All Questions
6
questions
0
votes
2
answers
118
views
Positive definite (inner product)
In my linear algebra course, we defined the positive definite of the inner product where $\langle z,z\rangle \ge 0$. My professor stated that because of this $\langle z,z\rangle \notin\mathbb{C}$?
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- 782
1
vote
1
answer
144
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A vector space contains $\mathbb{R}$ but have scalar product defined differently than vector product
Suppose we have a vector space with the underlying field being $\mathbb{R}$. Just out of curiosity, what are some examples of vector space $(V,+,\cdot)$, where $\mathbb{R} \subsetneq V$, but these ...
- 1,690
1
vote
1
answer
222
views
What is $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{R},\mathbb{R})$?
We already know that $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q}) = \mathbb{Q}$. Why? Because each homomorphism $f$ is uniquely determined by the value $f(1)$ and we can then calculate for ...
- 2,767
3
votes
1
answer
433
views
Why is $\mathbb{R}/\mathbb{Z}$ not an $\mathbb{R}$-vector space?
This is an embarrassing question which might seem elementary and possibly silly, but its suddenly confusing me. Clearly I'm missing something very obvious.
Take the structure $\mathbb{R}/\mathbb{Z}$. ...
- 2,517
2
votes
1
answer
99
views
If A is a square matrix of size n with real entries, with $A = A^{p+1}$, then $rank(A) + rank (I_n - A^p) = n$
Let A be a square matrix of size n with real entries, $n \geq 2$, with $A = A^{p+1}$, $p \geq 2 $, then $$rank(A) + rank (I_n - A^p) = n$$
If p is prime, in addition, $$rank (I_n - A)=rank (I_n - A^...
1
vote
1
answer
31
views
Dividing with imaginary numbers, simplifying
Alright, so I have $8-\frac{6i}{3i}$.
I multiplied by the conjugate of $3i$, and got $-18-\frac{24i}{9}$.
This is the part that confuses me, because I don't know how to divide this. Can I divide ...
- 11