• Support PF! Buy your school textbooks, materials and every day products Here!

Linear algebra: trace and dual space exercise

  • Thread starter mahler1
  • Start date
  • #1
222
0
Homework Statement .

Let ##A \in \mathbb C^{m\times n}##. Prove that tr##(A^*A)=0## if and only if ##A^*A=0## (here ##0## obviously means the zero matrix).


The attempt at a solution.

By definition of the trace of a matrix, the implication ← is obvious. I am having problems proving the other implication: first of all, I have doubts about the meaning of ##A^*##, if ##f:\mathbb C^n \to \mathbb C^m## is a linear transformation and ##A## is the associated matrix to ##f##, then ##f^t:{(\mathbb C^m)}^* \to {(\mathbb C^n)}^*## defined as ##f^t(\phi)=\phi \circ f## for all ##\phi \in {(\mathbb C^m)}^*##, so, I suppose ##A^*## just means the matrix which represents ##f^t##, am I right?

I have no idea how to prove ##tr(A^*A)=0 \implies A^*A=0##. I would appreciate any suggestions.
 

Answers and Replies

  • #2
hilbert2
Science Advisor
Insights Author
Gold Member
1,306
404
Try it first with a ##2\times 2## matrix. What are the diagonal elements of ##A^{*}A## if ##A=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)##? Why does ##Tr(A^{*}A)=0## imply that ##a,b,c## and ##d## are all zero? (I'm supposing that the ##"*"## means hermitian conjugate). Then try to extend the solution to more general matrices.
 
  • #3
222
0
Try it first with a ##2\times 2## matrix. What are the diagonal elements of ##A^{*}A## if ##A=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)##? Why does ##Tr(A^{*}A)=0## imply that ##a,b,c## and ##d## are all zero? (I'm supposing that the ##"*"## means hermitian conjugate). Then try to extend the solution to more general matrices.
Thanks for clearing up what it means ##A^*##. I want to check if this is correct:

By definition, the entry of the product matrix ##(A^*A)_{ij}=\sum_{k=1}^m (a^*)_{ik}a_{kj}##, but ##(a^*)_{ik}=\overline a_{ki}##, so ##\sum_{k=1}^m (a^*)_{ik}a_{kj}= \sum_{k=1}^m \overline a_{ki} a_{kj}##.

Now, ##tr(A^*A)=\sum_{p=1}^n (A^*A)_{pp}=\sum_{p=1}^n \sum_{k=1}^m (a^*)_{pk}a_{kp}=\sum_{p=1}^n \sum_{k=1}^m \overline a_{kp} a_{kp}=\sum_{p=1}^n \sum_{k=1}^m |a_{kp}|^2##.

Since ##tr(A^*A)=0 \implies \sum_{p=1}^n \sum_{k=1}^m |a_{kp}|^2=0 \implies a_{kp}=0 \space \forall 1\leq k \leq m, 1\leq p \leq n##. From here it follows ##A=0 \implies A^*A=0##.
 
  • #4
hilbert2
Science Advisor
Insights Author
Gold Member
1,306
404
That looks correct. The idea is simply that if a sum of real non-negative numbers is zero, then every term in the sum must be zero.
 

Related Threads for: Linear algebra: trace and dual space exercise

  • Last Post
Replies
2
Views
706
  • Last Post
Replies
7
Views
4K
Replies
0
Views
3K
Replies
11
Views
2K
Replies
4
Views
766
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
3
Views
1K
Top