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I am taking a graduate Matrix Analysis class and can not get this answer. I have googled, read the textbook but nothing.

A matrix $A\in M_n$ is a square root of $B \in M_n$ if $A^2=B$. Show that every diagonalizable $B\in M_n$ has a square root. Does $$\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$$ matrix have a square root? Why?

Arnaud D.
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Linda H
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  • Does what matrix have a square root? – Angina Seng Jun 23 '17 at 15:08
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    What field are you working over? The first part is false over $\Bbb R$. – Arnaud D. Jun 23 '17 at 15:10
  • You can start from the definition. Assume that $A = \begin{bmatrix}a_1&a_2\a_3&a_4\end{bmatrix}$ See what the product $AA$ becomes and try to make the $a_1,a_2,\cdots$ fit the 0s and 1s – mathreadler Jun 23 '17 at 15:10
  • You can find a list of relevant questions there : https://math.stackexchange.com/questions/347952/how-to-find-a-matrix-square-root-with-all-real-entries-if-it-exists/348868#348868 – Arnaud D. Jun 23 '17 at 15:13

4 Answers4

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You can clearly see that the matrix B represents a Jordan block and it's minimal polynomial is $t^2$. If you consider now the Kernel of the matrix, you see that its dimension is $1$. So you have that the algebrical multiplicity of your eigenvalue $\lambda =0$ is $2$ but the geometrical multiplicity, which represents the dimension of the eigenvectors space, is $1$. But a matrix is diagonalizable if and only if the geometrical multiplicity of an eigenvalue is equal to its algebrical multiplicity. So the matrix isn't diagonalizable.

Alberto Andrenucci
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If $$A^2=B=\pmatrix{0&1\\0&0}$$ then $AB=BA$. If $AB=BA$ then $$A=\pmatrix{x&y\\0&x}.$$ Do any of these square to $A$?

Angina Seng
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Let $A\in M_2(\Bbb C)$ such that $$A^2= \begin{pmatrix}0&1\\0&0\end{pmatrix}=B.$$

If $x$ is an eigenvalue of $A$, then $x^2$ is an eigenvalues of $B$. Therefore, all eigenvalues of $A$ are zero. If $A$ is diagonalisable, then $A=0$ (diagonalisable matrix with nothing but zero eigenvalues is zero).

Hence $A$ is not diagonalisable. Since it is a $2\times 2$ matrix, its normal Jordan form would be equal to $\begin{pmatrix}0&1\\0&0\end{pmatrix}$. This implies that the normal Jordan form of $A^2$ is a zero matrix, hence $A^2=0$, which obviously contradicts the identity $A^2=B$.

The final conclusion is that $B$ does not have a square root.

TZakrevskiy
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There are two problems given here. What are you asking about? For the first, "show that every diagonal matrix has a square root", that's not true unless the underlying field is the field of complex numbers. If the field is the complex numbers, then it is easy to see that the square root of $\begin{pmatrix}a & 0 \\ 0 & b \end{pmatrix}$ is $\begin{pmatrix}\sqrt{a} & 0 \\ 0 & \sqrt{b}\end{pmatrix}$.

To show that $\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$ does not have a square root, look at the generic $\begin{pmatrix}a & b \\ c & d \end{pmatrix}$.

$\begin{pmatrix}a & b \\ c & d \end{pmatrix}^2a= \begin{pmatrix}a^2+ bc & ab+ bd \\ ac+ cd & bc+ d^2\end{pmatrix}$. In order that this be equal to $\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$, we must have $a^2+ bc= 0$, $ab+ bd= 1$, $ac+ cd= 0$, and $bc+ d^2= 0$. The first and fourth equations give $a^2- d^2= (a- d)(a+ d)= 0$ so that either a- d= 0 or a+ d= 0.

If a= d then $ac+ cd= 2ac= 0$ so either a= 0 or c= 0. With a= 0, the equations become bc= 0, bd= 1, cd= 0 an $bc+ d^2= 0$. If b= 0, bd= 0 not 1 so we must have c= 0. But then d= 0 so that it cannot satisfy bd= 1 again.

If a= -d then $ab+ bd= ab- ab= 0$, not 1 so that is not possible.

user247327
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