MHB Which one of these two statements is wrong ?

  • Thread starter Thread starter Yankel
  • Start date Start date
Yankel
Messages
390
Reaction score
0
a. every set of n linearly independent vectors are basis of a vector space with dimension n

b. An invertible matrix is necessarily diagonalizable

they both seems wrong to me, but only one suppose to be.

'a' sounds wrong because the n vectors must also span the vector space

and 'b' because I don't see the relation between invertible and diagonalizable
 
Physics news on Phys.org
Yankel said:
'a' sounds wrong because the n vectors must also span the vector space
Which vector space?
 
Yankel said:
a. every set of n linearly independent vectors are basis of a vector space with dimension n

b. An invertible matrix is necessarily diagonalizable

they both seems wrong to me, but only one suppose to be.

'a' sounds wrong because the n vectors must also span the vector space

and 'b' because I don't see the relation between invertible and diagonalizable

n linearly independent vectors span a vector space with n dimensions.
Since the vector space has only n dimensions, those vectors necessarily must span that vector space.

For 'b' you should try to find a counter example.
Can you think of a matrix that is not diagonalizable, but still has a non-zero determinant?
 
to extend serena statement.

If D is a diagnizable matrix then there exists an invertible nxn matrix P
such that D = $P^{-1}AP$ where A is the matrix with the eigen values of D on its diagonals.

Now take D = $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. whose eigen values are 1,1. Now if D is diagnoziable there would exist a nxn invertible matrix P such that

D = $P^{-1}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}P$, now 4 $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is just the identity matrix.
so D = $P^{-1}P = I$, since $D \not = I$, it is clear that D is not diagnizable.

Yet D is invertible Det(D) = 1.
 
Last edited:
thanks everyone for your answers, it's very helpful.

I still don't fully understand why when the dimension is n the vector must also span.

I will try to look for some further material on it

thanks
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top