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March 17th, 2017, 05:16 PM   #1
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Conceptualising a 2x2 matrix

Hi all

I am having trouble conceptualising what a 2x2 matrix implies. For example, can the matrix:

A = 2 4
3 7

be represented on a plane as two vectors? I'm thinking about this in the context of a vector being multiplied by this matrix repeatedly, and going to infinity.

Thanks for your help=)
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March 17th, 2017, 07:41 PM   #2
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Ooh what a good question. Imagine the unit square in the plane, defined by $\{(x, ~y) \in \mathbb R^2 : 0 \leq x, y \leq 1\}$.

Now think of the region of the plane defined by the vectors $(2,3)$ and $(4,7)$. In other words draw the arrow from the origin to $(2,3)$, and the vector from the origin to $(4,7)$, and shade in the paralellogram they define.

Now the matrix

$
A=
\left[ {\begin{array}{cc}
2 & 4 \ \\ 3 & 7 \ \end{array} } \right]
$

is the matrix of the linear transformation relative to the standard basis that maps the unit square to the shaded paralellogram.

Now depending on what class you're in, this may or may not make sense right now. But it's a great thing to know. And the determinant of the matrix? It's just the area of the parallelogram.

How can you see this? Multiply the matrix by the standard basis vectors

$
\left[
\begin{array}{c}
1\\
0\\
\end{array}
\right]
$

and

$
\left[
\begin{array}{c}
0\\
1\\
\end{array}
\right]
$

and you'll see how the matrix transforms the standard basis. And everything in between, ends up in between.
Thanks from romsek and ChristineJW

Last edited by Maschke; March 17th, 2017 at 07:47 PM.
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March 17th, 2017, 08:37 PM   #3
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Thanks Maschke! Also the this is the first time I've heard that the area of the parallelogram being equal to the determinant. Very useful!
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March 17th, 2017, 08:46 PM   #4
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Quote:
Originally Posted by ChristineJW View Post
Thanks Maschke! Also the this is the first time I've heard that the area of the parallelogram being equal to the determinant. Very useful!
You're very welcome. Also note that if the area of the parallelogram is zero, that is if the transformation collapses the square to a line, then the determinant is zero and the transformation's not invertible. The collapse of area and the collapse of dimension are two aspects of the same thing.
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