0
$\begingroup$

13. Suppose $V$ and $W$ are finite-dimensional vector spaces and $T:V \to W$ is an isomorphism. Then there exist bases $\mathcal{B}$ and $\mathcal{C}$, for $V$ and $W$ respectively, such that $[T]_{\mathcal{C},\mathcal{B}}$ is the identity matrix.

14. Let $T:V\to\mathbb{R}$ be a linear transformation. Suppose $\{v_1,\dots,v_n\}$ is a basis for $\ker(T)$. Suppose also that $v \in V$, $v \ne 0$, is not in $\ker(T)$. Prove $\{v,v_1,\dots,v_n\}$ is a basis for $V$.

15. Show that any linear transformation $T:V \to W$ may be written as a sum of linear transformations $T = T_1 + \cdots + T_k$ for some $k$, where each $T_i$ is a linear transformation of rank $1$.

Hey guys, I have a couple of questions I need help with. It'd be great if I could get any sort of help/hints! Thanks.

  • 0
    [MathJax](http://docs.mathjax.org/en/latest/index.html) is used on this site to render [LaTeX](http://meta.math.stackexchange.com/questions/5020/tex-latex-mathjax-basic-tutorial-and-quick-reference) to display math equations and expressions. Please copy the problems to this site using MathJax. If you need help, drop by the [chat room](http://chat.stackexchange.com/rooms/36/mathematics) and ask for help.2012-12-10

2 Answers 2

1

For 14 recall that $\mathbb{R}$ is a vector space over itself with dimension $1$ . what is $dim(Im(T))$ ?

For 15 recall there is a matrix $A$ s.t $Tv=Av$ for all $v$.

1
  1. Vector space isomorphisms are invertible, and all invertible linear transformations are change-of-basis transformations for appropriate bases. Under these bases, $T$ would be represented by the identity matrix.

  2. By the rank-nullity theorem, rank $T$ + nullity $T$ = dim $V$. Since Im$(T)$ is $R$, rank $T=1$. Clearly the vectors $\{v,v_1,\ldots,v_n\}$ are independent, and the rank-nullity theorem tells us $n+1=\dim V$, so this is a basis.

  3. Write $T$ as a matrix, then write this matrix as a sum of matrices with only one nonzero column. A matrix with only one nonzero column has one pivot column, so it defines a linear transformation of rank 1.

  • 0
    @user4939 Any change of basis transformation has a matrix whose columns are the original basis in the new basis. Within a basis, the matrix whose columns are the basis vectors is the identity in the basis. An alternative way to do this is to apply row operations until T is the identity matrix, then look at the matrix of the composition of those operations. Since T is an isomorphism you are guaranteed that you can do this.2012-12-10