Which one of these two statements is wrong ?

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Discussion Overview

The discussion revolves around the validity of two statements regarding linear algebra: one concerning the relationship between linearly independent vectors and the basis of a vector space, and the other regarding the diagonalizability of invertible matrices. The scope includes conceptual clarifications and mathematical reasoning.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants argue that statement 'a' is wrong because n linearly independent vectors must also span the vector space to be a basis.
  • Others clarify that if there are n linearly independent vectors, they necessarily span a vector space of dimension n.
  • Some participants express uncertainty about the relationship between invertibility and diagonalizability in statement 'b', suggesting that a counterexample may be needed.
  • A participant provides a specific example of a matrix that is invertible but not diagonalizable, illustrating the point that not all invertible matrices are diagonalizable.
  • Another participant expresses a lack of understanding regarding why n vectors must span the vector space when the dimension is n.

Areas of Agreement / Disagreement

Participants generally disagree on the correctness of the two statements, with multiple competing views on the implications of linear independence and diagonalizability. The discussion remains unresolved regarding the validity of the statements.

Contextual Notes

Some assumptions about the definitions of vector spaces and the properties of matrices may be implicit in the discussion. The relationship between linear independence, spanning sets, and the conditions for diagonalizability are not fully resolved.

Yankel
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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
 
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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
 

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