I had two true/false questions on my exam of which I missed.

$1)$ The map $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $T(x)=x+e_1$ is a linear transformation.

I know this to be false, because I got it wrong and put true. Linear transformations are something I was extremely fuzzy on going into this exam. However, with this problem I believe I simply overlooked the $x$ because I have never seen another problem setup like this. Is the "x" what makes this false?

$2)$ Suppose $A$ and $B$ are matrices such that the product $AB$ is defined. Then the columns of $AB$ are linear combinations of the columns of $B$.

I know this one to be false as well, for the same reason. What I know is that If $A$ is an $m \times n$ matrix with columns $a_1,\ldots,a_n$, and if $x$ is in $\mathbb{R}^n$, then the product of $A$ and $x$, denoted by $Ax$, is the linear combinations of the columns of $A$ using the corresponding entries in $x$ as weights.

Can someone further explain reasons for these being false.



It is the $e_1$ which is the problem. If $T$ was to be a linear transformation, then it would have to satisfy $T(0)=0$. However, $T(0)=e_1\neq{0}$.


The columns of $AB$ will be linear combinations of the columns of $A$, by the fact that you mention. Specifically, if $b_1,\ldots,b_n$ are the columns of $B$, then the columns of $AB$ will be $Ab_1, \ldots, Ab_n$, and all of these are linear combinations of the columns of $A$ (with the $b_{i}'s$ as weights).


I'll just address your first question. Given a vector space $V$ and a vector space $W$, we say that a function $T:V\to W$ is linear when $$T(av_1+bv_2)=aT(v_1)+bT(v_2)$$ for all $v_1,v_2\in V$ and all scalars $a,b$ (in you case, most likely taken to be real or complex numbers). In other words, $T$ is a linear map when it "respects" the operation of adding two vectors together, as well as the operation of scaling a vector by a number.

Thus, if the map in your question were linear, we'd have to have (for example) $$(2,2)=2\cdot (1,1)=2\cdot T(0,1)=T(2\cdot (0,1))=T((0,2))=(1,2),$$ or using the letter-names for the vectors, $$2e_1+2e_2=2\cdot(e_1+e_2)=2\cdot T(e_2)=T(2e_2)=e_1+2e_2.$$ But this is wrong, so we conclude that the map $T$ is not linear.

  • $\begingroup$ +1: although I prefer my very in-detail analysis, nice outline to actually explain. :) $\endgroup$ – gnometorule Mar 4 '13 at 2:03

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