# Linear Algebra (linear transformations)

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.

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## 2 Answers

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 $v$.

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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.

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Why $Im(T)$ is $\mathbb{R}$ in your solution ? – Belgi Dec 10 '12 at 8:53
Thank you, that makes a lot more sense. :) – user4939 Dec 10 '12 at 8:56
Hmm, I just thought about it. Why would T be represented by the identity matrix under the bases? – user4939 Dec 10 '12 at 9:01
@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. – Gyu Eun Lee Dec 10 '12 at 20:00
@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. – Gyu Eun Lee Dec 10 '12 at 20:05