2
$\begingroup$

Suppose that $V$ is a vector space and $x_1,x_2,\dots,x_n$ is a basis for $V$ and $T:V\rightarrow V$ is a linear transformation such that

$$T(x_1)=x_2\;,\; T(x_2)=x_3\;,\;\dots\;,\;T(x_{n-1)}=x_n\;,\;T(x_n)=0$$
Then find $n(T)$.

I have solved this for a special case assuming that the basis is the standard basis of $V$ but don't know how to solve it in the general case where the basis of $V$ is arbitrary. So suppose the basis is:

$$x_1=\begin{pmatrix}1\\0\\ \vdots\\0\\0\end{pmatrix}\qquad x_2=\begin{pmatrix}0\\1\\ \vdots\\0\\0\end{pmatrix}\qquad\dots\qquad x_{n-1}=\begin{pmatrix}0\\0\\ \vdots \\1\\0\end{pmatrix}\qquad x_n=\begin{pmatrix}0\\0\\ \vdots\\0\\1\end{pmatrix}$$

So if we write the transfotmation matrix $T_{n\times n}$, it will be of the form:

$$T=\begin{pmatrix}0&0&0&\cdots&0&0&0\\ 1&0&0&\cdots&0&0&0\\ 0&1&0&\cdots&0&0&0\\ \vdots&\vdots&\ddots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\cdots&0&0&0\\ 0&0&0&\cdots&1&0&0\\ 0&0&0&\cdots&0&1&0\end{pmatrix}$$
You see the $1^{th}$ row and $n^{th}$ column of $T$ is zero and the remaining submatrix is a $(n-1)\times (n-1)$ identity matrix with $determinant = 1$
From the definition we now that the $rank$ of a matrix is the size of the bigest submatrix with nonzero determinant so $r(T)=n-1$ we know that

$$r(T)+n(T)=number\, of\, columns$$ so $$n(T)=1$$
Could you please help me solve the problem in the general case where the basis $x_1,x_2,\dots x_n$ is chosen arbitrarily and is not essentially the standard basis of the vector space $V$ ?

$\endgroup$
4
$\begingroup$

When does $T$ map to $0$?

Consider $a = a_1x_1 + a_2x_2 + ... + a_nx_n$ such that $T(a) = 0 = 0x_1 + ... + 0x_n$. Then, \begin{equation} T(a_1x_1) = a_1T(x_1) = a_1x_2 \\ T(a_2x_2) = a_2T(x_2) = a_2x_3 \\ ... \\ T(a_{n-1}x_{n-1}) = a_{n-1}T(x_{n-1}) = a_{n-1}x_n \\ T(a_nx_n) = a_nT(x_n) = 0 \end{equation} Equation coefficients we have $a_1 = ... = a_{n-1} = 0$ and $a_n$ is arbitrary. Thus, the null space of T is the span of $\{x_n\}$.

$\endgroup$
4
$\begingroup$

What you did is not really a special case. With $x_1,...,x_n$ as basis, the standard matrix can be obtained by writing $T(x_i)$ as columns. It is exactly the matrix you have. So the nullity is $1$.

$\endgroup$
3
$\begingroup$

If you just want to compute the nullity, then you can notice that the range is generated by $x_2,x_3,\dots,x_n$, because

  1. a spanning set for the range is always $\{T(v_1),T(v_2),\dots,T(v_n)\}$ for any basis $\{v_1,\dots,v_n\}$ of the domain

  2. removing the zero vector from a spanning set leaves a spanning set

So $\{x_2,\dots,x_n\}$ is a basis of the range (because it's a linearly independent spanning set) and the rank-nullity theorem says $$ \operatorname{null}(T)=n-\operatorname{rank}(T)=n-(n-1)=1 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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