# Find the square root of a 3x3 matrix

• Dec 6th 2009, 12:52 AM
Gchan
Find the square root of a 3x3 matrix
I apologize if my notation isn't clear, newbie to this forum

I'm trying to find out how to find the square root of a 3x3 matrix.

For A=

[1, 1, 1
0, 1, 1
0 , 0, 1]

I know that, in general, A^x = (P^-1) (D^x) (P) for some invertible P. In the case of linearly independent eigenvectors P should form a basis of A's eigenspace. But, the eigenvalues of A here are all 1, and only has one eigenvector, [1, 0, 0] and its scalar multiples. So that method isn't going to work.

There is a method using Spectral Decomposition that I don't fully understand. It starts with the equation

for A nxn, eigenvalues v1....vs, multiplicities m1....ms, then there exists n uniquely defined consituent matrices E i,k: i = 1...s, k = 0.... m-1
s.t. for any analytic function f(x) we have

f (A) = (s sigma i = 1) (mi sigma k =0) f^(k) (vi) E i,k

Anyways if you can decode that it seems to me you can arrive at the constituent matrices of A by the following equations

(A-I) (A-I) = 0 + 0 + 2 E 1,0
(A-I) (A-I) = 2 E 1,2
which works out to
[ 0, 0, 1/2
0, 0, 0
0, 0, 0 ]

A-I = E 1,1
which is of course
[ 0, 1, 1
0, 0, 1
0, 0, 0]

and finally

I = E1,0

So we have 3 constituent matrices for A, let's say

X E1,0 + Y E 1,1 + Z E 1,2

It turns out for values X=1, Y= 1/2, and Z = -1/4 you get

[ 1, 1/2, 3/8
0, 1, 1/2
0, 0, 1]

whose square is A. So somehow (I don't know) we have to use the constituent matrices in a linear equation to general the square root of A. How to get the values of X,Y,Z I do not know.
• Dec 6th 2009, 04:27 AM
NonCommAlg
Quote:

Originally Posted by Gchan
I apologize if my notation isn't clear, newbie to this forum

I'm trying to find out how to find the square root of a 3x3 matrix.

For A=

[1, 1, 1
0, 1, 1
0 , 0, 1]

I know that, in general, A^x = (P^-1) (D^x) (P) for some invertible P. In the case of linearly independent eigenvectors P should form a basis of A's eigenspace. But, the eigenvalues of A here are all 1, and only has one eigenvector, [1, 0, 0] and its scalar multiples. So that method isn't going to work.

There is a method using Spectral Decomposition that I don't fully understand. It starts with the equation

for A nxn, eigenvalues v1....vs, multiplicities m1....ms, then there exists n uniquely defined consituent matrices E i,k: i = 1...s, k = 0.... m-1
s.t. for any analytic function f(x) we have

f (A) = (s sigma i = 1) (mi sigma k =0) f^(k) (vi) E i,k

Anyways if you can decode that it seems to me you can arrive at the constituent matrices of A by the following equations

(A-I) (A-I) = 0 + 0 + 2 E 1,0
(A-I) (A-I) = 2 E 1,2
which works out to
[ 0, 0, 1/2
0, 0, 0
0, 0, 0 ]

A-I = E 1,1
which is of course
[ 0, 1, 1
0, 0, 1
0, 0, 0]

and finally

I = E1,0

So we have 3 constituent matrices for A, let's say

X E1,0 + Y E 1,1 + Z E 1,2

It turns out for values X=1, Y= 1/2, and Z = -1/4 you get

[ 1, 1/2, 3/8
0, 1, 1/2
0, 0, 1]

whose square is A. So somehow (I don't know) we have to use the constituent matrices in a linear equation to general the square root of A. How to get the values of X,Y,Z I do not know.

finding a square root of a $3 \times 3$ upper triangular matrix $A=[a_{ij}]$ is not hard (i'll assume that the entries on the diagonal of $A$ are real and positive). define the $3 \times 3$ matrix $B=[b_{ij}]$ by:

for $i=1,2,3$ let $b_{ii}=\sqrt{a_{ii}}.$ also define $b_{21}=b_{31}=b_{32}=0.$ finally let $b_{12}=\frac{a_{12}}{b_{11} + b_{22}}, \ b_{23}=\frac{a_{23}}{b_{22}+b_{33}}$ and $b_{13}=\frac{a_{13} - b_{12}b_{23}}{b_{11}+b_{33}}.$ it's easy to see that $B^2=A.$

of course this formula will also work for matrices with complex entries if you choose a square root for each $a_{ii}$ and if the denominators in $b_{12}, \ b_{23}$ and $b_{13}$ are non-zero. finally, since every

square matrix with complex entries is similar to an upper triangular matrix, you can use the above to find a square root of an arbitrary $3 \times 3$ matrix.
• Dec 6th 2009, 05:30 AM
Gchan
Thank you very much.

Is there a way to then determine the Jordan form of A?

If eigenvalues are 1, then J=

[ 1, 0, 0
0, 1, 0
0, 0, 1]
or
[1,1,0
0,1,0
0,0,1]
or
[1,1,0
0,1,1
0,0,1]
or
[1,0,0
0,1,1
0,0,1]

Can we determine which is the Jordan form without finding P st A = (P^-1) (J) (P) ?