Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F(m)=m^T m$ with $m$ a $n\times n$ matrix. I came across the statement $D_I F(m)=m^t+m$, where $D_iF$ means the derivative of $F$ at the identity matrix.

I cannot understand how this emerges. I can imagine taking the directional derivative of $F$ at $I$ by $$\lim_{\lambda\rightarrow0} \frac{F(I+\lambda m)-F(I)}{\lambda}=\lim_{\lambda\rightarrow0} \frac{\lambda (m^T+m)+\lambda^2m^T m}{\lambda}=m^T+m.$$

Is that the meaning of $D_I F(m)$? Is there a way to define a "general" derivative of a map from the space of matrices to the space of matrices?

share|cite|improve this question
up vote 0 down vote accepted

We have $$F(m+h)-F(m)=(m+h)^t(m+h)-m^tm=\color{red}{m^tm}+m^th+h^tm+h^th\color{red}{-m^tm}\\=m^th+h^tm+h^th.$$ As $\frac{\lVert h^th\rVert}{\lVert h\rVert}\leq \lVert h^t\rVert$ (taking any submultiplicative norm on the space of $n\times n$ matrices, and $h\mapsto m^th+h^tm$ is linear, we get that $D_mF(h)=m^th+h^tm$. Indeed, $$\lim_{\lVert h\rVert\to 0}\frac{\lVert F(m+h)-F(x)-D_mF(h)\rVert}{\lVert h\rVert}=0.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.