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Suppose A is a 3 × 5 matrix, and that $dim\ Nul\ A = 2$. Prove the following;

(a) There exists a basis for $R^3$ made up of some of the columns of $A$.

(b) The rows of $A$ form a linearly independent set.

Part A;

$5 = Dim\ Nul(A) + Rk(A), Rk(A) = 3$ (rank nullity theorem)

So with rank of 3, $rref(A)$ contains 3 pivot columns, and the corresponding columns of $A$ (also the column space of A) form a basis of $R^3$.

Part B;

If the rows are linearly independent, then $Ax = 0$ IFF $x = 0$.

However since $Dim\ Nul(A) = 2$, there must be two $x \ne 0$ such that $Ax=0$. (contradiction?)

Am I doing this properly? Both answers seem intuitive, and I can't tell if I've proven them sufficiently (if at all). Is there anything I can or should add to improve these proofs?

1 Answers 1

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Part A is alright, although there is no need to mention the reduced row echelon form of $A$. The rank of $A$ is the dimension of the column space of $A$ (the span of the columns). You have shown that $A$ has rank $3$ so the span of the five columns of $A$ is a three dimensional subspace of $\mathbb{R}^3$ which means this is the whole of $\mathbb{R}^3$. You can always throw away linearly depend elements from a spanning set and get a basis so you can find a basis of $\mathbb{R}^3$ that consists of three columns of $A$. In practice, in order to find it you can indeed row reduce and look at the pivots.

Part B lacks some explanation. What is (more or less) obvious is that if the columns of $A$ are linearly independent then $Ax = 0$ iff $x = 0$. The reason is that $Ax$ is a linear combination $C_1 x_1 + \dots + C_5 x_5$ of the columns of $A$ and not of the rows so if the columns are linearly independent, this combination is zero iff the coefficients $x_1 = \dots = x_5$ are zero. However, the column rank of a matrix $A$ (the dimension of the column space) is the same as the row rank of $A$ (the dimension of the row space) so in your case, the span of the three rows is three dimensional and hence they form a linearly independent set.