Can Non-Diagonalizable Matrices Have Square Roots?

In summary: The definition that I was thinking of is the one that HallsofIvy provided.Thanks HallsofIvy!In summary, the square root of a 2x2 matrix cannot be determined. If a matrix has a single eigenvalue, it can be put into "Jordan Normal Form".
  • #1
SometimesY
2
0
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!
 
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  • #2
SometimesY: Boy, I would like to see a 2x2 matrix that cannot be diagonalized!
 
  • #3
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
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
SteamKing said:
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
SteamKing said:
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 textbook or other place defines a "diagonal matrix" the way you say?
 
  • #8
I stand corrected.
 

What is a square root of a matrix?

A square root of a matrix is another matrix that, when multiplied by itself, produces the original matrix. In other words, it is a matrix that, when squared, results in the original matrix.

Is the square root of a matrix unique?

No, the square root of a matrix is not always unique. A matrix may have multiple square roots, just like a number can have multiple square roots.

How do you find the square root of a matrix?

The process of finding the square root of a matrix involves using matrix decomposition techniques, such as eigenvalue decomposition or singular value decomposition.

Can any matrix have a square root?

No, not all matrices have a square root. A matrix must be square and have non-negative eigenvalues in order to have a square root.

What is the significance of the square root of a matrix?

The square root of a matrix has various applications in linear algebra, such as in solving systems of linear equations and in data compression techniques. It also plays a crucial role in the study of matrix algebra and its properties.

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