,e_r,f_1,\ldots,f_{n-r})$ to standard basis. Let $A$ be the matrix for linear transformation $T$ with respect to standard basis. The matrix that I'm looking for is $C$, which is the matrix of linear transformation T with respect to basis $(e_1,\ldots,e_r,f_1,\ldots,f_{n-r})$. $C = B^{-1} * A$. Is this correct? – user1376993 May 6 '12 at 15:39
The matrix with respect to the basis $(e_1,\ldots,e_r,f_1,\ldots,f_{n-r})$ is very easy; it's a diagonal matrix that has $0$'s in the first $r$ positions (corresponding to the fact that the $e_i$ map to $0$) and $1$'s in the last $n-r$ positions (corresponding to the fact that $f_i$ map to themselves). If you want to find the matrix of this transformation with respect to the standard basis (which would depend on what $W$ actually is) you would use $B$, but the matrix would be $BCB^{-1}=A$, not what you wrote. – Arturo Magidin May 6 '12 at 19:06
Ok, got it, but I don't really need A for this proof, right? I can just say that given $[v]_B \in V$ the $T([v]_B) = 0$ only if $[v]_B \in W$? – user1376993 May 6 '12 at 19:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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