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One does not actually need the Jordan form to construct a square root in the case of non-zero eigenvalues, although simply knowing that it exists is quite helpful. Let $a_1$, ..., $a_n$ denote the non-zero complex eigenvalues. Then a square root $S$ is given by $$S=P(A),$$ where $P$ is any polynomial such that $$(d/dz)^m P(z) = (d/dz)^m \sqrt z$$ for all $z=a_k$ and $0 \le m\le d-1$, where $d$ is the dimension of the space. (Here we have used complex derivatives, and you can choose any branch of the complex square root you like.)

This follows from the Jordan form, which tells us that for any polynomial $Q$ the value of $Q(A)$ depends only on the values of $Q$ and its derivatives up to order $b_k-1\le d-1$ at each of the eigenvalues of $A$, where $b_k$ is the size of the maximal Jordan block for the corresponding eigenvalue $a_k$. For example $$Q\left( \left[ \begin{array} [c]{cccc}% \lambda & 1 & & \\ & \lambda & 1 & \\ & & \lambda & 1\\ & & & \lambda \end{array} \right] \right) =\left[ \begin{array} [c]{cccc}% Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac{1} {2!}Q^{\prime\prime}\left( \lambda\right) & \frac{1}{3!}Q^{\prime \prime\prime}\left( \lambda\right) \\ & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac {1}{2!}Q^{\prime\prime}\left( \lambda\right) \\ & & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) \\ & & & Q\left( \lambda\right) \end{array} \right]$$ In particular, $$S^2=(P(A))^2=(P^2)(A)=A,$$ by choosing $$Q=P^2.$$ [By construction, $Q(z)=z$ to order $d-1$ at the eigenvalues of A.]

In the infinite-dimensional case one has the holomorphic calculus (also called the Dunford calculus), although in he finite-dimensional case one needs only polynomials. Indeed, in the finite-dimensional case the Dunford calculus simply extends the identity above for applying a polynomial $Q$ to a $4 \times 4$ Jordan block to the case that $Q$ is a holomorphic function.

This is not an exhaustive solution even in the case of nonzero eigenvalues, since it only finds square roots which are polynomial functions of the original matrix. For example, all reflections are square roots of the identity.

One does not actually need the Jordan form to construct a square root in the case of non-zero eigenvalues, although simply knowing that it exists is quite helpful. Let $a_1$, ..., $a_n$ denote the non-zero complex eigenvalues. Then a square root $S$ is given by $$S=P(A),$$ where $P$ is any polynomial such that $$(d/dz)^m P(z) = (d/dz)^m \sqrt z$$ for all $z=a_k$ and $0 \le m\le d-1$, where $d$ is the dimension of the space. (Here we have used complex derivatives, and you can choose any branch of the complex square root you like. One may find higher-order roots or compute functions such as $\exp (A)$ similarly.)

This follows from the Jordan form, which tells us that for any polynomial $Q$ the value of $Q(A)$ depends only on the values of $Q$ and its derivatives up to order $b_k-1\le d-1$ at each of the eigenvalues of $A$, where $b_k$ is the size of the maximal Jordan block for the corresponding eigenvalue $a_k$. For example $$Q\left( \left[ \begin{array} [c]{cccc}% \lambda & 1 & & \\ & \lambda & 1 & \\ & & \lambda & 1\\ & & & \lambda \end{array} \right] \right) =\left[ \begin{array} [c]{cccc}% Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac{1} {2!}Q^{\prime\prime}\left( \lambda\right) & \frac{1}{3!}Q^{\prime \prime\prime}\left( \lambda\right) \\ & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac {1}{2!}Q^{\prime\prime}\left( \lambda\right) \\ & & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) \\ & & & Q\left( \lambda\right) \end{array} \right]$$ In particular, $$S^2=(P(A))^2=(P^2)(A)=A,$$ by choosing $$Q=P^2.$$ [By construction, $Q(z)=z$ to order $d-1$ at the eigenvalues of A.]

In the infinite-dimensional case one has the holomorphic calculus (also called the Dunford calculus), although in the finite-dimensional case one needs only polynomials. Indeed, in the finite-dimensional case the Dunford calculus simply extends the identity above for applying a polynomial $Q$ to a $4 \times 4$ Jordan block to the case that $Q$ is a holomorphic function.

This is not an exhaustive solution even in the case of nonzero eigenvalues, since it only finds square roots which are polynomial functions of the original matrix. For example, all reflections are square roots of the identity.

One does not actually need the Jordan form to construct a square root in the case of non-zero eigenvalues, although simply knowing that it exists is quite helpful. Let $a_1$, ..., $a_n$ denote the non-zero complex eigenvalues. Then a square root $S$ is given by $$S=P(A),$$ where $P$ is any polynomial such that $$(d/dz)^m P(z) = (d/dz)^m \sqrt z$$ for all $z=a_k$ and $m\le d-1$, where $d$ is the dimension of the space. (Here we have used complex derivatives, and you can choose any branch of the complex square root you like.)

This follows from the Jordan form, which tells us that for any polynomial $Q$ the value of $Q(A)$ depends only on the values of $Q$ and its derivatives up to order $b_k-1\le d-1$ at each of the eigenvalues of $A$, where $b_k$ is the size of the maximal Jordan block for the corresponding eigenvalue $a_k$. For example $$Q\left( \left[ \begin{array} [c]{cccc}% \lambda & 1 & & \\ & \lambda & 1 & \\ & & \lambda & 1\\ & & & \lambda \end{array} \right] \right) =\left[ \begin{array} [c]{cccc}% Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac{1} {2!}Q^{\prime\prime}\left( \lambda\right) & \frac{1}{3!}Q^{\prime \prime\prime}\left( \lambda\right) \\ & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac {1}{2!}Q^{\prime\prime}\left( \lambda\right) \\ & & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) \\ & & & Q\left( \lambda\right) \end{array} \right]$$ In particular, $$S^2=(P(A))^2=(P^2)(A)=A,$$ by choosing $$Q=P^2.$$ [By construction, $Q(z)=z$ to order $d-1$ at the eigenvalues of A.]

In the infinite-dimensional case one has the holomorphic calculus (also called the Dunford calculus), although in he finite-dimensional case one needs only polynomials. Indeed, in the finite-dimensional case the Dunford calculus simply extends the identity above for applying a polynomial $Q$ to a $4 \times 4$ Jordan block to the case that $Q$ is a holomorphic function.

This is not an exhaustive solution even in the case of nonzero eigenvalues, since it only finds square roots which are polynomial functions of the original matrix. For example, all reflections are square roots of the identity.

One does not actually need the Jordan form to construct a square root in the case of non-zero eigenvalues, although simply knowing that it exists is quite helpful. Let $a_1$, ..., $a_n$ denote the non-zero complex eigenvalues. Then a square root $S$ is given by $$S=P(A),$$ where $P$ is any polynomial such that $$(d/dz)^m P(z) = (d/dz)^m \sqrt z$$ for all $z=a_k$ and $0 \le m\le d-1$, where $d$ is the dimension of the space. (Here we have used complex derivatives, and you can choose any branch of the complex square root you like.)

This follows from the Jordan form, which tells us that for any polynomial $Q$ the value of $Q(A)$ depends only on the values of $Q$ and its derivatives up to order $b_k-1\le d-1$ at each of the eigenvalues of $A$, where $b_k$ is the size of the maximal Jordan block for the corresponding eigenvalue $a_k$. For example $$Q\left( \left[ \begin{array} [c]{cccc}% \lambda & 1 & & \\ & \lambda & 1 & \\ & & \lambda & 1\\ & & & \lambda \end{array} \right] \right) =\left[ \begin{array} [c]{cccc}% Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac{1} {2!}Q^{\prime\prime}\left( \lambda\right) & \frac{1}{3!}Q^{\prime \prime\prime}\left( \lambda\right) \\ & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac {1}{2!}Q^{\prime\prime}\left( \lambda\right) \\ & & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) \\ & & & Q\left( \lambda\right) \end{array} \right]$$ In particular, $$S^2=(P(A))^2=(P^2)(A)=A,$$ by choosing $$Q=P^2.$$ [By construction, $Q(z)=z$ to order $d-1$ at the eigenvalues of A.]

In the infinite-dimensional case one has the holomorphic calculus (also called the Dunford calculus), although in he finite-dimensional case one needs only polynomials. Indeed, in the finite-dimensional case the Dunford calculus simply extends the identity above for applying a polynomial $Q$ to a $4 \times 4$ Jordan block to the case that $Q$ is a holomorphic function.

This is not an exhaustive solution even in the case of nonzero eigenvalues, since it only finds square roots which are polynomial functions of the original matrix. For example, all reflections are square roots of the identity.

One does not actually need the Jordan form to construct a square root in the case of non-zero eigenvalues, although simply knowing that it exists is quite helpful. Let $a_1$, ..., $a_n$ denote the non-zero complex eigenvalues. Then a square root $S$ is given by $$S=P(A),$$ where $P$ is any polynomial such that $$(d/dz)^m P(z) = (d/dz)^m \sqrt z$$ for all $z=a_k$ and $m\le d-1$, where $d$ is the dimension of the space. (Here we have used complex derivatives, and you can choose any branch of the complex square root you like.)

This follows from the Jordan form, which tells us that for any polynomial $Q$ the value of $Q(A)$ depends only on the values of $Q$ and its derivatives up to order $b_k-1\le d-1$ at each of the eigenvalues of $A$, where $b_k$ is the size of the maximal Jordan block for the corresponding eigenvalue $a_k$. For example $$Q\left( \left[ \begin{array} [c]{cccc}% \lambda & 1 & & \\ & \lambda & 1 & \\ & & \lambda & 1\\ & & & \lambda \end{array} \right] \right) =\left[ \begin{array} [c]{cccc}% Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac{1} {2!}Q^{\prime\prime}\left( \lambda\right) & \frac{1}{3!}Q^{\prime \prime\prime}\left( \lambda\right) \\ & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac {1}{2!}Q^{\prime\prime}\left( \lambda\right) \\ & & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) \\ & & & Q\left( \lambda\right) \end{array} \right]$$ In particular, $$S^2=(P(A))^2=(P^2)(A)=A,$$ by choosing $$Q=P^2.$$ [By construction, $Q(z)=z$ to order $d-1$ at the eigenvalues of A.]

In the infinite-dimensional case one has the holomorphic calculus (also called the Dunford calculus), although in he finite-dimensional case one needs only polynomials. Indeed, in the finite-dimensional case the Dunford calculus simply extends the identity above for applying a polynomial $Q$ to a $4 \times 4$ Jordan block to the case that $Q$ is a holomorphic function.

One does not actually need the Jordan form to construct a square root in the case of non-zero eigenvalues, although simply knowing that it exists is quite helpful. Let $a_1$, ..., $a_n$ denote the non-zero complex eigenvalues. Then a square root $S$ is given by $$S=P(A),$$ where $P$ is any polynomial such that $$(d/dz)^m P(z) = (d/dz)^m \sqrt z$$ for all $z=a_k$ and $m\le d-1$, where $d$ is the dimension of the space. (Here we have used complex derivatives, and you can choose any branch of the complex square root you like.)

This follows from the Jordan form, which tells us that for any polynomial $Q$ the value of $Q(A)$ depends only on the values of $Q$ and its derivatives up to order $b_k-1\le d-1$ at each of the eigenvalues of $A$, where $b_k$ is the size of the maximal Jordan block for the corresponding eigenvalue $a_k$. For example $$Q\left( \left[ \begin{array} [c]{cccc}% \lambda & 1 & & \\ & \lambda & 1 & \\ & & \lambda & 1\\ & & & \lambda \end{array} \right] \right) =\left[ \begin{array} [c]{cccc}% Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac{1} {2!}Q^{\prime\prime}\left( \lambda\right) & \frac{1}{3!}Q^{\prime \prime\prime}\left( \lambda\right) \\ & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) & \frac {1}{2!}Q^{\prime\prime}\left( \lambda\right) \\ & & Q\left( \lambda\right) & Q^{\prime}\left( \lambda\right) \\ & & & Q\left( \lambda\right) \end{array} \right]$$ In particular, $$S^2=(P(A))^2=(P^2)(A)=A,$$ by choosing $$Q=P^2.$$ [By construction, $Q(z)=z$ to order $d-1$ at the eigenvalues of A.]

In the infinite-dimensional case one has the holomorphic calculus (also called the Dunford calculus), although in he finite-dimensional case one needs only polynomials. Indeed, in the finite-dimensional case the Dunford calculus simply extends the identity above for applying a polynomial $Q$ to a $4 \times 4$ Jordan block to the case that $Q$ is a holomorphic function.

This is not an exhaustive solution even in the case of nonzero eigenvalues, since it only finds square roots which are polynomial functions of the original matrix. For example, all reflections are square roots of the identity.