Is the matrix A a perfect square?

[Maybe_Memorie]

Homework Statement

Is the following matrix a perfect square? A matrix A is a perfect square if there is a matrix B such that A = B²

____1 1 0 0
A = 0 1 0 0
-----0 0 1 0
-----0 0 0 1

Homework Equations

The Attempt at a Solution

I haven't come across anything like this before, but here's what I'm thinking.

All eigenvalues of A are equal to 1, and the minimal polynomial is m(t) = (t-1)²
and the characteristic polynomial is p(t) = (t-1)⁴.


Both of these are perfect squares, so does that mean the matrix is a perfect square?

 

[Ray Vickson]

Homework Statement

Is the following matrix a perfect square? A matrix A is a perfect square if there is a matrix B such that A = B²

____1 1 0 0
A = 0 1 0 0
-----0 0 1 0
-----0 0 0 1

Homework Equations

The Attempt at a Solution

I haven't come across anything like this before, but here's what I'm thinking.

All eigenvalues of A are equal to 1, and the minimal polynomial is m(t) = (t-1)²
and the characteristic polynomial is p(t) = (t-1)⁴.Both of these are perfect squares, so does that mean the matrix is a perfect square?

Here is a hint: look at the function f(A) = sqrt(e*I + A), where I = identity matrix and e > 0 is a small parameter. You can compute f(A) from the fact that A is a Jordan form with one Jordan block of size 2 and two of size 1, and all eigenvalues are 1. Basically, if f(x) = sum c_n*x^n is an analytic function, then f(A) = c_0*I + sum_{n=1..infinity} c_n * A^n; I = A^0. Since A is so simple, you can get all the A^n easily and do the sum. After you have a nice final formula for f(A) you can see if it has a definite limit as e --> 0. Or, for a quicker way see, eg, Gantmacher, Theory of Matrices, or Google "analytic functions of matrices".

RGV

 

[I like Serena]

Hi Maybe_Memorie!

What you're thinking may be true or not.
I do not know (yet).
But without proof you cannot use it.

Here's another method.
If A is a perfect square, it should have one or more square roots.
One method to find a square root is to calculate a power series expansion for a broken power:
(I+X)^{1 \over 2}=I + {1 \over 2} X - {1 \over 8} X^2 + ...
Are you already familar with those?