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.

  • $\begingroup$ MathJax is used on this site to render LaTeX to display math equations and expressions. Please copy the problems to this site using MathJax. If you need help, drop by the chat room and ask for help. $\endgroup$
    – robjohn
    Dec 10, 2012 at 9:40

2 Answers 2


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. 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.

  • $\begingroup$ Why $Im(T)$ is $\mathbb{R}$ in your solution ? $\endgroup$
    – Belgi
    Dec 10, 2012 at 8:53
  • $\begingroup$ Thank you, that makes a lot more sense. :) $\endgroup$
    – user4939
    Dec 10, 2012 at 8:56
  • $\begingroup$ Hmm, I just thought about it. Why would T be represented by the identity matrix under the bases? $\endgroup$
    – user4939
    Dec 10, 2012 at 9:01
  • $\begingroup$ @Belgi Im(T)=R tells us that rank T = dim(Im(T))=1, which along with the assumptions tells us the dimension of V. Then since we have a set of independent vectors with the same number of vectors as the dimension, the set in question is in fact a basis. $\endgroup$ Dec 10, 2012 at 20:00
  • $\begingroup$ @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. $\endgroup$ Dec 10, 2012 at 20:05

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