# Linear Algebra and Complex Numbers

1. Apr 6, 2007

### SNOOTCHIEBOOCHEE

1. Complex analysis is the study of number z= x+iy where i^2=-1. can you find a way to represent complex numbers as 2x2 matrices

i honestly have no clue where to start with this one. we are one week through my linear algebra course.

the only possible thing i can thing of is det (x -yi
1 1) but that seems really wrong

Last edited: Apr 6, 2007
2. Apr 6, 2007

### Hootenanny

Staff Emeritus
Okay, how do we represent the number one in matrix form?

3. Apr 6, 2007

Google gave me some pretty fruitful results on this one.

4. Apr 6, 2007

### SNOOTCHIEBOOCHEE

the number 1 in matrix form is just [1]

5. Apr 6, 2007

### HallsofIvy

Staff Emeritus
No, we're talking about 2 by 2 matrices. 1 is the "multiplicative identity" for the real numbers. What is the multiplicative identity for 2 by 2 matrices?

Now think about i. Where would you put the 1's so multiplying the matrix by itself will give you the negative of the identity matrix?

Last edited: Apr 6, 2007
6. Apr 6, 2007

### SNOOTCHIEBOOCHEE

diagonol 1's i dunno how to use LAtex but its like [1 0;0 1] on matlab prolly the transpose of that for the second question your asking

7. Apr 6, 2007

### Hootenanny

Staff Emeritus
I meant in 2x2 form. I'll start for you, we can write the real component as;

$$\Re = x\cdot\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right]$$

In otherwords, the 2x2 identity matrix. Now, for the imaginary part we want a matrix which represent an anti-clockwise rotation by $\pi/2$ about the origin. Can you think of a matrix that does this?

EDIT: Halls strikes again.

8. Apr 6, 2007

### SNOOTCHIEBOOCHEE

[0 1
1 0] ??

i think that would do it

9. Apr 6, 2007

### Hootenanny

Staff Emeritus
That's very close but not quite. Note that
$$\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right]\times\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right] = \left[ \begin{array}{cc} 1 & 0 \\ 0& 1\end{array}\right] = I_2$$
You want;
$$\left[ \begin{array}{cc} a & b \\ c & d\end{array}\right]\times\left[ \begin{array}{cc} a & b \\ c & d\end{array}\right] = -\left[ \begin{array}{cc} 1 & 0 \\ 0& 1\end{array}\right]$$

10. Apr 6, 2007

### SNOOTCHIEBOOCHEE

[0 -1
1 0]

ok got it i think

11. Apr 6, 2007

### Hootenanny

Staff Emeritus
Looks good to me (Nice name by-the-way )

12. Apr 6, 2007

### SNOOTCHIEBOOCHEE

x [1 0;0 1] + y[0 -1; 1 0] ???????

thans jay and silent bob are fantanstci

13. Apr 6, 2007

### Hootenanny

Staff Emeritus
Yep, your correct. However, you can represent it as a single matrix thus;

$$\left[ \begin{array}{cc} x & -y \\ y & x\end{array}\right]$$

Both forms are correct.

14. Apr 6, 2007

### SNOOTCHIEBOOCHEE

You guys are absolute geniuses and i owe you my life. ty

15. Apr 6, 2007

### HallsofIvy

Staff Emeritus
Okay, I'll send you my bill!