Can Non-Diagonalizable Matrices Have Square Roots?

[SometimesY]

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!

 

[SteamKing]

SometimesY: Boy, I would like to see a 2x2 matrix that cannot be diagonalized!

 

[HallsofIvy]

Here's one:
\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}

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",
\begin{bmatrix}x & 1 \\ 0 & x\end{bmatrix}
where "x" is that single eigenvalue.

Now, if
\begin{bmatrix}a & b \\ c & d \end{bmatrix}
is a "square root" for that matrix, we must have
\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}

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

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

Then a^2+ bc= a^2= x and bc+ d^2= d^2= x.

Since a^2= d^2= x 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
\begin{bmatrix}x & 1 \\ 0 & x\end{bmatrix}
has two square roots:
\begin{bmatrix}\sqrt{x} & \frac{1}{2\sqrt{x}} \\ 0 & \sqrt{x}\end{bmatrix}
and
\begin{bmatrix}-\sqrt{x} & -\frac{1}{2\sqrt{x}} \\ 0 & -\sqrt{x}\end{bmatrix}

 

[SteamKing]

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.

 

[AlephZero]

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.

 

[SometimesY]

Thanks HallsofIvy! That's exactly what I was looking for. It means a lot!

 

[HallsofIvy]

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?

 

[SteamKing]

I stand corrected.