Rank of a matrix when adding new columns Now, let $X$ be an $m\times n $matrix, with a rank of $r$.
Suppose that another matrix, let's call it $Z$, such that $Z=(X\ \ \ \ Y)$ is a matrix of size $m\times k$, where of course, that $n<k$.  The first $n$ columns are the same as the columns of $X$. 
How can I prove that rank $Z\geq$ rank $X$?
Also, can someone provide me some examples with rank $Z=$ rank $X$?  And perhaps also examples with rank $Z>$ rank $X$?
Thanks for the assistance!  I'm a self-learner on the textbook by Artin, some clarification would be greatly appreciated!
 A: Artin defines the rank of a matrix $A$ as the dimension of the image with $A$ acting by left multiplication. I leave it to you to check that this definition is consistent with the definition of the rank as the dimension of the columnspace of $A$, i.e. the vector space $\mathrm{col}(A)$ spanned by the column vectors of $A$.
Notice that since the columns of $X$ forms a subset of the columns of $Z$, it follows that the vector space spanned by the columns of $X$ is a subspace of the vector space spanned by the columns of $Z$. In other words, $\mathrm{col}(X)$ is a subspace of $\mathrm{col}(Z)$. And what do we know about the dimensions of a subspace in relation to its containing vector space?
With this interpretation, it's straightforward to deduce when $\mathrm{rank}(Z) > \mathrm{rank}(X)$. This happens if and only if $\mathrm{col}(X)$ is a proper subspace of $\mathrm{col}(Z)$. Analagously, we have $\mathrm{rank}(Z)=\mathrm{rank}(X)$ if and only if we have $\mathrm{col}(Z) = \mathrm{col}(X)$. I leave it to you to figure out the exact conditions on the columns of $Y$ for these cases to happen.
A: You are asking to prove that the rank of a matrix does not decrease by adding extra columns. Since the rank is by definition the maximum number of linearly independent columns one can select, this is obvious. Since the rank can never exceed the number of rows, it is also clear that adding columns will not always increase the rank; it some point it just must remain the same (and it can remain unchanged even when it could still increase; just add a column that is linearly dependent on the previous columns). If it is not obvious to you that adding columns sometimes does increase the rank (this seems evident to me, as otherwise the rank would remain stuck at$~0$), consider adding the columns of the identity matrix one by one.
A: Not a full answer, merely hints.
Think about column rank, then $\text{rk} Z \ge \text{rk} X$ should be clear. (Linear independent columns of $X$ will still be independent in $Z$!)
Examples for equal and unequal rank:
$$X := \begin{pmatrix} 1 \\ 0 \end{pmatrix}, ~~~ Y_1 := X, ~~~
Y_2 := \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$$
Then $( X, Y_1 )$ has the same rank as $X$ while $(X,Y_2)$ has higher rank.
