Recipe

Compute a matrix logarithm or square root

Use when: you need \log A or \sqrt A for a diagonalizable matrix A

Goal: the matrix \log A or \sqrt A, correctly flagged as real, complex, or undefined based on A's eigenvalues

Taught in Matrix logarithms and square roots

1. 1

   Find A's eigenvalues and a full set of independent eigenvectors, then write A = PDP^{-1} with D diagonal.

   Example

   Take A=\begin{bmatrix}5&4\\1&2\end{bmatrix}. Its characteristic equation is \lambda^2-7\lambda+6=0, factoring as (\lambda-6)(\lambda-1)=0, so \lambda=6 or \lambda=1. Solving (A-6I)\mathbf v=\mathbf 0 gives eigenvector (4,1); solving (A-I)\mathbf v=\mathbf 0 gives eigenvector (1,-1). P=\begin{bmatrix}4&1\\1&-1\end{bmatrix}, \quad D=\begin{bmatrix}6&0\\0&1\end{bmatrix}, \quad P^{-1}=\begin{bmatrix}1/5&1/5\\1/5&-4/5\end{bmatrix}

   ⚡ Shortcut

   If A is already diagonal, skip this step entirely: D=A and P=I, so the remaining steps just apply f to each diagonal entry directly. Every worked example below starts from exactly this shortcut.

   Why: applying \log or \sqrt{\cdot} entrywise to D and conjugating back by P is only valid once A is written in exactly this factored form.

2. 2

   Look at every eigenvalue on D's diagonal and sort each one into exactly one of three cases before applying \log or \sqrt{\cdot} to it.

   Example

   Both eigenvalues of the running example, 6 and 1, are positive, so both fall under the first case: use the real-valued function directly.

   • eigenvalue is positive — apply the ordinary real-valued function directly.
   • eigenvalue is negative — apply the complex-valued formula instead (\log(-r)=\log r+i\pi, or \sqrt{-r}=i\sqrt r); the result is genuinely complex.
   • eigenvalue is exactly zero and f=\log — no logarithm exists for this matrix at all; stop here.
   • eigenvalue is exactly zero and f=\sqrt{\cdot} — \sqrt0=0 exactly; continue normally.

   Why: this determines which formula gets applied entrywise in the next step, and misclassifying even one eigenvalue produces a wrong result or a spurious complex answer.

3. 3

   Replace each diagonal entry of D with f applied to it, using the branch selected in Step 2 for that entry.

   Example

   \log6\approx1.792 and \log1=0, both real since both eigenvalues were positive. f(D)=\begin{bmatrix}1.792&0\\0&0\end{bmatrix}

   Why: this is the one step where the choice of function, and the branch from Step 2, actually gets used; everything before and after it is bookkeeping.

4. 4

   Multiply the three matrices back together in that order to get the final answer as one genuine matrix.

   Example

   f(A)=\begin{bmatrix}4&1\\1&-1\end{bmatrix}\begin{bmatrix}1.792&0\\0&0\end{bmatrix}\begin{bmatrix}1/5&1/5\\1/5&-4/5\end{bmatrix}\approx\begin{bmatrix}1.433&1.433\\0.358&0.358\end{bmatrix}

   Both entries come out equal since \log1=0 erases that eigenvector's contribution from every column.

   Why: the eigendecomposition only carries meaning once it's been reassembled; D or f(D) alone says nothing about A itself.

5. 5

   Apply the inverse operation to the answer (exponentiate a logarithm, or square a square root) and confirm the original A comes back.

   Example

   e^{f(A)} recovers A=\begin{bmatrix}5&4\\1&2\end{bmatrix} exactly, confirmed by reversing the same eigendecomposition: Pe^{f(D)}P^{-1}=PDP^{-1}=A.

   Why: this catches an arithmetic slip in the eigenvectors, the logarithm, or the matrix multiplication before it's mistaken for a genuine final answer.