Square Root of a Matrix

  • Thread starter SometimesY
  • Start date
  • #1

Main Question or Discussion Point

Hey all,

I've run into a snag with research and it is partly related to taking the square root of a matrix. I've seen related stuff concerning matrices that can be diagonalized, however the 2x2 matrix in question is not diagonalizable. Can the square root of such a matrix be determined? If so, how?

Thanks!
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
SometimesY: Boy, I would like to see a 2x2 matrix that cannot be diagonalized!
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,770
911
Here's one:
[tex]\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}[/tex]

Happy?

If a 2 by 2 matrix cannot be diagonalized, it must have a single eigenvalue and it can be put into "Jordan Normal Form",
[tex]\begin{bmatrix}x & 1 \\ 0 & x\end{bmatrix}[/tex]
where "x" is that single eigenvalue.

Now, if
[tex]\begin{bmatrix}a & b \\ c & d \end{bmatrix}[/tex]
is a "square root" for that matrix, we must have
[tex]\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix}= \begin{bmatrix}a^2+ bc & ab+ bd \\ ac+ dc & bc+ d^2\end{bmatrix}= \begin{bmatrix}x & 1 \\ 0 & x\end{bmatrix}[/tex]

That is, we must have [itex]a^2+ bc= x[/itex], [itex]ab+ bd= 1[/itex], [itex]ac+ dc= 0[/itex], and [itex]bc+ d^2= x[/itex].

From [itex]ac+ dc= c(a+ d)= 0[/itex] we must have either c=0 or a+ d= 0. But a+ d= 0 cannot satisfy [itex]ab+ bd= b(a+ d)= 1[/itex] so we must have c= 0.

Then [itex]a^2+ bc= a^2= x[/itex] and [itex]bc+ d^2= d^2= x[/itex].

Since [itex]a^2= d^2= x[/itex] we must have either a= d or a= -d. If a= -d then ab+ db= b(a+d)= 0 so that ab+ db= 1 is impossible. If a= d, then ab+ db= 2ab= 1 so that b= 1/(2a). That is, the matrix
[tex]\begin{bmatrix}x & 1 \\ 0 & x\end{bmatrix}[/tex]
has two square roots:
[tex]\begin{bmatrix}\sqrt{x} & \frac{1}{2\sqrt{x}} \\ 0 & \sqrt{x}\end{bmatrix}[/tex]
and
[tex]\begin{bmatrix}-\sqrt{x} & -\frac{1}{2\sqrt{x}} \\ 0 & -\sqrt{x}\end{bmatrix}[/tex]
 
  • #4
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
Halls of Ivy: Nice try. But your example is a matrix which is already diagonalized. I understand a diagonal matrix to be one where the main diagonal has all non-zero entries and only zero entries above or below the main diagonal.
 
  • #5
AlephZero
Science Advisor
Homework Helper
6,993
291
I understand a diagonal matrix to be one where the main diagonal has all non-zero entries and only zero entries above or below the main diagonal.
Sorry, but that is not the definition that everybody else uses.

The thing you described is a upper or lower triangular matrix. Upper and lower say which side of the diagonal has non-zero terms.
 
  • #6
Thanks HallsofIvy! That's exactly what I was looking for. It means a lot!
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,770
911
Halls of Ivy: Nice try. But your example is a matrix which is already diagonalized. I understand a diagonal matrix to be one where the main diagonal has all non-zero entries and only zero entries above or below the main diagonal.
Those are "upper triangular" and "lower triangular" matrices.
http://en.wikipedia.org/wiki/Triangular_matrix

"Diagonal" matrices, oddly enough, are those that are non-zero only on the main diagonal.
http://en.wikipedia.org/wiki/Diagonal_matrix

What text book or other place defines a "diagonal matrix" the way you say?
 
  • #8
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
I stand corrected.
 

Related Threads for: Square Root of a Matrix

  • Last Post
Replies
10
Views
918
Replies
2
Views
2K
  • Last Post
Replies
4
Views
12K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
8
Views
36K
  • Last Post
2
Replies
27
Views
7K
Top