0
$\begingroup$

May I ask you to explain to me the meaning of columns of the matrix in linear algebra please? How can I use it?

Also, in my notes (attached below) it says: "Columns of matrix tell us how the corresponding linear function acts on the basic vectors {e1,...,en} and, quite significantly, these completely determine the matrix."

But columns of the matrix does not represent any linear function. It represents the coefficients of the n-th variable. Isn't it?

Screenshot of the lecture notes

Thank you.

$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

Let $\vec v_1,\dots,\vec v_n$ denote the columns of $A$. Then if $\vec x = (x_1,\dots,x_n)$, we have $$ A\vec x = x_1 \vec v_1 + x_2 \vec v_2 + \cdots + x_n \vec v_n $$ In other words: a vector $x \in \Bbb R^n$ acts on the matrix $A$ by producing a linear combination of its columns.

What you have in mind is the row picture. If $\vec r_1, \dots \vec r_m$ denote the rows of $A$, then we have $$ A \vec x = \pmatrix{\vec r_1 \cdot \vec x\\ \vdots \\ \vec r_m \cdot \vec x} $$ Where $\cdot$ denotes the dot product.

It may be hard to see, but we have simply said the same thing in two different ways.

$\endgroup$
2
  • $\begingroup$ Thank you. I understand linear combination of columns. That makes sense. But if we take v1 as a single column of the matrix A. How can v1 tell me how the corresponding linear function acts on the basis vector? Indeed, if I had a row r1 that acts on the vector, that makes sense (you presented that in the second example). But in this case it talks about the columns. $\endgroup$
    – Li Cooper
    Mar 15, 2016 at 18:23
  • $\begingroup$ The basis vector for us is just $x = (1,0,\dots,0)$. Plugging into the formula, it's clear that $A(1,0,\dots,0) = \vec v_1$ $\endgroup$
    – Ben Grossmann
    Mar 15, 2016 at 18:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .